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		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27620</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27620"/>
		<updated>2017-04-07T23:08:55Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:dipole pot diff.jpg|300px|thumb|right]] &lt;br /&gt;
&lt;br /&gt;
1) If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
&lt;br /&gt;
up&lt;br /&gt;
 &lt;br /&gt;
left&lt;br /&gt;
 &lt;br /&gt;
right &lt;br /&gt;
&lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
&lt;br /&gt;
positive&lt;br /&gt;
&lt;br /&gt;
negative &lt;br /&gt;
&lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Shortcut to remember: &#039;&#039;&lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]] &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2) Three charged metal disks are arranged as shown on the right. The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2&amp;lt;math&amp;gt; m^2&amp;lt;/math&amp;gt; (this is the area of one side of the disk). Calculate the potential difference Vd-Va. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; D &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; G &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; H &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to &amp;lt;math&amp;gt;\vec{E_2}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Update Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Electric Field of Capacitors: &amp;lt;math&amp;gt; E_{cap} = \frac{Q}{A\epsilon_0} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = (Vb-Vh) + (Vg-Vd) = &amp;lt;math&amp;gt; -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
= &amp;lt;math&amp;gt;-\frac{1.2e-7C*0.003m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt; +  &amp;lt;math&amp;gt;-\frac{7e-8C*0.0015m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
= 14.4 V&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
3) Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2 pot diff.jpg|300px|thumb|right|Problem 4 Diagram]] &lt;br /&gt;
&lt;br /&gt;
4) A very long, thin glass rod of length 2R carries a uniformly distributed charge +q. A very large plastic disk of radius R, carrying a uniformly distributed charge -Q, is located a distance d from the rod, where d&amp;lt;&amp;lt;R.&lt;br /&gt;
&lt;br /&gt;
a. Predict the sign of the potential difference Vc-Va. Justify your answer.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
E and dl are both positive because they point to the right direction. According to the integral form of the potential difference formula for non-uniform field, &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; , the change in V from position A to C has to be negative.&lt;br /&gt;
&lt;br /&gt;
b. Calculate the potential difference Vc-Va. Location A is located a distance w fro the center of the rod. Location C is located at the disk&#039;s center a distance d from the rod. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Assume d&amp;lt;&amp;lt;R. &amp;lt;math&amp;gt; E_{disk} = \frac{Q}{2A\epsilon_0} &amp;lt;/math&amp;gt; &amp;amp; &amp;lt;math&amp;gt;E_{rod} = \frac{1}{4\pi\epsilon_0}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt; \frac{2Q}{Lr}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Vc-Va = &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E_{net}} {\bullet} \vec{l} &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; \triangle {V_{rod}} + \triangle {V_{disk}}&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{rod}} {\bullet} \vec{l}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{disk}} {\bullet} \vec{l}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{rod}}=-\int\limits_{w}^{d} \frac{1}{4\pi\epsilon_0} &amp;lt;/math&amp;gt;&amp;lt;math&amp;gt;\frac{2q}{xL} dx&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} [ln(x)\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} ln(\frac{d}{w})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{disk}}=-\int\limits_{w}^{d} \frac{Q}{2A\epsilon_0} dx&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0} [x\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0L} (d-w)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = -&amp;lt;math&amp;gt;[\frac{2q}{4\pi\epsilon_0L} ln(\frac{d}{w})&amp;lt;/math&amp;gt; + &amp;lt;math&amp;gt;\frac{Q}{2A\epsilon_0L} (d-w)]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
5) The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem7 pot diff.jpg|400px|thumb|right|Problem 6 Diagram]]&lt;br /&gt;
&lt;br /&gt;
6) The diagram shows a spherical shell with a uniformed surface charge of -Q and a capacitor with uniformed distributed +Q2 and -Q2. R1&amp;lt;&amp;lt;R2, d&amp;lt;&amp;lt;R2. Calculate the potential difference V2-V1 and V3-V2.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
V2-V1:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;E_{sphere} =0&amp;lt;/math&amp;gt; inside the sphere, so the only electric field acting along this distance is &amp;lt;math&amp;gt; E_{cap}&amp;lt;/math&amp;gt;. &amp;lt;math&amp;gt; E_{cap} = \frac{Q_2}{A\epsilon_0}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\triangle V_{cap} = V_2-V_1 = -\int\limits_{0}^{-R_1} \frac{Q_2}{2A\epsilon_0} dy&amp;lt;/math&amp;gt; =  &amp;lt;math&amp;gt;\frac{-Q}{\pi\epsilon_o} [x\rbrack_0^{-R_1} = \frac{Q_2R_1}{\pi{R_1^2}\epsilon_0}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
V3-V2:&lt;br /&gt;
&lt;br /&gt;
The net electric field is due to the sphere and the capacitor. &amp;lt;math&amp;gt; E_{sphere} = -\frac{1}{4\pi\epsilon_0} \frac{Q}{y^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\triangle V_{sphere} = -\int\limits_{d}^{R_1} \frac{1}{4\pi\epsilon_0} \frac{Q}{y^2} dy&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; -\frac{Q}{4\pi\epsilon_0} [\frac{1}{y}\rbrack_d^{R_1}&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; -\frac{Q}{4\pi\epsilon_0} (\frac{1}{R_1} - \frac{1}{d})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\triangle V_{cap} = -\int\limits_{R_1}^{d} \frac{Q}{A\epsilon_0} dy&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;-\frac{Q}{A\epsilon_0} (d-R_1) = -\frac{Q}{\pi{R_2^2}\epsilon_0} (d-R_1)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\triangle V_{sphere}+\triangle V_{cap} = V_3-V_2 = -\frac{Q}{4\pi\epsilon_0} (\frac{1}{R_1} - \frac{1}{d}) -\frac{Q}{\pi{R_2^2}\epsilon_0} (d-R_1)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27619</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27619"/>
		<updated>2017-04-07T23:07:42Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:dipole pot diff.jpg|300px|thumb|right]] &lt;br /&gt;
&lt;br /&gt;
1) If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
&lt;br /&gt;
up&lt;br /&gt;
 &lt;br /&gt;
left&lt;br /&gt;
 &lt;br /&gt;
right &lt;br /&gt;
&lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
&lt;br /&gt;
positive&lt;br /&gt;
&lt;br /&gt;
negative &lt;br /&gt;
&lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Shortcut to remember: &#039;&#039;&lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]] &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2) Three charged metal disks are arranged as shown on the right. The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2&amp;lt;math&amp;gt; m^2&amp;lt;/math&amp;gt; (this is the area of one side of the disk). Calculate the potential difference Vd-Va. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; D &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; G &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; H &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to &amp;lt;math&amp;gt;\vec{E_2}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Update Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Electric Field of Capacitors: &amp;lt;math&amp;gt; E_{cap} = \frac{Q}{A\epsilon_0} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = (Vb-Vh) + (Vg-Vd) = &amp;lt;math&amp;gt; -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
= &amp;lt;math&amp;gt;-\frac{1.2e-7C*0.003m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt; +  &amp;lt;math&amp;gt;-\frac{7e-8C*0.0015m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
= 14.4 V&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
3) Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2 pot diff.jpg|300px|thumb|right|Problem 4 Diagram]] &lt;br /&gt;
&lt;br /&gt;
4) A very long, thin glass rod of length 2R carries a uniformly distributed charge +q. A very large plastic disk of radius R, carrying a uniformly distributed charge -Q, is located a distance d from the rod, where d&amp;lt;&amp;lt;R.&lt;br /&gt;
&lt;br /&gt;
a. Predict the sign of the potential difference Vc-Va. Justify your answer.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
E and dl are both positive because they point to the right direction. According to the integral form of the potential difference formula for non-uniform field, &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; , the change in V from position A to C has to be negative.&lt;br /&gt;
&lt;br /&gt;
b. Calculate the potential difference Vc-Va. Location A is located a distance w fro the center of the rod. Location C is located at the disk&#039;s center a distance d from the rod. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Assume d&amp;lt;&amp;lt;R. &amp;lt;math&amp;gt; E_{disk} = \frac{Q}{2A\epsilon_0} &amp;lt;/math&amp;gt; &amp;amp; &amp;lt;math&amp;gt;E_{rod} = \frac{1}{4\pi\epsilon_0}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt; \frac{2Q}{Lr}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Vc-Va = &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E_{net}} {\bullet} \vec{l} &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; \triangle {V_{rod}} + \triangle {V_{disk}}&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{rod}} {\bullet} \vec{l}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{disk}} {\bullet} \vec{l}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{rod}}=-\int\limits_{w}^{d} \frac{1}{4\pi\epsilon_0} &amp;lt;/math&amp;gt;&amp;lt;math&amp;gt;\frac{2q}{xL} dx&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} [ln(x)\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} ln(\frac{d}{w})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{disk}}=-\int\limits_{w}^{d} \frac{Q}{2A\epsilon_0} dx&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0} [x\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0L} (d-w)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = -&amp;lt;math&amp;gt;[\frac{2q}{4\pi\epsilon_0L} ln(\frac{d}{w})&amp;lt;/math&amp;gt; + &amp;lt;math&amp;gt;\frac{Q}{2A\epsilon_0L} (d-w)]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
5) The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem7 pot diff.jpg|400px|thumb|right|Problem 6 Diagram]]&lt;br /&gt;
&lt;br /&gt;
6) The diagram shows a spherical shell with a uniformed surface charge of -Q and a capacitor with uniformed distributed +Q2 and -Q2. R1&amp;lt;&amp;lt;R2, d&amp;lt;&amp;lt;R2. Calculate the potential difference V2-V1 and V3-V2.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
V2-V1:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;E_{sphere} =0&amp;lt;/math&amp;gt; inside the sphere, so the only electric field acting along this distance is &amp;lt;math&amp;gt; E_{cap}&amp;lt;/math&amp;gt;. &amp;lt;math&amp;gt; E_{cap} = \frac{Q_2}{A\epsilon_0}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\triangle V_{cap} = V2-V1 = -\int\limits_{0}^{-R_1} \frac{Q_2}{2A\epsilon_0} dy&amp;lt;/math&amp;gt; =  &amp;lt;math&amp;gt;\frac{-Q}{\pi\epsilon_o} [x\rbrack_0^{-R_1} = \frac{Q_2R_1}{\pi{R_1^2}\epsilon_0}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
V3-V2:&lt;br /&gt;
&lt;br /&gt;
The net electric field is due to the sphere and the capacitor. &amp;lt;math&amp;gt; E_{sphere} = -\frac{1}{4\pi\epsilon_0} \frac{Q}{y^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\triangle V_{sphere} = -\int\limits_{d}^{R_1} \frac{1}{4\pi\epsilon_0} \frac{Q}{y^2} dy&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; -\frac{Q}{4\pi\epsilon_0} [\frac{1}{y}\rbrack_d^{R_1}&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; -\frac{Q}{4\pi\epsilon_0} (\frac{1}{R_1} - \frac{1}{d})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\triangle V_{cap} = -\int\limits_{R_1}^{d} \frac{Q}{A\epsilon_0} dy&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;-\frac{Q}{A\epsilon_0} (d-R_1) = -\frac{Q}{\pi{R_2^2}\epsilon_0} (d-R_1)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\triangle V_{sphere}+\triangle V_{cap} = V3-V2 = -\frac{Q}{4\pi\epsilon_0} (\frac{1}{R_1} - \frac{1}{d}) -\frac{Q}{\pi{R_2^2}\epsilon_0} (d-R_1)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27618</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27618"/>
		<updated>2017-04-07T23:05:54Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:dipole pot diff.jpg|300px|thumb|right]] &lt;br /&gt;
&lt;br /&gt;
1) If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
&lt;br /&gt;
up&lt;br /&gt;
 &lt;br /&gt;
left&lt;br /&gt;
 &lt;br /&gt;
right &lt;br /&gt;
&lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
&lt;br /&gt;
positive&lt;br /&gt;
&lt;br /&gt;
negative &lt;br /&gt;
&lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Shortcut to remember: &#039;&#039;&lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]] &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2) Three charged metal disks are arranged as shown on the right. The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2&amp;lt;math&amp;gt; m^2&amp;lt;/math&amp;gt; (this is the area of one side of the disk). Calculate the potential difference Vd-Va. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; D &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; G &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; H &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to &amp;lt;math&amp;gt;\vec{E_2}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Update Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Electric Field of Capacitors: &amp;lt;math&amp;gt; E_{cap} = \frac{Q}{A\epsilon_0} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = (Vb-Vh) + (Vg-Vd) = &amp;lt;math&amp;gt; -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
= &amp;lt;math&amp;gt;-\frac{1.2e-7C*0.003m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt; +  &amp;lt;math&amp;gt;-\frac{7e-8C*0.0015m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
= 14.4 V&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
3) Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2 pot diff.jpg|300px|thumb|right|Problem 4 Diagram]] &lt;br /&gt;
&lt;br /&gt;
4) A very long, thin glass rod of length 2R carries a uniformly distributed charge +q. A very large plastic disk of radius R, carrying a uniformly distributed charge -Q, is located a distance d from the rod, where d&amp;lt;&amp;lt;R.&lt;br /&gt;
&lt;br /&gt;
a. Predict the sign of the potential difference Vc-Va. Justify your answer.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
E and dl are both positive because they point to the right direction. According to the integral form of the potential difference formula for non-uniform field, &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; , the change in V from position A to C has to be negative.&lt;br /&gt;
&lt;br /&gt;
b. Calculate the potential difference Vc-Va. Location A is located a distance w fro the center of the rod. Location C is located at the disk&#039;s center a distance d from the rod. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Assume d&amp;lt;&amp;lt;R. &amp;lt;math&amp;gt; E_{disk} = \frac{Q}{2A\epsilon_0} &amp;lt;/math&amp;gt; &amp;amp; &amp;lt;math&amp;gt;E_{rod} = \frac{1}{4\pi\epsilon_0}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt; \frac{2Q}{Lr}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Vc-Va = &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E_{net}} {\bullet} \vec{l} &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; \triangle {V_{rod}} + \triangle {V_{disk}}&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{rod}} {\bullet} \vec{l}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{disk}} {\bullet} \vec{l}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{rod}}=-\int\limits_{w}^{d} \frac{1}{4\pi\epsilon_0} &amp;lt;/math&amp;gt;&amp;lt;math&amp;gt;\frac{2q}{xL} dx&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} [ln(x)\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} ln(\frac{d}{w})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{disk}}=-\int\limits_{w}^{d} \frac{Q}{2A\epsilon_0} dx&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0} [x\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0L} (d-w)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = -&amp;lt;math&amp;gt;[\frac{2q}{4\pi\epsilon_0L} ln(\frac{d}{w})&amp;lt;/math&amp;gt; + &amp;lt;math&amp;gt;\frac{Q}{2A\epsilon_0L} (d-w)]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
5) The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem7 pot diff.jpg|400px|thumb|right|Problem 6 Diagram]]&lt;br /&gt;
&lt;br /&gt;
6) The diagram shows a spherical shell with a uniformed surface charge of -Q and a capacitor with uniformed distributed +Q2 and -Q2. R1&amp;lt;&amp;lt;R2, d&amp;lt;&amp;lt;R2. Calculate the potential difference V2-V1 and V3-V2.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
V2-V1:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;E_{sphere} =0&amp;lt;/math&amp;gt; inside the sphere, so the only electric field acting along this distance is &amp;lt;math&amp;gt; E_{cap}&amp;lt;/math&amp;gt;. &amp;lt;math&amp;gt; E_{cap} = \frac{Q_2}{A\epsilon_0}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\triangle V_{cap} = -\int\limits_{0}^{-R_1} \frac{Q_2}{2A\epsilon_0} dy&amp;lt;/math&amp;gt; =  &amp;lt;math&amp;gt;\frac{-Q}{\pi\epsilon_o} [x\rbrack_0^{-R_1} = \frac{Q_2R_1}{\pi{R_1^2}\epsilon_0}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
V3-V2:&lt;br /&gt;
&lt;br /&gt;
The net electric field is due to the sphere and the capacitor. &amp;lt;math&amp;gt; E_{sphere} = -\frac{1}{4\pi\epsilon_0} \frac{Q}{y^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\triangle V_{sphere} = -\int\limits_{d}^{R_1} \frac{1}{4\pi\epsilon_0} \frac{Q}{y^2} dy&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; -\frac{Q}{4\pi\epsilon_0} [\frac{1}{y}\rbrack_d^{R_1}&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; -\frac{Q}{4\pi\epsilon_0} (\frac{1}{R_1} - \frac{1}{d})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\triangle V_{cap} = -\int\limits_{R_1}^{d} \frac{Q}{A\epsilon_0} dy&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;-\frac{Q}{A\epsilon_0} (d-R_1) = -\frac{Q}{\pi{R_2^2}\epsilon_0} (d-R_1)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27617</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27617"/>
		<updated>2017-04-07T22:35:36Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:dipole pot diff.jpg|300px|thumb|right]] &lt;br /&gt;
&lt;br /&gt;
1) If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
&lt;br /&gt;
up&lt;br /&gt;
 &lt;br /&gt;
left&lt;br /&gt;
 &lt;br /&gt;
right &lt;br /&gt;
&lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
&lt;br /&gt;
positive&lt;br /&gt;
&lt;br /&gt;
negative &lt;br /&gt;
&lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Shortcut to remember: &#039;&#039;&lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]] &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2) Three charged metal disks are arranged as shown on the right. The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2&amp;lt;math&amp;gt; m^2&amp;lt;/math&amp;gt; (this is the area of one side of the disk). Calculate the potential difference Vd-Va. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; D &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; G &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; H &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to &amp;lt;math&amp;gt;\vec{E_2}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Update Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Electric Field of Capacitors: &amp;lt;math&amp;gt; E_{cap} = \frac{Q}{A\epsilon_0} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = (Vb-Vh) + (Vg-Vd) = &amp;lt;math&amp;gt; -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
= &amp;lt;math&amp;gt;-\frac{1.2e-7C*0.003m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt; +  &amp;lt;math&amp;gt;-\frac{7e-8C*0.0015m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
= 14.4 V&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
3) Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2 pot diff.jpg|300px|thumb|right|Problem 4 Diagram]] &lt;br /&gt;
&lt;br /&gt;
4) A very long, thin glass rod of length 2R carries a uniformly distributed charge +q. A very large plastic disk of radius R, carrying a uniformly distributed charge -Q, is located a distance d from the rod, where d&amp;lt;&amp;lt;R.&lt;br /&gt;
&lt;br /&gt;
a. Predict the sign of the potential difference Vc-Va. Justify your answer.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
E and dl are both positive because they point to the right direction. According to the integral form of the potential difference formula for non-uniform field, &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; , the change in V from position A to C has to be negative.&lt;br /&gt;
&lt;br /&gt;
b. Calculate the potential difference Vc-Va. Location A is located a distance w fro the center of the rod. Location C is located at the disk&#039;s center a distance d from the rod. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Assume d&amp;lt;&amp;lt;R. &amp;lt;math&amp;gt; E_{disk} = \frac{Q}{2A\epsilon_0} &amp;lt;/math&amp;gt; &amp;amp; &amp;lt;math&amp;gt;E_{rod} = \frac{1}{4\pi\epsilon_0}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt; \frac{2Q}{Lr}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Vc-Va = &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E_{net}} {\bullet} \vec{l} &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; \triangle {V_{rod}} + \triangle {V_{disk}}&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{rod}} {\bullet} \vec{l}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{disk}} {\bullet} \vec{l}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{rod}}=-\int\limits_{w}^{d} \frac{1}{4\pi\epsilon_0} &amp;lt;/math&amp;gt;&amp;lt;math&amp;gt;\frac{2q}{xL} dx&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} [ln(x)\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} ln(\frac{d}{w})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{disk}}=-\int\limits_{w}^{d} \frac{Q}{2A\epsilon_0} dx&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0} [x\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0L} (d-w)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = -&amp;lt;math&amp;gt;[\frac{2q}{4\pi\epsilon_0L} ln(\frac{d}{w})&amp;lt;/math&amp;gt; + &amp;lt;math&amp;gt;\frac{Q}{2A\epsilon_0L} (d-w)]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
5) The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[[File:Problem7 pot diff.jpg|300px|thumb|right|Problem 7 Diagram]]&lt;br /&gt;
&lt;br /&gt;
7) The diagram shows a spherical shell with a uniformed surface charge of -Q and a capacitor with uniformed distributed +Q2 and -Q2. R1&amp;lt;&amp;lt;R2, d&amp;lt;&amp;lt;R2. Calculate the potential difference V2-V1 and V3-V2.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27608</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27608"/>
		<updated>2017-04-07T22:07:23Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]] &lt;br /&gt;
&lt;br /&gt;
1) If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
&lt;br /&gt;
up&lt;br /&gt;
 &lt;br /&gt;
left&lt;br /&gt;
 &lt;br /&gt;
right &lt;br /&gt;
&lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
&lt;br /&gt;
positive&lt;br /&gt;
&lt;br /&gt;
negative &lt;br /&gt;
&lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]] &lt;br /&gt;
&lt;br /&gt;
2) Three charged metal disks are arranged as shown on the right. The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2&amp;lt;math&amp;gt; m^2&amp;lt;/math&amp;gt; (this is the area of one side of the disk). Calculate the potential difference Vd-Va. &lt;br /&gt;
&lt;br /&gt;
Answer: &lt;br /&gt;
Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; D &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; G &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; H &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to &amp;lt;math&amp;gt;\vec{E_2}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Update Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Electric Field of Capacitors: &amp;lt;math&amp;gt; E_{cap} = \frac{Q}{A\epsilon_0} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = (Vb-Vh) + (Vg-Vd) = &amp;lt;math&amp;gt; -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
= &amp;lt;math&amp;gt;-\frac{1.2e-7C*0.003m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt; +  &amp;lt;math&amp;gt;-\frac{7e-8C*0.0015m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
= 14.4 V&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
3) Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
Answer:&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2 pot diff.jpg|300px|thumb|right|Problem 4 Diagram]] &lt;br /&gt;
&lt;br /&gt;
2) A very long, thin glass rod of length 2R carries a uniformly distributed charge +q. A very large plastic disk of radius R, carrying a uniformly distributed charge -Q, is located a distance d from the rod, where d&amp;lt;&amp;lt;R.&lt;br /&gt;
&lt;br /&gt;
a. Predict the sign of the potential difference Vc-Va. Justify your answer.&lt;br /&gt;
&lt;br /&gt;
Answer:&lt;br /&gt;
E and dl are both positive because they point to the right direction. According to the integral form of the potential difference formula for non-uniform field, &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; , the change in V from position A to C has to be negative.&lt;br /&gt;
&lt;br /&gt;
b. Calculate the potential difference Vc-Va. Location A is located a distance w fro the center of the rod. Location C is located at the disk&#039;s center a distance d from the rod. &lt;br /&gt;
&lt;br /&gt;
Answer:&lt;br /&gt;
Assume d&amp;lt;&amp;lt;R. &amp;lt;math&amp;gt; E_{disk} = \frac{Q}{2A\epsilon_0} &amp;lt;/math&amp;gt; &amp;amp; &amp;lt;math&amp;gt;E_{rod} = \frac{1}{4\pi\epsilon_0}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt; \frac{2Q}{Lr}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Vc-Va = &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E_{net}} {\bullet} \vec{l} &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; \triangle {V_{rod}} + \triangle {V_{disk}}&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{rod}} {\bullet} \vec{l}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{disk}} {\bullet} \vec{l}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{rod}}=-\int\limits_{w}^{d} \frac{1}{4\pi\epsilon_0} &amp;lt;/math&amp;gt;&amp;lt;math&amp;gt;\frac{2q}{xL} dx&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} [ln(x)\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} ln(\frac{d}{w})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{disk}}=-\int\limits_{w}^{d} \frac{Q}{2A\epsilon_0} dx&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0} [x\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0L} (d-w)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = -&amp;lt;math&amp;gt;[\frac{2q}{4\pi\epsilon_0L} ln(\frac{d}{w})&amp;lt;/math&amp;gt; + &amp;lt;math&amp;gt;\frac{Q}{2A\epsilon_0L} (d-w)]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
5) The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
Answer:&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27606</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27606"/>
		<updated>2017-04-07T22:04:03Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]] &lt;br /&gt;
&lt;br /&gt;
1) If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
&lt;br /&gt;
up&lt;br /&gt;
 &lt;br /&gt;
left&lt;br /&gt;
 &lt;br /&gt;
right &lt;br /&gt;
&lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
&lt;br /&gt;
positive&lt;br /&gt;
&lt;br /&gt;
negative &lt;br /&gt;
&lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]] &lt;br /&gt;
&lt;br /&gt;
2) Three charged metal disks are arranged as shown on the right. The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2&amp;lt;math&amp;gt; m^2&amp;lt;/math&amp;gt; (this is the area of one side of the disk). Calculate the potential difference Vd-Va. &lt;br /&gt;
&lt;br /&gt;
Answer: &lt;br /&gt;
Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; D &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; G &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; H &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to &amp;lt;math&amp;gt;\vec{E_2}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Update Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Electric Field of Capacitors: &amp;lt;math&amp;gt; E_{cap} = \frac{Q}{A\epsilon_0} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = (Vb-Vh) + (Vg-Vd) = &amp;lt;math&amp;gt; -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
= &amp;lt;math&amp;gt;-\frac{1.2e-7C*0.003m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt; +  &amp;lt;math&amp;gt;-\frac{7e-8C*0.0015m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
= 14.4 V&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
3) Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
Answer:&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2 pot diff.jpg|300px|thumb|right|Problem 4 Diagram]] &lt;br /&gt;
&lt;br /&gt;
2) A very long, thin glass rod of length 2R carries a uniformly distributed charge +q. A very large plastic disk of radius R, carrying a uniformly distributed charge -Q, is located a distance d from the rod, where d&amp;lt;&amp;lt;R.&lt;br /&gt;
&lt;br /&gt;
a. Predict the sign of the potential difference Vc-Va. Justify your answer.&lt;br /&gt;
&lt;br /&gt;
Answer:&lt;br /&gt;
E and dl are both positive because they point to the right direction. According to the integral form of the potential difference formula for non-uniform field, &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; , the change in V from position A to C has to be negative.&lt;br /&gt;
&lt;br /&gt;
b. Calculate the potential difference Vc-Va. Location A is located a distance w fro the center of the rod. Location C is located at the disk&#039;s center a distance d from the rod. &lt;br /&gt;
&lt;br /&gt;
Answer:&lt;br /&gt;
Assume d&amp;lt;&amp;lt;R. &amp;lt;math&amp;gt; E_{disk} = \frac{Q}{2A\epsilon_0} &amp;lt;/math&amp;gt; &amp;amp; &amp;lt;math&amp;gt;E_{rod} = \frac{1}{4\pi\epsilon_0}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt; \frac{2Q}{Lr}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Vc-Va = &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E_{net}} {\bullet} \vec{l} &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; \triangle {V_{rod}} + \triangle {V_{disk}}&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{rod}} {\bullet} \vec{l}&amp;lt;/math&amp;gt; &amp;lt;math&amp;gt;-\int\limits_{i}^{f} \vec{E_{disk}} {\bullet} \vec{l}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{rod}}=-\int\limits_{w}^{d} \frac{1}{4\pi\epsilon_0} &amp;lt;/math&amp;gt;&amp;lt;math&amp;gt;\frac{2q}{xL} dx&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} ln(x)\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-2q}{4\pi\epsilon_0L} ln(\frac{d}{w})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V_{disk}}=-\int\limits_{w}^{d} \frac{Q}{2A\epsilon_0} dx&amp;lt;/math&amp;gt;= &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0} [x\rbrack^d_w &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\frac{-Q}{2A\epsilon_0L} (d-w)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27603</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27603"/>
		<updated>2017-04-07T21:34:15Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]] &lt;br /&gt;
&lt;br /&gt;
1) If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
&lt;br /&gt;
up&lt;br /&gt;
 &lt;br /&gt;
left&lt;br /&gt;
 &lt;br /&gt;
right &lt;br /&gt;
&lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
&lt;br /&gt;
positive&lt;br /&gt;
&lt;br /&gt;
negative &lt;br /&gt;
&lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]] &lt;br /&gt;
&lt;br /&gt;
2) Three charged metal disks are arranged as shown on the right. The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2&amp;lt;math&amp;gt; m^2&amp;lt;/math&amp;gt; (this is the area of one side of the disk). Calculate the potential difference Vd-Va. &lt;br /&gt;
&lt;br /&gt;
Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; D &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; G &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; H &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to &amp;lt;math&amp;gt;\vec{E_2}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Update Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Electric Field of Capacitors: &amp;lt;math&amp;gt; E_cap = \frac{Q}{A\epsilon_0} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = (Vb-Vh) + (Vg-Vd) = &amp;lt;math&amp;gt; -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
= &amp;lt;math&amp;gt;-\frac{1.2e-7C*0.003m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt; +  &amp;lt;math&amp;gt;-\frac{7e-8C*0.0015m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
= 14.4 V&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
3) Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Problem2 pot diff.jpg|500px|thumb|right|Problem 4 Diagram]] &lt;br /&gt;
&lt;br /&gt;
2) A very long, thin glass rod of length 2R carries a uniformly distributed charge +q. A very large plastic disk of radius R, carrying a uniformly distributed charge -Q, is located a distance d from the rod, where d&amp;lt;&amp;lt;R.&lt;br /&gt;
&lt;br /&gt;
a. Predict the sign of the potential difference Vc-Va. Justify your answer.&lt;br /&gt;
&lt;br /&gt;
E and dl are both positive because they point to the right direction. According to the integral form of the potential difference formula for non-uniform field, &amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; , the change in V from position a to c has to be negative.&lt;br /&gt;
&lt;br /&gt;
b. Calculate the potential difference Vc-Va. Location A is located a distance w fro the center of the rod. Location C is located at the disk&#039;s center a distance d from the rod. &lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27602</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27602"/>
		<updated>2017-04-07T21:20:38Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]] &lt;br /&gt;
&lt;br /&gt;
1) If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
&lt;br /&gt;
up&lt;br /&gt;
 &lt;br /&gt;
left&lt;br /&gt;
 &lt;br /&gt;
right &lt;br /&gt;
&lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
&lt;br /&gt;
positive&lt;br /&gt;
&lt;br /&gt;
negative &lt;br /&gt;
&lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]] &lt;br /&gt;
&lt;br /&gt;
2) Three charged metal disks are arranged as shown on the right. The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2 &amp;lt;math&amp;gt; m^2&amp;lt;/math&amp;gt; (this is the area of one side of the disk). Calculate the potential difference Vd-Va. &lt;br /&gt;
&lt;br /&gt;
Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; D &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; G &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; H &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to &amp;lt;math&amp;gt;\vec{E_2}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Update Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Electric Field of Capacitors: &amp;lt;math&amp;gt; E_cap = \frac{Q}{A\epsilon_0} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = (Vb-Vh) + (Vg-Vd) = &amp;lt;math&amp;gt; -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
= &amp;lt;math&amp;gt;-\frac{1.2e-7C*0.003m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt; +  &amp;lt;math&amp;gt;-\frac{7e-8C*0.0015m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
= 14.4 V&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27601</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27601"/>
		<updated>2017-04-07T21:18:40Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]] &lt;br /&gt;
&lt;br /&gt;
1) If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
&lt;br /&gt;
up&lt;br /&gt;
 &lt;br /&gt;
left&lt;br /&gt;
 &lt;br /&gt;
right &lt;br /&gt;
&lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
&lt;br /&gt;
positive&lt;br /&gt;
&lt;br /&gt;
negative &lt;br /&gt;
&lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]] &lt;br /&gt;
&lt;br /&gt;
2) Calculate the potential difference Vd-Va. &lt;br /&gt;
&lt;br /&gt;
Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; D &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; G &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; H &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to &amp;lt;math&amp;gt;\vec{E_2}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Update Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Electric Field of Capacitors: &amp;lt;math&amp;gt; E_cap = \frac{Q}{A\epsilon_0} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = (Vb-Vh) + (Vg-Vd) = &amp;lt;math&amp;gt; -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
= &amp;lt;math&amp;gt;-\frac{1.2e-7C*0.003m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt; +  &amp;lt;math&amp;gt;-\frac{7e-8C*0.0015m}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
= 14.4 V&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27600</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27600"/>
		<updated>2017-04-07T21:17:15Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]] &lt;br /&gt;
&lt;br /&gt;
1) If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
&lt;br /&gt;
up&lt;br /&gt;
 &lt;br /&gt;
left&lt;br /&gt;
 &lt;br /&gt;
right &lt;br /&gt;
&lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
&lt;br /&gt;
positive&lt;br /&gt;
&lt;br /&gt;
negative &lt;br /&gt;
&lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]] &lt;br /&gt;
&lt;br /&gt;
2) Calculate the potential difference Vd-Va. &lt;br /&gt;
&lt;br /&gt;
Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; D &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; G &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; H &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to &amp;lt;math&amp;gt;\vec{E_2}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Update Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Electric Field of Capacitors: &amp;lt;math&amp;gt; E_cap = \frac{Q}{A\epsilon_0} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Vc-Va = (Vb-Vh) + (Vg-Vd) = &amp;lt;math&amp;gt; -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
= &amp;lt;math&amp;gt;-\frac{1.2e-7C*3mm}{2m^2*8.85e-12 C^2/Nm^2}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27599</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27599"/>
		<updated>2017-04-07T21:02:03Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]]&lt;br /&gt;
&lt;br /&gt;
1) If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
up &lt;br /&gt;
left &lt;br /&gt;
right &lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
positive&lt;br /&gt;
negative &lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]] &lt;br /&gt;
&lt;br /&gt;
2) Calculate the potential difference Vd-Va. &lt;br /&gt;
&lt;br /&gt;
Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; D &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; G &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; H &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to &amp;lt;math&amp;gt;\vec{E_2}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Update Path: A &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; B &amp;lt;math&amp;gt;\rightarrow&amp;lt;/math&amp;gt; C&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Vc-Va = (Vb-Vh) + (Vg-Vd) = &amp;lt;math&amp;gt; -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=File:IMG_0370.JPG&amp;diff=27598</id>
		<title>File:IMG 0370.JPG</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=File:IMG_0370.JPG&amp;diff=27598"/>
		<updated>2017-04-07T20:43:43Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: Qqu3 uploaded a new version of &amp;amp;quot;File:IMG 0370.JPG&amp;amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=File:IMG_0370.JPG&amp;diff=27597</id>
		<title>File:IMG 0370.JPG</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=File:IMG_0370.JPG&amp;diff=27597"/>
		<updated>2017-04-07T20:43:25Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27596</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27596"/>
		<updated>2017-04-07T20:42:12Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
[[File:Example.jpg]]&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]]&lt;br /&gt;
&lt;br /&gt;
Refer to the figure above to answer the following question:&lt;br /&gt;
&lt;br /&gt;
If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
up &lt;br /&gt;
left &lt;br /&gt;
right &lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
positive&lt;br /&gt;
negative &lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27299</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27299"/>
		<updated>2017-04-03T16:59:30Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.&lt;br /&gt;
&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]]&lt;br /&gt;
&lt;br /&gt;
Refer to the figure above to answer the following question:&lt;br /&gt;
&lt;br /&gt;
If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
up &lt;br /&gt;
left &lt;br /&gt;
right &lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
positive&lt;br /&gt;
negative &lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27298</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27298"/>
		<updated>2017-04-03T16:58:22Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steps to calculate Potential Difference in a Non-Uniform Field:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1. Pick a path from initial to final location where the electric field can be evaluated at every location.&lt;br /&gt;
2. Divide the path into smaller pieces, with each pieces having its unique electric field and &amp;lt;math&amp;gt; \vec{l} &amp;lt;/math&amp;gt;.&lt;br /&gt;
3. Identify the electric field of the piece.&lt;br /&gt;
4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece&lt;br /&gt;
5. Move on to calculate the next piece.&lt;br /&gt;
6. Add contribution of all pieces together to get the net potential difference in non-uniform field.&lt;br /&gt;
&lt;br /&gt;
For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]]&lt;br /&gt;
&lt;br /&gt;
Refer to the figure above to answer the following question:&lt;br /&gt;
&lt;br /&gt;
If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
up &lt;br /&gt;
left &lt;br /&gt;
right &lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
positive&lt;br /&gt;
negative &lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27297</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27297"/>
		<updated>2017-04-03T16:43:52Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;In order to apply this general formula for potential difference to a point charge:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]]&lt;br /&gt;
&lt;br /&gt;
Refer to the figure above to answer the following question:&lt;br /&gt;
&lt;br /&gt;
If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
up &lt;br /&gt;
left &lt;br /&gt;
right &lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
positive&lt;br /&gt;
negative &lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27296</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27296"/>
		<updated>2017-04-03T16:43:19Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. The elctric field of the object can be that of a point charge, long rod, ,disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;In order to apply this general formula for potential difference to a point charge:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]]&lt;br /&gt;
&lt;br /&gt;
Refer to the figure above to answer the following question:&lt;br /&gt;
&lt;br /&gt;
If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
up &lt;br /&gt;
left &lt;br /&gt;
right &lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
positive&lt;br /&gt;
negative &lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27295</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27295"/>
		<updated>2017-04-03T16:39:31Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;In order to apply this general formula for potential difference to a point charge:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]]&lt;br /&gt;
&lt;br /&gt;
Refer to the figure above to answer the following question:&lt;br /&gt;
&lt;br /&gt;
If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
up &lt;br /&gt;
left &lt;br /&gt;
right &lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
positive&lt;br /&gt;
negative &lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27294</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27294"/>
		<updated>2017-04-03T16:38:04Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} &amp;lt;/math&amp;gt; with being the intial location and f being the final location. And the sum of all the contribution of potential difference can be written as: &amp;lt;math&amp;gt; {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} &amp;lt;/math&amp;gt;. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;In order to apply this general formula for potential difference to a point charge:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]]&lt;br /&gt;
&lt;br /&gt;
Refer to the figure above to answer the following question:&lt;br /&gt;
&lt;br /&gt;
If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
up &lt;br /&gt;
left &lt;br /&gt;
right &lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
positive&lt;br /&gt;
negative &lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27293</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27293"/>
		<updated>2017-04-03T16:30:05Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; {\triangle V}=-\int\limits_{i}^{f}Ed {\bullet} l &amp;lt;/math&amp;gt; being the intial location and f being the final location. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;In order to apply this general formula for potential difference to a point charge:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]]&lt;br /&gt;
&lt;br /&gt;
Refer to the figure above to answer the following question:&lt;br /&gt;
&lt;br /&gt;
If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
up &lt;br /&gt;
left &lt;br /&gt;
right &lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
positive&lt;br /&gt;
negative &lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27292</id>
		<title>Potential Difference of Point Charge in a Non-Uniform Field</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Difference_of_Point_Charge_in_a_Non-Uniform_Field&amp;diff=27292"/>
		<updated>2017-04-03T16:16:10Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by dmengesha3 Dina&#039;&#039;&#039; Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be: &lt;br /&gt;
&amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; -- i being the intial location and f being the final location. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;In order to apply this general formula for potential difference to a point charge:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we carry out the integration we find that Vf- Vi = &amp;lt;math&amp;gt;\vartriangle(9x10^9)\frac{Q}{r}&amp;lt;/math&amp;gt; which can also be written &amp;lt;math&amp;gt;((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This equation can be used to determine the potential difference of a point charge between any two different locations.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Refer to this Website:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html&lt;br /&gt;
&lt;br /&gt;
To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.&lt;br /&gt;
&lt;br /&gt;
Exercises&lt;br /&gt;
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther stil&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the magnitude of the arrows decreases as the distance from the point charge increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Repeat the steps of #1 but replace the positive point charge with a negative point charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;You should notice the same trend occur&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Click &amp;quot;Show E-field&amp;quot; on the right menu bar and move the point charge around the plot&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
4. Move the point charge closer to the observation location that measures potential difference&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Notice that the closer the point charge is to this location the higher the magnitude of the potential difference&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
[[File:Physicswikiproblem1.jpg]]&lt;br /&gt;
&lt;br /&gt;
Refer to the figure above to answer the following question:&lt;br /&gt;
&lt;br /&gt;
If the path is along the line i to f what direction is the electric field of the dipole? &lt;br /&gt;
&lt;br /&gt;
down &lt;br /&gt;
up &lt;br /&gt;
left &lt;br /&gt;
right &lt;br /&gt;
Cannot determine &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Answer: To the left. The electric field would point toward the negative end of the dipole&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Is the potential difference Vf-Vi &lt;br /&gt;
positive&lt;br /&gt;
negative &lt;br /&gt;
zero&lt;br /&gt;
&lt;br /&gt;
Answer: It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative. &lt;br /&gt;
&lt;br /&gt;
Shortcut to remember: &lt;br /&gt;
E and dl same direction --&amp;gt; negative potential difference&lt;br /&gt;
E and dl opposite direction--&amp;gt;positive potential difference&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Calculate the change in potential moving from location &amp;lt;math&amp;gt;(9x10^-9) m &amp;lt;/math&amp;gt; to location &amp;lt;math&amp;gt; (9.7x10^-9) m &amp;lt;/math&amp;gt; from a proton.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the final location is: (9.7x10^-9) m&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
the intial location is: (9x10^-9) m &lt;br /&gt;
&lt;br /&gt;
So if we plug these values into the above equation we found by integrating Edr. &lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&amp;lt;math&amp;gt;((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Which yields a potential difference of -0.0115 V.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location &amp;lt;x,y,z&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Working it through&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Remember that potential difference &amp;lt;math&amp;gt; \textstyle\int\limits_{i}^{f}-Edl &amp;lt;/math&amp;gt; thus Electric field can be calculated as the &amp;lt;math&amp;gt;\frac {-dv}{dx}&amp;lt;/math&amp;gt; So one would take the derivative of the potential difference with respect to each variable giving:&lt;br /&gt;
&lt;br /&gt;
E= &amp;lt;-14y,-14x-6,4&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Connectedness===&lt;br /&gt;
 How is this connected to something you are interested in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Connections in Chemistry&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 How is it connected to your major?&lt;br /&gt;
&lt;br /&gt;
I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy &amp;lt;math&amp;gt;\vartriangle G &amp;lt;/math&amp;gt; of a reaction in order to understand the reaction&#039;s energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &#039;&#039;&#039;Industrial Application&#039;&#039;&#039;&lt;br /&gt;
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
[[File:volta.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta. &lt;br /&gt;
&lt;br /&gt;
Refer to:&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Alessandro_Volta&lt;br /&gt;
&lt;br /&gt;
To read more about Volta&#039;s early works and life.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Calculating Electric Field of a Point Charge&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Point_Charge&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Potential Difference in Uniform Electric Field&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[[Connectedness]]&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[History]]&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Volt&lt;br /&gt;
&lt;br /&gt;
http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Loop&amp;diff=27291</id>
		<title>Magnetic Field of a Loop</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Loop&amp;diff=27291"/>
		<updated>2017-04-03T16:00:03Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Claimed by Jeffrey Mullavey; &#039;&#039;&#039;CLAIMED BY JORDAN MARSHALL TO EDIT FALL 2016&#039;&#039;&#039;&lt;br /&gt;
Claimed by Qianyi Qu Spring 2017&lt;br /&gt;
== Creation of a Magnetic Field around a Current-Carrying Loop==&lt;br /&gt;
&lt;br /&gt;
Like other magnetic field patterns, a magnetic field can be created through motion of charge through a loop. The system is not considered to be in equilibrium, therefore there is a movement of a mobile sea of electrons, which causes an electric current in the wire. Ideal loops are considered to be circular, so for the sake of calculations, a perfectly circular loop will be used. Thus, the conventional current is directed clockwise or counterclockwise through the loop, and depending on the direction of the flow of current, the magnetic field on the axis through the center of the loop will either go in the positive or negative direction of the axis, as shown below. Formulas have been derived to assist in the calculation of these magnetic fields.&lt;br /&gt;
&lt;br /&gt;
==Calculation of Magnetic Field==&lt;br /&gt;
&lt;br /&gt;
The magnetic field created by a loop is easiest to calculate on axis. This means drawing a line though the center of the loop perpendicular to the plane of the loop. This axis is commonly referred to as the &amp;quot;z-axis.&amp;quot; The magnetic field is calculated by integrating across the bounds of the loop (0 to 2 pi), but can also be approximated with great accuracy using a derived formula. Calculation of magnetic field off of this axis is much more difficult, and usually requires the assistance of computer software. For this reason, only the calculation on axis will be addressed.&lt;br /&gt;
&lt;br /&gt;
===Magnitude of Magnetic Field===&lt;br /&gt;
&lt;br /&gt;
When calculating the magnetic field at a point on the z-axis, one can use the following formula:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec B=\frac{\mu_0}{4 \pi} \frac{2I \pi R^2}{(z^2 + R^2)^{3/2}} \text{ ,where R is the radius of the circular loop, and z is the distance from the center of the loop} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This allows for the calculation of the magnitude in units of Teslas.&lt;br /&gt;
&lt;br /&gt;
If the distance from the center of the loop is much greater than the radius of the loop, an approximation can be made. The formula can be simplified to&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec B=\frac{\mu_0}{4 \pi} \frac{2I \pi R^2}{z^3} \text{ ,where z is much greater than R} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Direction of Magnetic Field===&lt;br /&gt;
&lt;br /&gt;
[[File: Mag Field in Loop.PNG]]&lt;br /&gt;
&lt;br /&gt;
The right hand rule can be used to find the direction of the magnetic field at a given point. Putting your right fingers in the direction of the conventional current, and curling them over the vector r will allow your thumb to point in the direction of the magnetic field. For example, a conventional current, I, running counterclockwise would produce a field pointing out of the page on the z-axis. This rule holds regardless of where the observational location is on the center axis.&lt;br /&gt;
&lt;br /&gt;
[[File: RHR7.PNG]]&lt;br /&gt;
&lt;br /&gt;
The magnetic field pattern for locations outside the ring points in the opposite direction, in accordance with the right hand rule. The right fingers still point in the direction of the current, however the observational vector perpendicular to the perimeter of the loop is now in the opposite direction. Thus, the magnetic field is in the opposite direction.&lt;br /&gt;
&lt;br /&gt;
===Multiple Loops===&lt;br /&gt;
&lt;br /&gt;
To provide a higher magnitude in the magnetic field, many practical applications require the addition of multiple loops to &amp;quot;magnify&amp;quot; the effect of the magnetic field. In many scenarios, electric current will run through a formation of a number of loops, N, to satisfy this increasing need for a larger magnetic field. In this case, the magnitude of the induced magnetic field can be found by calculating the field produced by one loop and multiplying it by the number of loops.&lt;br /&gt;
&lt;br /&gt;
This equation for the multiple loop configuration of the magnetic field of a loop will follow the equation below.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec B=\frac{\mu_0}{4 \pi} \frac{2I \pi R^2}{(z^2 + R^2)^{3/2}} N  &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Example===&lt;br /&gt;
 &lt;br /&gt;
Point C is located 2 meters away from the center of a loop of current. The loop has 5 Amps of current flowing around it and has a radius of .1 meters. What is the magnetic field at point C?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;b&amp;gt; ANSWER &amp;lt;/b&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec B=\frac{\mu_0}{4 \pi} \frac{2I \pi R^2}{z^3} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Using this equation, you can plug in the numbers.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec B=\frac{\mu_0}{4 \pi} \frac{2(5) \pi (.1)^2}{2^3} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
B = 3.93e-9 T&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
Magnetic fields from electric loops are observed often in science and industrial applications. For example, a solenoid is often modeled as a bunch of loops contributing to a collective magnetic field. It is important to be able to model the field, since many scientific applications of magnetism involve circular loops. The calculations can be compared to experiments done in the lab. &lt;br /&gt;
&lt;br /&gt;
This validation of theory can be extended to industrial and technological innovations seen throughout different career paths. Take, for example, the thriving CubeSat business. CubeSats are essentially miniaturized satellites (sometimes weighing between 8 and 4 kg) that perform very specific scientific missions. The payloads carried on these satellites often require a specific attitude with respect to the Earth/Sun in order to complete communications with a ground station below or to continue to receive solar power from the sun. To do this, many CubeSats often employ the use of Magnetorquer rods. These rods are essentially a a solenoid (i.e. current-carrying loop with several thousand loops, N) that surrounds a metal rod (sometimes Stainless Steel). When current flows through the coils, a magnetic field is induced along the axis of the rod. The magnetic field continues to interact with the Earth&#039;s magnetic field, producing a torque on the spacecraft. When multiple magnetorquer rods are used on different axes of the spacecraft, the CubeSat can have complete control it&#039;s attitude to continue the mission as necessary.&lt;br /&gt;
&lt;br /&gt;
In all concentrations of engineering, electric and magnetic properties are important. For example, these concepts can be applied to the synthesis and manufacturing of conductors and carbon nanotubes. In many electrical experiments, it is important to understand how objects will be impacted by the creation of an induced magnetic field.&lt;br /&gt;
&lt;br /&gt;
==See Also==&lt;br /&gt;
&lt;br /&gt;
There are quite a few youtube videos out there that may explain the process for calculating the magnetic field on the axis through a loop carrying a current. One of those such videos can be found [https://www.youtube.com/watch?v=lN296gUXkl4 here]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
Matter and Interactions Vol. II&lt;br /&gt;
&lt;br /&gt;
[[Category:Fields]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Angular_Impulse&amp;diff=23660</id>
		<title>Angular Impulse</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Angular_Impulse&amp;diff=23660"/>
		<updated>2016-11-17T03:30:37Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Angular impulse represents the effect of a moment of force, or torque (&amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\tau&amp;lt;/math&amp;gt;), acting on a system over a certain period of time (&amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\Delta t&amp;lt;/math&amp;gt;). Angular impulse helps indicate the direction that the system will rotate in (clockwise or counterclockwise) since it is associated with change in velocity and acceleration.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Angular impulse is the torque acting over some time interval, or the change in angular momentum. Angular momentum can be changed by an angular impulse. There is no common symbol for angular momentum like how &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\vec{F}&amp;lt;/math&amp;gt; is for force and &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\vec{p}&amp;lt;/math&amp;gt; is for momentum, and as a result it is almost always referred to as &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\Delta\vec{L}&amp;lt;/math&amp;gt;, since it is equal to the change in angular momentum &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\vec{L}&amp;lt;/math&amp;gt;, just like how linear impulse (&amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;J&amp;lt;/math&amp;gt;) is equal to the change in linear momentum, &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\Delta\vec{p}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The angular impulse is equal to the net cross product of a force vector, &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\vec{F}&amp;lt;/math&amp;gt;, applied at a particular location a vector distance &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\vec{d}&amp;lt;/math&amp;gt; from a pivot point times a specified time interval &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\Delta t&amp;lt;/math&amp;gt;. This is also equal to the net torque &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\sum{\vec{\tau}}&amp;lt;/math&amp;gt; times a specified time interval &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\Delta t&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\Delta \vec{L} = \sum{(\vec{F}\times\vec{d})}⋅\Delta t = \sum{\vec{\tau}}⋅\Delta t&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The angular impulse is equal to the moment of inertia &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;I&amp;lt;/math&amp;gt; times the change in angular velocity &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\Delta\vec{\omega}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\Delta \vec{L} = I\Delta\vec{\omega} = I\vec{\omega_f} - I\vec{\omega_i}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==== Angular Momentum Principle ====&lt;br /&gt;
The angular momentum principle directly involves angular impulse as shown in the image below:&lt;br /&gt;
[[File:Netangularimpulse.png]]&lt;br /&gt;
&lt;br /&gt;
Both sides are equal to the net angular impulse for a system.&lt;br /&gt;
&lt;br /&gt;
==== Units ====&lt;br /&gt;
The units for angular impulse are the same as those for angular momentum: &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;kg⋅m^2/s&amp;lt;/math&amp;gt; or &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;N⋅m⋅s&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Angular impulse is often used to update angular momentum when there is a torque acting on an object. Much like how force times time (impulse) is used to update momentum by adding it to an initial momentum in order to obtain the final momentum when the force is constant, angular impulse can also be used to find final angular momentum or final angular velocity. This can be done by adding angular impulse to an initial angular momentum in a while loop and setting that equal to the final angular momentum. Below is part of a simple code example of a while loop that will update the final angular momentum by adding angular impulse. The final angular momentum is angm_f, the initial angular momentum is angm_i, torquenet is the net torque, deltat is some pre-defined time step, and t is time.&lt;br /&gt;
&lt;br /&gt;
[[File:Angm.png]]&lt;br /&gt;
&lt;br /&gt;
In order to find final angular velocity, one could simply divide the final angular momentum by the moment of inertia (a constant) within the while loop after updating the final angular momentum (angm_f in the example) and before updating the time (t in the example).&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
The moment of inertia of an upright solid cylinder is 22.5 m ⋅ r^2. The cylinder is rotated from rest and has a final angular velocity of 5 rad/s. What is the angular impulse of the cylinder?&lt;br /&gt;
&lt;br /&gt;
====Solution====&lt;br /&gt;
We are given the moment of inertia, final angular velocity, and deduce that the initial angular velocity is 0 rad/s since it began rotating from rest. As a result, we can find the angular velocity by multiplying the moment of inertia with the change in angular velocity, or 5 rad/s minus 0 rad/s times 22.5 m ⋅ r^2. This gives us the angular impulse, 112.5 N ⋅ m ⋅ s.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A net force of 40 N is applied to the rim of a spinning wheel for 2 seconds. The radius of the wheel is 20 cm. Find the angular impulse that is applied to this wheel.&lt;br /&gt;
&lt;br /&gt;
====Solution====&lt;br /&gt;
First, we need to find the torque acting on the wheel. Torque is equal to the applied force times the radius, or 40 N ⋅ .2 meters which equals 8 N ⋅ m.&lt;br /&gt;
Currently, we know the torque applied as well as the duration of its application. Therefore, we can find the angular impulse, which is the applied torque times the duration, or (8 N ⋅ m) ⋅ (2 s) which equals 16 N ⋅ m ⋅ s, the angular impulse.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
A point on a solid disk on the xy-plane with radius 5 cm rotates with a constant initial frequency of 5 revolutions per second. The mass of the disk is 2 kilograms. Someone applies a force to the wheel suddenly, giving it a constant angular acceleration of 6 rad/s^2 for 3 seconds. Use the angular impulse to find the final angular velocity.&lt;br /&gt;
====Solution====&lt;br /&gt;
This problem involves several steps. It is clear that if we are going to find the angular impulse, we either need to find the torque times the time interval or the change in angular velocity.&lt;br /&gt;
For finding the angular velocity, we are given the initial frequency and radius of the disk, so since angular velocity is equal to 2 times pi times the frequency, we get 10⋅pi rad/s for the initial angular velocity.&lt;br /&gt;
&lt;br /&gt;
Now that we have the initial angular velocity, we need to somehow find the final angular velocity. Since angular impulse is the change in angular velocity, we can use this to determine what the final angular velocity is.&lt;br /&gt;
&lt;br /&gt;
Angular impulse is equal to torque times time. Remember that the torque is equal to the moment of inertia times angular acceleration. We are told that we are dealing with a solid disk, so we need to find the moment of inertia which is mr^2, or 2 kg ⋅ (.05 m)^2, which gives us a moment of inertia of .005 kg ⋅ m^2. Since we are given the angular acceleration and have the moment of inertia, we can now find the torque, which is equal to (.005 kg ⋅ m^2) ⋅ (6 rad/s^2), or .03 N⋅m.&lt;br /&gt;
&lt;br /&gt;
Since we now have the torque and the duration that it is applied is given, we can now find the angular impulse. We multiple the torque by time, or (.03 N⋅m) ⋅ (3 s), and get .09 N⋅m⋅s as our angular impulse.&lt;br /&gt;
&lt;br /&gt;
Now that we have the angular impulse, we can find the final angular velocity. Before that, we need our initial angular momentum, which we get by multiplying the initial angular velocity by the moment of inertia. This gives us .05 pi or about .157 N⋅m⋅s. We have the angular impulse, so we add this to the initial angular momentum and get our final angular momentum, about .247 N⋅m⋅s. Then we divide by the moment of inertia to get the final angular velocity, about 49.42 rad/s.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
Angular impulse is present in so many things in daily life, from wheels turning on a bicycle to turning the steering wheel in a car and even a person just spinning around in place. Personally, I&#039;m really interested in computers, desktop computers in particular. This topic relates to the turning of fans in my case, on my graphics card, and on my processor, so angular momentum is critical when it can mean a negative one would result in drastically lower fan speeds that would make a computer overheat or a postive one would result in an increase in fan speed which would likely result in the fans being really noisy and annoying.&lt;br /&gt;
&lt;br /&gt;
As a Computer Science major, angular momentum relates to actual computers in the example I gave previously. Not only that, but in the branch of artificial intelligence, if robots are involved then angular impulse can be critical in their circular motion.&lt;br /&gt;
&lt;br /&gt;
Angular impulse has numerous industrial applications, being critical in any rotating device, like cars (wheels, steering wheels), generators, and even in water/wind mills which can provide hydroelectric/wind power.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
[http:///www.physicsbook.gatech.edu/Torque Torque]&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsbook.gatech.edu/The_Angular_Momentum_Principle The Angular Momentum Principle]&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsbook.gatech.edu/The_Moments_of_Inertia Moments of Inertia]&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsbook.gatech.edu/Angular_Velocity Angular Velocity]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[https://www.youtube.com/watch?v=KWrVvLxUhlA Angular Momentum Impulse Video]&lt;br /&gt;
&lt;br /&gt;
[http://webpages.charter.net/griche/pt/u8s2prb.htm Angular Impulse Practice]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
Chapter 11 of [https://books.google.com/books?id=Gz4HBgAAQBAJ&amp;amp;pg=PA544&amp;amp;lpg=PA544&amp;amp;dq=matter+and+interactions+4th+edition+torque&amp;amp;source=bl&amp;amp;ots=ShdH7G8bcV&amp;amp;sig=uEQbxhpX3-UqcQf4ilXjp2reG5s&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwin-JTU6MTJAhWDQCYKHUoLA00Q6AEIMjAD#v=onepage&amp;amp;q&amp;amp;f=false Matter &amp;amp; Interactions 4th Edition]&amp;lt;div&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
[https://en.wikipedia.org/wiki/List_of_equations_in_classical_mechanics#Derived_dynamic_quantities List of Equations in Classical Mechanics]&amp;lt;div&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
[http://wweb.uta.edu/faculty/ricard/Classes/KINE-3301/Notes/Lesson-13.html Lesson 13 Angular Impulse]&amp;lt;div&amp;gt;&amp;lt;/div&amp;gt;&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23576</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23576"/>
		<updated>2016-11-15T07:39:35Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Main Idea */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Edited by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (&amp;lt;math&amp;gt;\Delta p = 0&amp;lt;/math&amp;gt;), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
[[File:Merrygoround01.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
[[File:FullSizeRender (7).jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23575</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23575"/>
		<updated>2016-11-15T07:37:51Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Main Idea */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Edited by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (&amp;lt;math&amp;gt;\delta p = 0&amp;lt;/math&amp;gt;), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
[[File:Merrygoround01.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
[[File:FullSizeRender (7).jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23574</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23574"/>
		<updated>2016-11-15T07:37:09Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Edited by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
[[File:Merrygoround01.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
[[File:FullSizeRender (7).jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23573</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23573"/>
		<updated>2016-11-15T07:23:22Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Difficult */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
[[File:Merrygoround01.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
[[File:FullSizeRender (7).jpg]]&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23572</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23572"/>
		<updated>2016-11-15T06:50:13Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Simple */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
[[File:Merrygoround01.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23571</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23571"/>
		<updated>2016-11-15T06:41:51Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Simple */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
[[File:FullSizeRender (4).jpg]]&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23570</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23570"/>
		<updated>2016-11-15T06:41:26Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Middling */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23569</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23569"/>
		<updated>2016-11-15T06:29:48Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Middling */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg|upright=0.90|]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23568</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23568"/>
		<updated>2016-11-15T06:29:24Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Middling */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg|upright=0.60|]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23567</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23567"/>
		<updated>2016-11-15T06:27:49Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Middling */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg|upright|]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23566</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23566"/>
		<updated>2016-11-15T06:26:15Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Simple */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016.pdf]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23565</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23565"/>
		<updated>2016-11-15T06:25:10Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Simple */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg]]&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016.pdf]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23564</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23564"/>
		<updated>2016-11-15T06:24:40Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Simple */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016-page-001.jpg|thumb|Problem 2]]&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016.pdf]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23563</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23563"/>
		<updated>2016-11-15T06:19:44Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016.pdf]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23562</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23562"/>
		<updated>2016-11-15T06:18:55Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Middling */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
[[File:Note Nov 15, 2016.pdf|thumb|Problem 2]]&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23561</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23561"/>
		<updated>2016-11-15T05:12:49Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: Tarzan with a mass of 90kg is hanging from a vine that is 8m long. At the instant when he makes an 30 degree angle between the vine and the vertical, what is the rate of change of the magnitude of Tarzan&#039;s momentum?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Example&#039;&#039;: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23560</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23560"/>
		<updated>2016-11-15T05:07:05Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
Ex: A child whose mass is 30 kg sits on a merry-go-round at a distance of 3m from the center. The merry-go-round makes one revolution every 8s. What are the magnitude and direction of the net force acting on the child?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Ex: The Ferris Wheel has a radius r and rotates counter-clockwise with a constant angular rate ω. A person with mass m is riding in the gondola at an angle of θ = 40◦ from the x-axis as shown. What is the magnitude of the parallel and perpendicular components of the contact forces?&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23559</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23559"/>
		<updated>2016-11-15T04:51:39Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Simple */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23558</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23558"/>
		<updated>2016-11-15T04:50:52Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Simple */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23557</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23557"/>
		<updated>2016-11-15T04:49:02Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately. For example, if an object is traveling in a circular path at constant speed, we can neglect the parallel component of &amp;lt;math&amp;gt; \frac{d\vec{p}{dt}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; \vec{F}_{net}&amp;lt;/math&amp;gt; and consider only the perpendicular components, such as tension, gravitational, and contact forces. &lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A oriented coordinates system must be created to accommodate this change and it will be aligned with &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23556</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23556"/>
		<updated>2016-11-15T04:37:58Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
If a person on the Ferris Wheel is not positioned on the x- or y-axis, we cannot assume that the gravitational and contact forces are acting the same way as when the person is sitting on the Cartesian coordinate system. The reason we chose this coordinate system is because it coincides with the directions of the parallel and perpendicular forces. A new coordinates system must be created to accommodate the change and it will only be depended on &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23555</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23555"/>
		<updated>2016-11-15T04:20:51Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = {\frac{d\vec{p}}{dt}} = {\frac{d\vec{{|}p{|}}}{dt}}\hat p + {\frac{d\hat p}{dt}}{\vec{{|}p{|}}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23554</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23554"/>
		<updated>2016-11-15T04:10:42Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; [[File:Forces .jpg|thumb|Parallel and Perpendicular Components of Force in a Circular Path]]&lt;br /&gt;
&lt;br /&gt;
Note that &amp;lt;math&amp;gt; \vec{F}_\perp&amp;lt;/math&amp;gt; always points inward of a circular path. &amp;lt;math&amp;gt; \vec{F}_\parallel &amp;lt;/math&amp;gt; gives the change in magnitude of the object&#039;s momentum, and &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; gives the change in direction of the object&#039;s momentum.&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23553</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23553"/>
		<updated>2016-11-15T03:29:35Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Parallel and Perpendicular Forces */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
[[File:Example.jpg|thumb|]]&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23552</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23552"/>
		<updated>2016-11-15T03:18:33Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Parallel and Perpendicular Forces */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
[[File:forces.jpg]200px|thumb|right|Components of Forces]]]&lt;br /&gt;
&lt;br /&gt;
[[File:forces.jpg]]&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
&lt;br /&gt;
[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
&lt;br /&gt;
Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
&lt;br /&gt;
===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
&lt;br /&gt;
[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
&lt;br /&gt;
===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23551</id>
		<title>Curving Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Curving_Motion&amp;diff=23551"/>
		<updated>2016-11-15T03:17:58Z</updated>

		<summary type="html">&lt;p&gt;Qqu3: /* Parallel and Perpendicular Forces */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;claimed by Aayush Kumar&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Editing by Qianyi Qu (qqu3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
== Main Idea ==&lt;br /&gt;
&lt;br /&gt;
In the previous section, we introduced the idea of conservation of momentum. By choosing multiple particles as the system, we can assume that the forces exerted by the particles in the system are reciprocal of one another, based on Newton&#039;s Third Law of Motion. The total change in momentum is zero if the system&#039;s forces all canceled out and no forces in the surrounding are acting on the system. This special case, where the momentum of the system remains constant (deltap = 0), can be implicated in many situations that helps us to identify forces exerted on the system. However, in the case when the momentum of the system changes with time, which means velocity and/or direction of motion are nonconstant and additional terms are needed to define the magnitude and direction of the momentum of the objects in the system. The momentum of an object is nonconstant when it is traveling along a curved path, regardless of whether or not it&#039;s speed along the curve is changing because the direction also changes. Curving motion can be analyzed using the properties of the parallel and perpendicular components of net Force as well as understanding the motion&#039;s &amp;quot;Kissing Circle&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Parallel and Perpendicular Forces ==&lt;br /&gt;
&lt;br /&gt;
The parallel force lies along the direction of the momentum vectors and is calculated as &amp;lt;math&amp;gt; \vec{F}_\parallel = {|}\vec{F}{|}cos\theta \hat p&amp;lt;/math&amp;gt;. Since &amp;lt;math&amp;gt; \vec{F} &amp;lt;/math&amp;gt; is the sum of the parallel and perpendicular components, &amp;lt;math&amp;gt; \vec{F}_\perp &amp;lt;/math&amp;gt; can be obtain simply by taking the difference between the two: &amp;lt;math&amp;gt; \vec{F}_\perp = \vec{F} - \vec{F}_\parallel &amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
[[File:forces.jpg]200px|thumb|right|Components of Forces]]]&lt;br /&gt;
&lt;br /&gt;
==Understanding the Components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt;==&lt;br /&gt;
&lt;br /&gt;
When the change in momentum with respect to time is nonzero, we will consider both the parallel and perpendicular components of momentum and their relationship to the net force. If an object&#039;s speeding up or slowing down, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel \ne 0 &amp;lt;/math&amp;gt;. If an objects is turning while traveling along a path, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp \ne 0 &amp;lt;/math&amp;gt;. The parallel component accounts for the change in speed of the motion, while the perpendicular component accounts for the change in direction of the motion. Here is a list of useful formulas:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}} = {\frac{d\vec{p}}{dt}}_\parallel + {\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_\parallel + \vec{F}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel = {\frac{d\vec{{|}p{|}}}{dt}}\hat p = \vec{F}_\parallel &amp;lt;/math&amp;gt;      (Change in magnitude of momentum)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = \vec{F}_\perp &amp;lt;/math&amp;gt;     (Change in direction of momentum)&lt;br /&gt;
&lt;br /&gt;
===Perpendicular Component and the Kissing Circle===&lt;br /&gt;
The perpendicular component of momentum can be evaluate further more so it becomes &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp = {\frac{d\hat p}{dt}}{\vec{{|}p{|}}} = {\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n= \frac{m v^{2}}{R}\hat n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The term &amp;lt;math&amp;gt;{\vec{{|}p{|}}} \frac{{|}\vec{v}{|}}{R} \hat n &amp;lt;/math&amp;gt; can be understood by drawing a imaginary circle that touches or &#039;kisses&#039; some point on a curved path to represent the instantaneous curvature of an object&#039;s trajectory, where R is the radius of the circle. For a smooth, continuous curving motion, &amp;lt;math&amp;gt; \frac{{|}\vec{v}{|}}{R} &amp;lt;/math&amp;gt; tells us the degree of which the object is turning and the rate of change of direction,&amp;lt;math&amp;gt; {\frac{d\hat p}{dt}} &amp;lt;/math&amp;gt;. The &#039;Kissing Circle&#039; indicates the direction of the perpendicular component of momentum, which points towards the center of the kissing circle and it&#039;s denoted by &amp;lt;math&amp;gt;\hat n&amp;lt;/math&amp;gt;.&lt;br /&gt;
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[[File:Osculating circle.svg|Kissing Circle]]&lt;br /&gt;
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===Parallel Component and Tangential Properties of Curving Motion===&lt;br /&gt;
The parallel component of momentum always points in the direction of motion, denoted by &amp;lt;math&amp;gt;\hat p&amp;lt;/math&amp;gt; and it&#039;s tangent to the path of the motion. The term &amp;lt;math&amp;gt;{\frac{d\vec{{|}p{|}}}{dt}}&amp;lt;/math&amp;gt; describes the change in velocity of the objects with respect to time (since &amp;lt;math&amp;gt;\vec{p} = m\vec{v}) &amp;lt;/math&amp;gt; and it corresponds to the change in the magnitude of the momentum.&lt;br /&gt;
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There are two scenarios of curving motion that we will entertain here, one where an object is not changing speed along its curved path and another where the object speeds up or slows down along this curved path. We will start with the constant speed scenario:&amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel=0&amp;lt;/math&amp;gt;&lt;br /&gt;
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When an object&#039;s tangential velocity is constant throughout the entire motion, we can say that the parallel component of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is zero, as any change in momentum is solely for altering the direction of the object&#039;s momentum as opposed to its magnitude. To understand this in more mathematical terms, the only way the magnitude of a momentum vector can be changed is if some component of the force acting on the object lies in the direction of the objects motion. Otherwise, the net force must be in the perpendicular direction towards the center of the kissing circle, like shown below:&lt;br /&gt;
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[[File:Vector-diagram.png|Kissing Circle]]&lt;br /&gt;
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Here we see that the tangential velocities indicated by the red arrows are all of the same magnitude, and the only thing altering them is the centripetal acceleration, which leads into our discussion of centripetal and centrifugal forces later. Now we will consider cases where &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is nonzero. In these cases, the speed of the object along it&#039;s curved path is changing and our net force vector has a component that is parallel to the object&#039;s tangential motion.&lt;br /&gt;
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Calculating &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt;&lt;br /&gt;
Given a instantaneous force acting on the object as well as it&#039;s initial momentum, we can calculate the final momentum using the Momentum Principle and then observe any change in magnitude between these two vector quantities. Here, &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; is equal to the difference between the magnitudes of the initial and final momentum divided by the change in time. From this, we can calculate &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\perp&amp;lt;/math&amp;gt; by subtracting &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_\parallel&amp;lt;/math&amp;gt; from the net change in momentum. &lt;br /&gt;
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===A Computational Model Example===&lt;br /&gt;
Here&#039;s a glowscript animation utilizing this method of calculating the parallel and perpendicular components of &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}&amp;lt;/math&amp;gt; is linked below:&lt;br /&gt;
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[http://www.glowscript.org/#/user/aayush.kumarmail/folder/Private/program/SpaceVoyageandCurvingMotion Curving Motion with a Satellite Projectile]&lt;br /&gt;
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===Centripetal and Centrifugal Forces===&lt;br /&gt;
The Centripetal and Centrifugal Forces of an object in motion demonstrate reciprocity in that the Centripetal Force acts inward towards the center of the kissing circle whereas the centrifugal force is the object&#039;s tendency to fly outward.&lt;br /&gt;
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===Simple===&lt;br /&gt;
It&#039;s fairly simple identifying cases of curving motion, in which case a person can easily apply &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt; to a situation immediately.&lt;br /&gt;
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===Middling===&lt;br /&gt;
Occasionally it can be difficult to gauge any components that influence the object&#039;s tangential velocity and effectively speed up or slow down the object by altering the magnitude of the momentum vector.&lt;br /&gt;
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===Difficult===&lt;br /&gt;
It can be fairly difficult to understand when there are other factors that influence the perpendicular component of the change of momentum, or when an individual needs to identify another influential force in that axis. For instance, in the Ferris Wheel example, when asked about having to find the necessary velocity for the rider to feel weightless, the individual needs to understand that there&#039;s a surface contact force that is also in play and in this case needs to be 0, in which case gravitational force is equal to &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
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==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
Curving motion is heavily employed when observing planetary orbits, something I find highly fascinating. Although in reality these orbits aren&#039;t perfectly circular as we often make them out to be, they still heavily demonstrate the ideas of curving motion and especially &amp;lt;math&amp;gt;{\frac{m v^{2}}{R}}&amp;lt;/math&amp;gt;!&lt;br /&gt;
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#How is it connected to your major?&lt;br /&gt;
As of now I am a CS major, and using vPython to model curving motion animations among many things helps apply programming ideas such as iteration and variable update.&lt;br /&gt;
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#Is there an interesting industrial application?&lt;br /&gt;
Centripetal and Centrifugal forces can be found in many industrial applications, as curved motion is a very frequent occurrence when observing interactions between various objects. For instance, a centrifuge using these mechanics helps astronauts train for higher effective gravity.&lt;br /&gt;
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[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Qqu3</name></author>
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