Point Charge: Difference between revisions

Revision as of 01:58, 9 April 2017

This page is all about the Electric Field due to a Point Charge.CLAIMED BY Richard Granger 4-8-2017 == Electric Field==

The Main Idea

Every charge creates its own electric field. The two known elementary charges are the proton and electron, both of which are modeled as point charges. The electric field for point charges points radially outward in the case of the proton, and radially inward in the case of the electron. The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

A Mathematical Model of Electric Field due to Point Charge

The Electric Field of a Point Charge can be found by using the following formula:

[math]\displaystyle{ \vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r }[/math]

Which is derived from Gauss's Law, modelling the point charge as a sphere.

[math]\displaystyle{ {\epsilon_0} }[/math] is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately [math]\displaystyle{ 8.854*10^{-12}\frac{C^2}{N m^2} }[/math]

[math]\displaystyle{ \frac{1}{4 \pi \epsilon_0 } }[/math] is known as Coulomb's Constant and is approximately [math]\displaystyle{ 8.987*10^{9}\frac{N m^2}{C^2} }[/math]

r is the magnitude of the distance between the observation location and the source location , q is the charge of the particle and [math]\displaystyle{ \hat r }[/math] is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by [math]\displaystyle{ F = Eq }[/math] where E is the electric field and q is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

[math]\displaystyle{ F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r }[/math]

By solving for the electric field in [math]\displaystyle{ F = Eq }[/math] modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

[math]\displaystyle{ E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r }[/math]

Below is a visual representation of electric field lines due to a positive and negative point charge

A Computational Model

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of [math]\displaystyle{ \frac{1}{r^{2}} }[/math] as the observation location gets closer to the proton.

Examples

Simple

There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

Step 1: Find [math]\displaystyle{ \hat r }[/math]

Find [math]\displaystyle{ \vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = \lt 4,-3,1\gt }[/math]m

Calculate the magnitude of [math]\displaystyle{ \vec r }[/math]. ([math]\displaystyle{ \sqrt{4^2+(-3)^2+1^2}=\sqrt{26} }[/math]

From [math]\displaystyle{ \vec r }[/math], find the unit vector [math]\displaystyle{ \hat{r}. }[/math] [math]\displaystyle{ \lt \frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}\gt }[/math]

Step 2: Find the magnitude of the Electric Field

[math]\displaystyle{ E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} }[/math]

Step 3: Multiply the magnitude of the Electric Field by [math]\displaystyle{ \hat{r} }[/math] to find the Electric Field

[math]\displaystyle{ E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*\lt \frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}\gt = \lt -4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}\gt N/C }[/math]

Middling

A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is [math]\displaystyle{ \lt 0.124, 0.188, 0.109\gt }[/math] N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and [math]\displaystyle{ \hat{r} }[/math] Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,[math]\displaystyle{ E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} }[/math] one can find the magnitude and sign of the charge.

Step 1: Find [math]\displaystyle{ \vec r_{obs} - \vec r_{particle} }[/math].

[math]\displaystyle{ \vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = \lt 0.19,0.29,0.17\gt m }[/math]

To find [math]\displaystyle{ \vec r_{mag} }[/math], find the magnitude of [math]\displaystyle{ \lt 0.19,0.29,0.17\gt }[/math]

[math]\displaystyle{ \sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39 }[/math]

Step 2: Find the magnitude of the Electric Field

[math]\displaystyle{ E= \lt 0.124, 0.188, 0.109\gt N/C }[/math]

[math]\displaystyle{ E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25 }[/math]

Step 3: Find q by rearranging the equation for [math]\displaystyle{ E_{mag} }[/math]

[math]\displaystyle{ E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} }[/math]

By rearranging this equation we get

[math]\displaystyle{ q= {4 pi * epsilon_0 } *{r^2}*E_{mag} }[/math]

[math]\displaystyle{ q= {1/(9*10^9)} *{0.39^2}*0.25 }[/math]

[math]\displaystyle{ q= + 4.3*10^{-12} C }[/math]

Difficult

The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is [math]\displaystyle{ \lt -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}\gt }[/math] N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

[math]\displaystyle{ E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} }[/math]

to find the rmag value. To find [math]\displaystyle{ \hat r }[/math] we can find the direction of the electric field as that is obviously going to be in the same direction as [math]\displaystyle{ \hat r }[/math]. Then, once we find [math]\displaystyle{ \hat r }[/math], all that is left to do is multiply [math]\displaystyle{ \hat r }[/math] by rmag and that will give us the [math]\displaystyle{ r }[/math] vector. We can then find the location of the particle as we know [math]\displaystyle{ r=r_{observation}-r_{particle} }[/math]

Step 1: Find the magnitude of the Electric field.

[math]\displaystyle{ F = Eq }[/math]

[math]\displaystyle{ \lt 5.5*10^{3}, -7.6*10^{3}, 0\gt = E * -2mC }[/math]

[math]\displaystyle{ E = \frac{\lt -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}\gt }{-2mC} = \lt 2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}\gt }[/math] N/C

Step 2:: Find [math]\displaystyle{ \vec r_{obs} - \vec r_{particle} }[/math].

[math]\displaystyle{ \vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = \lt 3.98 , 3.98 , 3.98\gt m }[/math]

To find [math]\displaystyle{ \vec r_{mag} }[/math], find the magnitude of [math]\displaystyle{ \lt 3.98 , 3.98 , 3.98\gt }[/math]

[math]\displaystyle{ \sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9 }[/math]

Step 3: Find the magnitude of the Electric Field

[math]\displaystyle{ E= \lt 2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}\gt }[/math] N/C

[math]\displaystyle{ E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6} }[/math]

Step 4: Find q by rearranging the equation for [math]\displaystyle{ E_{mag} }[/math]

[math]\displaystyle{ E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} }[/math]

By rearranging this equation we get

[math]\displaystyle{ q= {4 pi * epsilon_0 } *{r^2}*E_{mag} }[/math]

[math]\displaystyle{ q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} }[/math]

[math]\displaystyle{ q= + 0.253 C }[/math]

Connectedness

1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the Superposition Principle are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

History

Charles de Coulomb

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

@@ Line 4: / Line 4: @@
 ==The Main Idea==
-Looking at one of the four fundamental interactions of physics, electromagnetism, you can derive the expression for Coulombs Force Law. Using this expression, there is just a simple equation that is used to derive the the exact value of the electric field. The electric field of a point charge can be directly coordinated with a couple of independent variables. The variables that are directly coordinated with the value of an electric field is the direction of the particle and the charge of the particle. Electric field of a point charge is also inversely proportional to distance.
+Every charge creates its own electric field. The two known elementary charges are the proton and electron, both of which are modeled as point charges. The electric field for point charges points radially outward in the case of the proton, and radially inward in the case of the electron. The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.
@@ Line 16: / Line 16: @@
 <math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>
-<math>Where \frac{1}{4 \pi \epsilon_0 } \text{is a constant representing the permittivity of free space, and is approximately }  9*10^{9} \frac{N m^2}{C^2} </math>
+Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.
-, r is the magnitude of the distance between the observation location and the source location
-, q is the charge of the particle
-and <math>\hat r \text { is the direction of the distance from the source location to the observation point.} </math>
-It is very easy to derive Coulombs Law using the equation for the electric field of a point charge, as all you have to do is multiply the value obtained from the Electric Field of a point charge by the charge of the second particle.
+<math>{\epsilon_0} </math>  is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>
-Below is a visual representation of electric fields due to a positive and negative point charge
+<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>
+'''''r''''' is the magnitude of the distance between the observation location and the source location
+, '''''q''''' is the charge of the particle
+and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.
+The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:
+<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>
+By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:
+<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>
+Below is a visual representation of electric field lines due to a positive and negative point charge
 [[File:Point_charge.GIF]]
@@ Line 33: / Line 45: @@
 ===A Computational Model===
-Here is a link to some code which can help visualize and compute the Electric Field due to an electron at different points. You can adjust the values of the observation locations in order to observe how the effects of distance, both magnitude and direction, can affect the electric field.
+Below is a link to some code which can help visualize the Electric Field due to a proton.
-https://trinket.io/glowscript/8d66c87ced
-<iframe src="https://trinket.io/embed/glowscript/8d66c87ced" width="100%" height="356" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen></iframe>
+https://trinket.io/glowscript/725d552305
+Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.
@@ Line 46: / Line 57: @@
 ===Simple===
-There is an electron at <0,0,0>m. Calculate the electric field at <4,-3,1>m.
+There is an electron at the origin. Calculate the electric field at (4,-3,1)m.
 '''Step 1:''' Find <math>\hat r</math>
-Find <math>\vec r_{obs} - \vec r_{electron}   (<4,-3,1> - <0,0,0> = <4,-3,1> </math>m
+Find <math>\vec r_{obs} - \vec r_{electron}   ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m
-Calculate the magnitude of r.  (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>
+Calculate the magnitude of <math>\vec r</math>.  (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>
-From r, find the unit vector <math>\hat{r}.</math> <math>  <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>
+From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math>  <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>
 '''Step 2:''' Find the magnitude of the Electric Field
@@ Line 62: / Line 73: @@
 '''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field
-E= <math>  \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}>=<-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}>   N/C    </math>
+<math>E =  \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}>   N/C    </math>
 ===Middling===
-A particle of unknown charge is located at <-.21, .02, .11> m. Its electric field at point <-.02, .31, .28> m is <.124, .188, .109> N/C . Find the value of the particles charge.
+A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.
-While this problem looks challenging, lets think about it for a second. We are given both an observation location and a source location, so we can find both r and <math>\hat{r}</math>  Since we are given the value of the electric field, we can simply find the magnitude of the electric field. Then, we can use the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>  to find the charge.
+Given both an observation location and a source location, one can find both r and <math>\hat{r}</math>  Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>  one can find the magnitude and sign of the charge.
 '''Step 1:'''
-Find <math>\vec r_{obs} - \vec r_{particle}   <-.02,.31,.28> m - <-.21,.02,.11> m = <.19,.29,.17> m </math>
+Find <math>\vec r_{obs} - \vec r_{particle} </math>.
+<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>
+To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>
+<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>
-To find r_{mag}, we simply find the magnitude of <.19,.29,.17>.
-(<math>\sqrt{.19^2+.29^2+.17^2}=\sqrt{.1491}</math>= .39
+'''Step 2:''' Find the magnitude of the Electric Field
-'''Step 2:''' Find the magnitude of the Electric Field
+<math>E= <0.124, 0.188, 0.109> N/C</math>
-E= <.124, .188, .109> N/C
+<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>
-E_{mag} = (<math>\sqrt{.124^2+.188^2+.109^2}=\sqrt{.0626}</math>=.25
-'''Step 3:''' Find q by rearranging the equation for E_{mag}
+'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>
 <math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>
@@ Line 90: / Line 106: @@
 By rearranging this equation we get
-<math> q= {4 pi epsilon_0 } *{r^2}*E_{mag} </math>
+<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>
-<math> q= {1/(9*10^9)} *{.39^2}*.25 </math>
+<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>
-q= 4.3*10^-12 C
+<math> q= + 4.3*10^{-12} C </math>
 ===Difficult===
-The electric field at a location <.5.4, -1.8, 7.2>m has been found to be <5.5*10^3, -7.6*10^3, 0> N/C. There is a particle with a charge of -2nC in the surrounding area. What is the location of the particle?
+The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?
+Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.
-Again, while at first sight the problem may look tough, it is really not that difficult once you break it down. Since we are given the vector value for electric field, we can also find the magnitude of the electric field. Once we find the magnitude of the electric field we can use the equation
 <math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>
@@ Line 110: / Line 128: @@
 Find the magnitude of the Electric field.
-<math> E</math>= (5.5*10^3, -7.6*10^3, 0) N/C
+<math> F = Eq </math>
-<math> E_{mag} </math> = <math>\sqrt{(5.5*10^3)^2+(-7.6*10^3)^2+0^2}=\sqrt{.8.8*10^7}= 9381</math>
+<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>
-'''Step 2:''':
+<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C
-Rearrange the equation for the magnitude of electric field and solve for rmag.
-<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>
-<math> r_{mag}^2= \frac{1}{4 \pi \epsilon_0 } \frac{q}{ E_{mag}} </math>
-<math> r_{mag}^2= \frac{1}{4 \pi \epsilon_0 } \frac{2nC}{ 9381} </math>
+'''Step 2:''':
+Find <math>\vec r_{obs} - \vec r_{particle} </math>.
-<math> r_{mag}^2= .00192 </math>
+<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>
-<math> r_{mag}= .04m </math>
-'''Step 3:'''
+To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>
-Find the direction of the electric field
-Because the source charge is negative, <math>\hat r</math> <math>=-\hat E</math>
+<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>
-<math>\hat E</math> <math>= E/ E_{mag} </math>
-<math>\hat E</math> <math>= (<5.5*10^3, -7.6*10^3, 0> / 9381) N/C </math>
+'''Step 3:''' Find the magnitude of the Electric Field
-<math>\hat r</math> <math>=-\hat E</math> = <-.56, .81,0>
+<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C
-'''Step 4:'''
+<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>
-Find <math> r</math>
-<math> r</math> = <math>r_{mag}</math>  *<math>\hat r</math>
-<math> r</math>  = <math> .04m * <-.56, .81,0></math>
+'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>
-<math> r</math> = <math><-.02, .03, 0> m</math>
+<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>
-'''Step 5:'''
+By rearranging this equation we get
-Find <math> r_{particle}</math>
-<math> r</math>  <math>= r_{observation}-</math>  <math> r_{particle}</math>
+<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>
-<math> r_{particle}</math> <math>= r_{observation}-</math>  <math> r</math>
+<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>
-<math> r_{particle}</math> <math>= <.5.4, -1.8, 7.2>-</math>  <math><-.02, .03, 0></math>
+<math> q= + 0.253 C </math>
-<math> r_{particle}</math> <math>=<5.42, -1.83, 0></math>
@@ Line 164: / Line 172: @@
 .How is this topic connected to something that you are interested in?
-I am really interested in the electric field of a point particle because it seems as if this one fundamental idea has so connections and relations throughout physics 2. Understanding this basic principle has helped me learn about much harder topics in magnetism.
+This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.
 .How is it connected to your major?
-I am a BME major and this is related to me because there are several instances where I will have to work with electric fields and circuits and this will be the foundation of what I do/learn.
+As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.
 .Is there an interesting industrial application?
-An interesting application is that electric fields of point charges can be used to find forces, both electric and magnetic forces. Then you can predict the motion of various particles looking at the forces that are acting upon them. .
+The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.
 ==History==
@@ Line 186: / Line 195: @@
 [[Electric Field]] <br>
-[[Electric Force]]
+[[Electric Force]] <br>
+[[Superposition Principle]]
 ===Further reading===

Point Charge: Difference between revisions

Revision as of 01:58, 9 April 2017

Contents

The Main Idea

A Mathematical Model of Electric Field due to Point Charge

A Computational Model

Examples

Simple

Middling

Difficult

Connectedness

History

See also

Further reading

External links

References

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