Physics Book - User contributions [en]

Superposition Principle

2021-11-29T18:37:15Z

Jnation8: /* A Mathematical Model */

'''Claimed By: Jared Nation'''

== The Big Picture ==
The superposition principle is based on the idea that in a closed system, an object receives a net force equal to the sum of all outside forces acting on it. This is commonly used to calculate the net electric field or magnetic field on an object. More specifically, the Superposition Principle states that the net result of multiple vectors acting on a given place and time is equal to the vector sum of each individual vector. For example, think about two waves colliding. When the waves collide, the height of the combined wave is the sum of the height of the two waves. This principle can be seen throughout nature and theory. A simple example can be seen below where Vector 3 is the sum of Vectors 1 and 2.

[[File:69832.jpg|400px|Imaage: 400 pixels]]
===The Wave Model===

Superposition is easy to think about in the form of waves. Let's say, you are at the beach watching the ocean. A huge 5 foot wave comes towards the land and is about to crash. However, just as this happens, a smaller 1 foot tall wave forms due to the undercurrent and goes out to the ocean. Before the waves crash, they meet. At this meeting point, the combined waves become 6 feet tall (1 + 5). Why is this? This is because of superposition.

[[File:Wav sup.jpeg|500px|center]]

Let's now think about an orchestra. Before the concert starts, individual instruments practice on stage. Sitting in the audience, you can hear the individual instruments. However, the sound of that one instrument does not nearly compare the the sound of the entire orchestra. When the entire orchestra begins to play, the volume is much louder. This is because of the superposition principle. Each instrument produces sound waves. When an entire orchestra performs at once, the sound waves from each instrument are all summed together. The superposition principle is not only seen in waves. The superposition principle can be seen in forces, electric fields, magnetic fields, etc. However, it is easier to understand when looking at waves that everyone notices.

[[File:Standing_wave_2.gif]]

This gif depicts the superposition principal. The peaks and the throughs (the bottom of the waves) are summed together as the different waves overlap. Just as the waves of the ocean and the sound waves of the orchestra summed together, the waves in this gif sum together.

===A Mathematical Model===

The Superposition Principle is derived from the properties of additivity and homogeneity for linear systems which are defined in terms of a scalar value of a by the following equations:

<math>F(x_1 + x_2) = F(x_1) + F(x_2)\quad Additivity</math>

<math>aF(x) = F(ax)\quad Homogeneity</math>
[[File:Superposition.gif|thumb|Example of the Superposition of Electric Fields using charges with changing locations]]

The principle can be applied to any linear system and can be used to find the net result of functions, vectors, vector fields, etc. For the topic of introductory physics, it will mainly apply to vectors and vector fields such as electric forces and fields.

If given a number of vectors passing through a certain point, the resultant vector is given by simply adding all the the vectors at that point. For example, for a number of uniform electric fields passing though a single point, the resulting electric field at that point is given by:

<math>\vec{E} = \vec{E}_{1} + \vec{E}_{2} +...+ \vec{E}_{n} = \sum_{i=1}^n\vec{E}_{i}</math>

This same concept can be applied to electric and magnetic fields and forces. This is helpful when dealing with scenarios in which the effect that multiple point charges have on each other is in an area void of other fields. By adding together each of the contributing field vectors at a specific point you can figure out the net contribution that all of the present charges have on the overall field experienced at that location.

You can calculate the electric field of a single point charge by using the following equation:
<math>\vec{E} = \frac{1}{4 \pi \epsilon_0}\sum_{i=1}^n\frac{q_i}{r_i^2}\hat{r_i}</math>

Where:

<math>\frac{1}{4 \pi \epsilon_0}</math> is a constant that equals <math>9e9</math>.

<math>q_i</math> is the charge of the particle you are studying. The sign of the particle matters (positive for proton, negative for electron.)

<math>r_i</math> is the magnitude of the distance between the particle and the observation location.

<math>\hat{r_i}</math> is the vector pointing from the particle to the observation location.

Conceptually is is also important to note that an object cannot exert a force on itself, so the sum of all of the forces does not include the object itself.

==Examples==

When attempting to solve these problems be sure to show all steps in your solution and include diagrams whenever possible.

===Simple - Two Point Charges and One Dimension===
[[File:SuperpositionExample.png]]

If <math>Q_1</math> equals 1e-4 C and <math>Q_2</math> equals 1e-5C, what is the net electric field at the midpoint of both charges?

<div class="toccolours mw-collapsible mw-collapsed">

'''''Click for Solution'''''
<div class="mw-collapsible-content">
We know that the distance of each charge to the midpoint is <math>1m</math>. Since this value is the magnitude of the distance it will serve as our <math>r_i</math> value for both of the electric field formulas.

Our <math>\hat{r_1}</math> value for <math>E_1</math> will be <math>1</math> and the <math>\hat{r_2}</math> value for <math>E_2</math> will be <math>-1</math>, if you use the midpoint as your observation point in the x-direction.

Now that we have all the necessary values to use the electric field formula, we can plug them in.

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{q_1}{r_1^2}\hat{r_1} = (9e9)\frac{1e-4C}{1m^2}(1m) = 9e5\frac{N}{C}</math>
<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{q_2}{r_2^2}\hat{r_2} = (9e9)\frac{1e-5C}{1m^2}(-1m) = -9e4\frac{N}{C}</math>

By this point, you have the individual contributions of each charge. To get the net contribution of all the charges, you only need to add them.

<math>\vec{E_1}+\vec{E_2}=9e5\frac{N}{C}-9e4\frac{N}{C}= 8.1e5\frac{N}{C}</math>

This means that the net electric field will point towards the right with a magnitude of <math>8.1e5\frac{N}{C}</math>.
</div>
</div>
=== Medium - Three Point charges and two Dimensions ===
[[File:Superposfile2.JPG]]

----
Using k for the constant term
<math> E = Q1 * rhat/ (r)^2 </math>

<math> F = q3 * E </math>

<math> r = (4d, -3d,0> - (0,3d,)) = (4d,-6d,0) </math>

<math> k * Q3 * Q1 * rhat/ (r)^2 </math>

<math> k* Q^2 * 1/(52d^2) * (2/13 * sqrt(13),-3/13 sqrt(13),0) </math>

=== Medium - Three Point charges and two Dimensions===
[[File:Superpos file.JPG]]

----

What is the magnitude of the net force on the charge -q'?

To begin, we recognize that this problem will be using the superposition principle. It is additionally
important to recognize that objects can't exert forces on themselves. Keeping these in mind, The force exerted on -q' will be a combination of the forces exerted by 3q and negative q.

For this problem, let's consider -q' to be q3, 3q to be q1, and -1 to be q2 and, For the sake of simplicity, we are going to call our constant 'k'

Let's find the forces separately then add them together:

<math> k * (q1*q3)/(r)^2 * rhat </math>

Now that we have the general form of the equation, we plug in variables.

<math> F1 = k * 3q(-q')/(sqrt(3) * l)^2 * (sin(theta),cos(theta),0) </math>
<math>

= k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) </math>

Great! Now we have the first force from q1, now we need to find the force from q2

<math> F2 = k * (q2*q3)/ (r)^2 * rhat </math>

<math> = k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>

Now that we've found both forces, we add them together.

<math> Fnet = F1+F2 </math>
<math> = k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) + k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>
<math> = k * (q*q')/(l^2) * (-2sin(theta),0,0) </math>

Once we were able to recognize our problem, it became as simple as plugging in variables and using algebra to solve the problem.

Additionally, if we were asked "What is the direction of the force on the charge -q'?"

Looking back at our work, we can see that our vertical components cancel out, leaving
<math> -2sin(theta) </math>
Indicating our direction is negative x.

We could get to this conclusion without looking at the math by understanding that electric field falls off in a way that resembles
<math> 1/(r)^2 </math>

For q1, it is sqrt(3) times farther than q2, and 3 times the charge. Which is equivalent to
<math> 3*q/(sqrt(3))^2 </math>
Which is equivalent to q2 in the vertical direction. In the horizontal direction, q2 is slightly closer, meaning that the total net charge should be in the negative x direction.

===Medium - Two Point Charges and Two Dimensions===
[[File:BrooksEx1.png]]

If <math>Q_1</math> and <math>Q_2</math> are positive point charges with a charge of e, what is the net electric field at point P?

<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
To begin this problem, the first step is to find <math>r_1</math> and <math>r_2</math>, the vectors from the charges to point P as well as their magnitudes:

<math>\vec{r_1} = 3\hat{i}+2\hat{j}-(0\hat{i}+0\hat{j})\Rightarrow\vec{r_1} = 3\hat{i}+2\hat{j}</math>

<math>||\vec{r_1}|| = \sqrt{3^2 + 2^2} =\sqrt{13}</math>

<math>\vec{r_2} = 3\hat{i}+2\hat{j}-(4\hat{i}+0\hat{j})\Rightarrow\vec{r_2} = -1\hat{i}+2\hat{j}</math>

<math>||\vec{r_2}|| = \sqrt{-1^2 + 2^2} =\sqrt{5}</math>

Using these in the equation for an electric field from a point charge, you get:

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{Q_1}{||r_1||^2}\hat{r_1}=\frac{1}{4 \pi \epsilon_0}\frac{e}{13}<\frac{3}{\sqrt{13}}\hat{i}+\frac{2}{\sqrt{13}}\hat{j}> = <9.21E-11\hat{i}+6.14E-11\hat{j}> </math>N/C

<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{Q_2}{||r_2||^2}\hat{r_2}=\frac{1}{4 \pi \epsilon_0}\frac{e}{5}<\frac{-1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j}> = <-1.29E-10\hat{i}+2.58E-10\hat{j}> </math>N/C

Then, simply add the two electric fields together:

<math>\vec{E}=\vec{E_1}+\vec{E_2} = <-3.69E-11\hat{i}+3.19E-10\hat{j}></math>
</div>
</div>

===Difficult - Five Point Charges and Three Dimensions===

[[File:brooksEx2.png]]

If all point charges have a charge of e, what the the net electric field present at point L?
<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
This problem is similar to the previous examples but this one includes the z axis and more points. Since there are 5 points and you have already had some practice, you will only see the procedure for the first one. We will still work through the whole problem. Again, first each vector and magnitude:

<math>\vec{d_5} = 0\hat{i}+0\hat{j}+0\hat{k}-(2\hat{i}-1\hat{j}-1\hat{k})-\Rightarrow\vec{d_5} = -2\hat{i}+1\hat{j}+1\hat{k}</math>

<math>||\vec{d_5}|| = \sqrt{(-2)^2 + 1^2 + 1^2} = 2</math>

Do the same for each of the other point charges and plug them into the electric field formula:

<math>\vec{E_5} = \frac{1}{4 \pi \epsilon_0}\frac{Q_5}{||d_5||^2}\hat{d_5}=\frac{1}{4 \pi \epsilon_0}\frac{e}{4}<\frac{-2}{2}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}> = <-3.6E-10\hat{i}+1.8E-10\hat{j}+1.8E-10\hat{k}></math>N/C

Follow the same steps for the other electric fields and add them all together to get your final answer:

<math>\vec{E_1} = <1.44E-9\hat{i}+0\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_2} = <0\hat{i}-1.44E-9\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_3} = <-1.29E-10\hat{i}+0\hat{j}+2.58E-10\hat{k}></math>N/C

<math>\vec{E_4} = <0\hat{i}+0\hat{j}-1.44E-9\hat{k}></math>N/C

<math>\vec{E} = \vec{E_1}+\vec{E_2}+\vec{E_3}+\vec{E_4}+\vec{E_5}=<9.51E-10\hat{i}-1.26E-9\hat{j}+1.00E-9\hat{k}></math>N/C
</div>
</div>

==Connectedness==
[[File:Superposition Example.PNG|thumb|left|A diagram of a circuit you will be able to analyze further on by using the superposition of currents.]]
The superposition principle is critical as you progress to more complicated topics in Physics. This principle simplifies many topics that you will encounter further ahead, by assuming that their net contribution will be equal to the addition of all the individual ones. You will use superposition with electric fields and forces, as well as magnetic fields and forces. In addition, if in a problem you are only given the final net contribution of any type of field, you can inversely solve it by constructing equations that relate the individual formulas. You can then plug in values, and subtract in order to find the magnitudes of each individual contributor. This will also enable you to better analyze different types of circuits.

However, the superposition principle also aids outside of the world of abstract electromagnetism. It is also useful for static problems, which you might encounter if you are a Civil or Mechanical Engineer, or systems and circuitry, which you might encounter if you are a Biomedical engineer.

==History==
[[File:Danielbernoulli.jpg|thumb|right|Daniel Bernoulli]]
The superposition principle was supposedly first stated by [[Daniel Bernoulli]]. Bernoulli was a famous scientist whose primary work was in fluid mechanics and statistics. If you recognize the name, he is primarily remembered for discovering the Bernoulli Principle. In 1753, he stated that, "The general motion of a vibrating system is given by a superposition of its proper vibrations." This idea was at first rejected by some other popular scientists until it became widely accepted due to the work of Joseph Fourier. Fourier was a famous scientist known for his work on the Fourier Series, which is used in heat transfer and vibrations, as well as his discovery of the Greenhouse Effect. Fournier’s support legitimized Bernoulli’s claims and has made a huge impact on history.

== See also ==

===Next Topics===
Now that you have a solid understanding of the Superposition Principle, you can move on to these topics:
*[[Electric Field]]
**[[Point Charge]]
**[[Electric Dipole]]
*[[Electric Force]]
*[[Magnetic Field]]
*[[Magnetic Force]]
*[[Superposition Principle]]
===Further reading===

Books, Articles or other print media on this topic

[http://www.acoustics.salford.ac.uk/feschools/waves/super.php Superposition of Waves]

[http://whatis.techtarget.com/definition/Schrodingers-cat Schrodingers' Cat from a Superposition Point of View]

===External Links===

[https://www.youtube.com/watch?v=p3Xugztl9Ho Visual Explanation of Superposition in Electric Fields]

[https://www.youtube.com/watch?v=mdulzEfQXDE Electric Fields and Superposition Principle Crash Course Video]

[https://www.youtube.com/watch?v=S1TXN1M9t18 Instructional video on how to calculate the net electric field using the superposition principle]

[https://www.youtube.com/watch?v=lCM5dql_ul0 Additional Video on calculating superposition principle]

[https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1479978&tag=1 Using superposition principle to analyze solar cells]

==References==
[http://physics-help.info/physicsguide/electricity/electric_field.shtml Electric Fields]

[https://en.wikipedia.org/wiki/Superposition_principle Superposition Principle]

[https://en.wikipedia.org/wiki/Daniel_Bernoulli Daniel Bernoulli Biography]

[https://en.wikipedia.org/wiki/Joseph_Fourier Joseph Fourier Biography]
[[Category:Fields]]

Superposition Principle

2021-11-29T18:10:18Z

Jnation8: /* The Big Picture */

'''Claimed By: Jared Nation'''

== The Big Picture ==
The superposition principle is based on the idea that in a closed system, an object receives a net force equal to the sum of all outside forces acting on it. This is commonly used to calculate the net electric field or magnetic field on an object. More specifically, the Superposition Principle states that the net result of multiple vectors acting on a given place and time is equal to the vector sum of each individual vector. For example, think about two waves colliding. When the waves collide, the height of the combined wave is the sum of the height of the two waves. This principle can be seen throughout nature and theory. A simple example can be seen below where Vector 3 is the sum of Vectors 1 and 2.

[[File:69832.jpg|400px|Imaage: 400 pixels]]
===The Wave Model===

Superposition is easy to think about in the form of waves. Let's say, you are at the beach watching the ocean. A huge 5 foot wave comes towards the land and is about to crash. However, just as this happens, a smaller 1 foot tall wave forms due to the undercurrent and goes out to the ocean. Before the waves crash, they meet. At this meeting point, the combined waves become 6 feet tall (1 + 5). Why is this? This is because of superposition.

[[File:Wav sup.jpeg|500px|center]]

Let's now think about an orchestra. Before the concert starts, individual instruments practice on stage. Sitting in the audience, you can hear the individual instruments. However, the sound of that one instrument does not nearly compare the the sound of the entire orchestra. When the entire orchestra begins to play, the volume is much louder. This is because of the superposition principle. Each instrument produces sound waves. When an entire orchestra performs at once, the sound waves from each instrument are all summed together. The superposition principle is not only seen in waves. The superposition principle can be seen in forces, electric fields, magnetic fields, etc. However, it is easier to understand when looking at waves that everyone notices.

[[File:Standing_wave_2.gif]]

This gif depicts the superposition principal. The peaks and the throughs (the bottom of the waves) are summed together as the different waves overlap. Just as the waves of the ocean and the sound waves of the orchestra summed together, the waves in this gif sum together.

===A Mathematical Model===

The Superposition Principle is derived from the properties of additivity and homogeneity for linear systems which are defined in terms of a scalar value of a by the following equations:

<math>F(x_1 + x_2) = F(x_1) + F(x_2)\quad Additivity</math>

<math>aF(x) = F(ax)\quad Homogeneity</math>
[[File:Superposition.gif|thumb|Example of the Superposition of Electric Fields using charges with changing locations]]

The principle can be applied to any linear system and can be used to find the net result of functions, vectors, vector fields, etc. For the topic of introductory physics, it will mainly apply to vectors and vector fields such as electric forces and fields.

If given a number of vectors passing through a certain point, the resultant vector is given by simply adding all the the vectors at that point. For example, for a number of uniform electric fields passing though a single point, the resulting electric field at that point is given by:

<math>\vec{E} = \vec{E}_{1} + \vec{E}_{2} +...+ \vec{E}_{n} = \sum_{i=1}^n\vec{E}_{i}</math>

This same concept can be applied to electric forces as well as to magnetic fields and forces. This is more useful when dealing with the effect that multiple point charges have on each other is an area of void of other electric fields. By adding together each of the contributing electric field vectors at a specific point you can figure out the net contribution that all of the present charges have on the overall electric field experienced at said location.

You can calculate the electric field of a single point charge by using the following equation:
<math>\vec{E} = \frac{1}{4 \pi \epsilon_0}\sum_{i=1}^n\frac{q_i}{r_i^2}\hat{r_i}</math>

Where:

<math>\frac{1}{4 \pi \epsilon_0}</math> is a constant that equals <math>9e9</math>.

<math>q_i</math> is the charge of the particle you are studying.

<math>r_i</math> is the magnitude of the distance between the particle and the observation location.

<math>\hat{r_i}</math> is the vector pointing from the particle to the observation location.

Conceptually is is also important to note that an object cannot exert a force on itself, so the sum of all of the forces does not include the object itself.

==Examples==

When attempting to solve these problems be sure to show all steps in your solution and include diagrams whenever possible.

===Simple - Two Point Charges and One Dimension===
[[File:SuperpositionExample.png]]

If <math>Q_1</math> equals 1e-4 C and <math>Q_2</math> equals 1e-5C, what is the net electric field at the midpoint of both charges?

<div class="toccolours mw-collapsible mw-collapsed">

'''''Click for Solution'''''
<div class="mw-collapsible-content">
We know that the distance of each charge to the midpoint is <math>1m</math>. Since this value is the magnitude of the distance it will serve as our <math>r_i</math> value for both of the electric field formulas.

Our <math>\hat{r_1}</math> value for <math>E_1</math> will be <math>1</math> and the <math>\hat{r_2}</math> value for <math>E_2</math> will be <math>-1</math>, if you use the midpoint as your observation point in the x-direction.

Now that we have all the necessary values to use the electric field formula, we can plug them in.

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{q_1}{r_1^2}\hat{r_1} = (9e9)\frac{1e-4C}{1m^2}(1m) = 9e5\frac{N}{C}</math>
<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{q_2}{r_2^2}\hat{r_2} = (9e9)\frac{1e-5C}{1m^2}(-1m) = -9e4\frac{N}{C}</math>

By this point, you have the individual contributions of each charge. To get the net contribution of all the charges, you only need to add them.

<math>\vec{E_1}+\vec{E_2}=9e5\frac{N}{C}-9e4\frac{N}{C}= 8.1e5\frac{N}{C}</math>

This means that the net electric field will point towards the right with a magnitude of <math>8.1e5\frac{N}{C}</math>.
</div>
</div>
=== Medium - Three Point charges and two Dimensions ===
[[File:Superposfile2.JPG]]

----
Using k for the constant term
<math> E = Q1 * rhat/ (r)^2 </math>

<math> F = q3 * E </math>

<math> r = (4d, -3d,0> - (0,3d,)) = (4d,-6d,0) </math>

<math> k * Q3 * Q1 * rhat/ (r)^2 </math>

<math> k* Q^2 * 1/(52d^2) * (2/13 * sqrt(13),-3/13 sqrt(13),0) </math>

=== Medium - Three Point charges and two Dimensions===
[[File:Superpos file.JPG]]

----

What is the magnitude of the net force on the charge -q'?

To begin, we recognize that this problem will be using the superposition principle. It is additionally
important to recognize that objects can't exert forces on themselves. Keeping these in mind, The force exerted on -q' will be a combination of the forces exerted by 3q and negative q.

For this problem, let's consider -q' to be q3, 3q to be q1, and -1 to be q2 and, For the sake of simplicity, we are going to call our constant 'k'

Let's find the forces separately then add them together:

<math> k * (q1*q3)/(r)^2 * rhat </math>

Now that we have the general form of the equation, we plug in variables.

<math> F1 = k * 3q(-q')/(sqrt(3) * l)^2 * (sin(theta),cos(theta),0) </math>
<math>

= k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) </math>

Great! Now we have the first force from q1, now we need to find the force from q2

<math> F2 = k * (q2*q3)/ (r)^2 * rhat </math>

<math> = k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>

Now that we've found both forces, we add them together.

<math> Fnet = F1+F2 </math>
<math> = k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) + k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>
<math> = k * (q*q')/(l^2) * (-2sin(theta),0,0) </math>

Once we were able to recognize our problem, it became as simple as plugging in variables and using algebra to solve the problem.

Additionally, if we were asked "What is the direction of the force on the charge -q'?"

Looking back at our work, we can see that our vertical components cancel out, leaving
<math> -2sin(theta) </math>
Indicating our direction is negative x.

We could get to this conclusion without looking at the math by understanding that electric field falls off in a way that resembles
<math> 1/(r)^2 </math>

For q1, it is sqrt(3) times farther than q2, and 3 times the charge. Which is equivalent to
<math> 3*q/(sqrt(3))^2 </math>
Which is equivalent to q2 in the vertical direction. In the horizontal direction, q2 is slightly closer, meaning that the total net charge should be in the negative x direction.

===Medium - Two Point Charges and Two Dimensions===
[[File:BrooksEx1.png]]

If <math>Q_1</math> and <math>Q_2</math> are positive point charges with a charge of e, what is the net electric field at point P?

<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
To begin this problem, the first step is to find <math>r_1</math> and <math>r_2</math>, the vectors from the charges to point P as well as their magnitudes:

<math>\vec{r_1} = 3\hat{i}+2\hat{j}-(0\hat{i}+0\hat{j})\Rightarrow\vec{r_1} = 3\hat{i}+2\hat{j}</math>

<math>||\vec{r_1}|| = \sqrt{3^2 + 2^2} =\sqrt{13}</math>

<math>\vec{r_2} = 3\hat{i}+2\hat{j}-(4\hat{i}+0\hat{j})\Rightarrow\vec{r_2} = -1\hat{i}+2\hat{j}</math>

<math>||\vec{r_2}|| = \sqrt{-1^2 + 2^2} =\sqrt{5}</math>

Using these in the equation for an electric field from a point charge, you get:

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{Q_1}{||r_1||^2}\hat{r_1}=\frac{1}{4 \pi \epsilon_0}\frac{e}{13}<\frac{3}{\sqrt{13}}\hat{i}+\frac{2}{\sqrt{13}}\hat{j}> = <9.21E-11\hat{i}+6.14E-11\hat{j}> </math>N/C

<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{Q_2}{||r_2||^2}\hat{r_2}=\frac{1}{4 \pi \epsilon_0}\frac{e}{5}<\frac{-1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j}> = <-1.29E-10\hat{i}+2.58E-10\hat{j}> </math>N/C

Then, simply add the two electric fields together:

<math>\vec{E}=\vec{E_1}+\vec{E_2} = <-3.69E-11\hat{i}+3.19E-10\hat{j}></math>
</div>
</div>

===Difficult - Five Point Charges and Three Dimensions===

[[File:brooksEx2.png]]

If all point charges have a charge of e, what the the net electric field present at point L?
<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
This problem is similar to the previous examples but this one includes the z axis and more points. Since there are 5 points and you have already had some practice, you will only see the procedure for the first one. We will still work through the whole problem. Again, first each vector and magnitude:

<math>\vec{d_5} = 0\hat{i}+0\hat{j}+0\hat{k}-(2\hat{i}-1\hat{j}-1\hat{k})-\Rightarrow\vec{d_5} = -2\hat{i}+1\hat{j}+1\hat{k}</math>

<math>||\vec{d_5}|| = \sqrt{(-2)^2 + 1^2 + 1^2} = 2</math>

Do the same for each of the other point charges and plug them into the electric field formula:

<math>\vec{E_5} = \frac{1}{4 \pi \epsilon_0}\frac{Q_5}{||d_5||^2}\hat{d_5}=\frac{1}{4 \pi \epsilon_0}\frac{e}{4}<\frac{-2}{2}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}> = <-3.6E-10\hat{i}+1.8E-10\hat{j}+1.8E-10\hat{k}></math>N/C

Follow the same steps for the other electric fields and add them all together to get your final answer:

<math>\vec{E_1} = <1.44E-9\hat{i}+0\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_2} = <0\hat{i}-1.44E-9\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_3} = <-1.29E-10\hat{i}+0\hat{j}+2.58E-10\hat{k}></math>N/C

<math>\vec{E_4} = <0\hat{i}+0\hat{j}-1.44E-9\hat{k}></math>N/C

<math>\vec{E} = \vec{E_1}+\vec{E_2}+\vec{E_3}+\vec{E_4}+\vec{E_5}=<9.51E-10\hat{i}-1.26E-9\hat{j}+1.00E-9\hat{k}></math>N/C
</div>
</div>

==Connectedness==
[[File:Superposition Example.PNG|thumb|left|A diagram of a circuit you will be able to analyze further on by using the superposition of currents.]]
The superposition principle is critical as you progress to more complicated topics in Physics. This principle simplifies many topics that you will encounter further ahead, by assuming that their net contribution will be equal to the addition of all the individual ones. You will use superposition with electric fields and forces, as well as magnetic fields and forces. In addition, if in a problem you are only given the final net contribution of any type of field, you can inversely solve it by constructing equations that relate the individual formulas. You can then plug in values, and subtract in order to find the magnitudes of each individual contributor. This will also enable you to better analyze different types of circuits.

However, the superposition principle also aids outside of the world of abstract electromagnetism. It is also useful for static problems, which you might encounter if you are a Civil or Mechanical Engineer, or systems and circuitry, which you might encounter if you are a Biomedical engineer.

==History==
[[File:Danielbernoulli.jpg|thumb|right|Daniel Bernoulli]]
The superposition principle was supposedly first stated by [[Daniel Bernoulli]]. Bernoulli was a famous scientist whose primary work was in fluid mechanics and statistics. If you recognize the name, he is primarily remembered for discovering the Bernoulli Principle. In 1753, he stated that, "The general motion of a vibrating system is given by a superposition of its proper vibrations." This idea was at first rejected by some other popular scientists until it became widely accepted due to the work of Joseph Fourier. Fourier was a famous scientist known for his work on the Fourier Series, which is used in heat transfer and vibrations, as well as his discovery of the Greenhouse Effect. Fournier’s support legitimized Bernoulli’s claims and has made a huge impact on history.

== See also ==

===Next Topics===
Now that you have a solid understanding of the Superposition Principle, you can move on to these topics:
*[[Electric Field]]
**[[Point Charge]]
**[[Electric Dipole]]
*[[Electric Force]]
*[[Magnetic Field]]
*[[Magnetic Force]]
*[[Superposition Principle]]
===Further reading===

Books, Articles or other print media on this topic

[http://www.acoustics.salford.ac.uk/feschools/waves/super.php Superposition of Waves]

[http://whatis.techtarget.com/definition/Schrodingers-cat Schrodingers' Cat from a Superposition Point of View]

===External Links===

[https://www.youtube.com/watch?v=p3Xugztl9Ho Visual Explanation of Superposition in Electric Fields]

[https://www.youtube.com/watch?v=mdulzEfQXDE Electric Fields and Superposition Principle Crash Course Video]

[https://www.youtube.com/watch?v=S1TXN1M9t18 Instructional video on how to calculate the net electric field using the superposition principle]

[https://www.youtube.com/watch?v=lCM5dql_ul0 Additional Video on calculating superposition principle]

[https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1479978&tag=1 Using superposition principle to analyze solar cells]

==References==
[http://physics-help.info/physicsguide/electricity/electric_field.shtml Electric Fields]

[https://en.wikipedia.org/wiki/Superposition_principle Superposition Principle]

[https://en.wikipedia.org/wiki/Daniel_Bernoulli Daniel Bernoulli Biography]

[https://en.wikipedia.org/wiki/Joseph_Fourier Joseph Fourier Biography]
[[Category:Fields]]

Superposition Principle

2021-11-29T18:06:19Z

Jnation8: /* The Big Picture */

'''Claimed By: Jared Nation'''

== The Big Picture ==
The superposition principle is based on the idea that in a closed system, an object receives a net force equal to the sum of all outside forces acting on it. This is commonly used to calculate the net electric field or magnetic field on an object. More specifically, the Superposition Principle states that the net result of multiple vectors acting on a given place and time is equal to the vector sum of each individual vector. For example, think about two waves colliding. When the waves collide, the height of the combined wave is the sum of the height of the two waves. This principle can be seen throughout nature and theory.
[[File:69832.jpg]]
===The Wave Model===

Superposition is easy to think about in the form of waves. Let's say, you are at the beach watching the ocean. A huge 5 foot wave comes towards the land and is about to crash. However, just as this happens, a smaller 1 foot tall wave forms due to the undercurrent and goes out to the ocean. Before the waves crash, they meet. At this meeting point, the combined waves become 6 feet tall (1 + 5). Why is this? This is because of superposition.

[[File:Wav sup.jpeg|500px|center]]

Let's now think about an orchestra. Before the concert starts, individual instruments practice on stage. Sitting in the audience, you can hear the individual instruments. However, the sound of that one instrument does not nearly compare the the sound of the entire orchestra. When the entire orchestra begins to play, the volume is much louder. This is because of the superposition principle. Each instrument produces sound waves. When an entire orchestra performs at once, the sound waves from each instrument are all summed together. The superposition principle is not only seen in waves. The superposition principle can be seen in forces, electric fields, magnetic fields, etc. However, it is easier to understand when looking at waves that everyone notices.

[[File:Standing_wave_2.gif]]

This gif depicts the superposition principal. The peaks and the throughs (the bottom of the waves) are summed together as the different waves overlap. Just as the waves of the ocean and the sound waves of the orchestra summed together, the waves in this gif sum together.

===A Mathematical Model===

The Superposition Principle is derived from the properties of additivity and homogeneity for linear systems which are defined in terms of a scalar value of a by the following equations:

<math>F(x_1 + x_2) = F(x_1) + F(x_2)\quad Additivity</math>

<math>aF(x) = F(ax)\quad Homogeneity</math>
[[File:Superposition.gif|thumb|Example of the Superposition of Electric Fields using charges with changing locations]]

The principle can be applied to any linear system and can be used to find the net result of functions, vectors, vector fields, etc. For the topic of introductory physics, it will mainly apply to vectors and vector fields such as electric forces and fields.

If given a number of vectors passing through a certain point, the resultant vector is given by simply adding all the the vectors at that point. For example, for a number of uniform electric fields passing though a single point, the resulting electric field at that point is given by:

<math>\vec{E} = \vec{E}_{1} + \vec{E}_{2} +...+ \vec{E}_{n} = \sum_{i=1}^n\vec{E}_{i}</math>

This same concept can be applied to electric forces as well as to magnetic fields and forces. This is more useful when dealing with the effect that multiple point charges have on each other is an area of void of other electric fields. By adding together each of the contributing electric field vectors at a specific point you can figure out the net contribution that all of the present charges have on the overall electric field experienced at said location.

You can calculate the electric field of a single point charge by using the following equation:
<math>\vec{E} = \frac{1}{4 \pi \epsilon_0}\sum_{i=1}^n\frac{q_i}{r_i^2}\hat{r_i}</math>

Where:

<math>\frac{1}{4 \pi \epsilon_0}</math> is a constant that equals <math>9e9</math>.

<math>q_i</math> is the charge of the particle you are studying.

<math>r_i</math> is the magnitude of the distance between the particle and the observation location.

<math>\hat{r_i}</math> is the vector pointing from the particle to the observation location.

Conceptually is is also important to note that an object cannot exert a force on itself, so the sum of all of the forces does not include the object itself.

==Examples==

When attempting to solve these problems be sure to show all steps in your solution and include diagrams whenever possible.

===Simple - Two Point Charges and One Dimension===
[[File:SuperpositionExample.png]]

If <math>Q_1</math> equals 1e-4 C and <math>Q_2</math> equals 1e-5C, what is the net electric field at the midpoint of both charges?

<div class="toccolours mw-collapsible mw-collapsed">

'''''Click for Solution'''''
<div class="mw-collapsible-content">
We know that the distance of each charge to the midpoint is <math>1m</math>. Since this value is the magnitude of the distance it will serve as our <math>r_i</math> value for both of the electric field formulas.

Our <math>\hat{r_1}</math> value for <math>E_1</math> will be <math>1</math> and the <math>\hat{r_2}</math> value for <math>E_2</math> will be <math>-1</math>, if you use the midpoint as your observation point in the x-direction.

Now that we have all the necessary values to use the electric field formula, we can plug them in.

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{q_1}{r_1^2}\hat{r_1} = (9e9)\frac{1e-4C}{1m^2}(1m) = 9e5\frac{N}{C}</math>
<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{q_2}{r_2^2}\hat{r_2} = (9e9)\frac{1e-5C}{1m^2}(-1m) = -9e4\frac{N}{C}</math>

By this point, you have the individual contributions of each charge. To get the net contribution of all the charges, you only need to add them.

<math>\vec{E_1}+\vec{E_2}=9e5\frac{N}{C}-9e4\frac{N}{C}= 8.1e5\frac{N}{C}</math>

This means that the net electric field will point towards the right with a magnitude of <math>8.1e5\frac{N}{C}</math>.
</div>
</div>
=== Medium - Three Point charges and two Dimensions ===
[[File:Superposfile2.JPG]]

----
Using k for the constant term
<math> E = Q1 * rhat/ (r)^2 </math>

<math> F = q3 * E </math>

<math> r = (4d, -3d,0> - (0,3d,)) = (4d,-6d,0) </math>

<math> k * Q3 * Q1 * rhat/ (r)^2 </math>

<math> k* Q^2 * 1/(52d^2) * (2/13 * sqrt(13),-3/13 sqrt(13),0) </math>

=== Medium - Three Point charges and two Dimensions===
[[File:Superpos file.JPG]]

----

What is the magnitude of the net force on the charge -q'?

To begin, we recognize that this problem will be using the superposition principle. It is additionally
important to recognize that objects can't exert forces on themselves. Keeping these in mind, The force exerted on -q' will be a combination of the forces exerted by 3q and negative q.

For this problem, let's consider -q' to be q3, 3q to be q1, and -1 to be q2 and, For the sake of simplicity, we are going to call our constant 'k'

Let's find the forces separately then add them together:

<math> k * (q1*q3)/(r)^2 * rhat </math>

Now that we have the general form of the equation, we plug in variables.

<math> F1 = k * 3q(-q')/(sqrt(3) * l)^2 * (sin(theta),cos(theta),0) </math>
<math>

= k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) </math>

Great! Now we have the first force from q1, now we need to find the force from q2

<math> F2 = k * (q2*q3)/ (r)^2 * rhat </math>

<math> = k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>

Now that we've found both forces, we add them together.

<math> Fnet = F1+F2 </math>
<math> = k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) + k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>
<math> = k * (q*q')/(l^2) * (-2sin(theta),0,0) </math>

Once we were able to recognize our problem, it became as simple as plugging in variables and using algebra to solve the problem.

Additionally, if we were asked "What is the direction of the force on the charge -q'?"

Looking back at our work, we can see that our vertical components cancel out, leaving
<math> -2sin(theta) </math>
Indicating our direction is negative x.

We could get to this conclusion without looking at the math by understanding that electric field falls off in a way that resembles
<math> 1/(r)^2 </math>

For q1, it is sqrt(3) times farther than q2, and 3 times the charge. Which is equivalent to
<math> 3*q/(sqrt(3))^2 </math>
Which is equivalent to q2 in the vertical direction. In the horizontal direction, q2 is slightly closer, meaning that the total net charge should be in the negative x direction.

===Medium - Two Point Charges and Two Dimensions===
[[File:BrooksEx1.png]]

If <math>Q_1</math> and <math>Q_2</math> are positive point charges with a charge of e, what is the net electric field at point P?

<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
To begin this problem, the first step is to find <math>r_1</math> and <math>r_2</math>, the vectors from the charges to point P as well as their magnitudes:

<math>\vec{r_1} = 3\hat{i}+2\hat{j}-(0\hat{i}+0\hat{j})\Rightarrow\vec{r_1} = 3\hat{i}+2\hat{j}</math>

<math>||\vec{r_1}|| = \sqrt{3^2 + 2^2} =\sqrt{13}</math>

<math>\vec{r_2} = 3\hat{i}+2\hat{j}-(4\hat{i}+0\hat{j})\Rightarrow\vec{r_2} = -1\hat{i}+2\hat{j}</math>

<math>||\vec{r_2}|| = \sqrt{-1^2 + 2^2} =\sqrt{5}</math>

Using these in the equation for an electric field from a point charge, you get:

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{Q_1}{||r_1||^2}\hat{r_1}=\frac{1}{4 \pi \epsilon_0}\frac{e}{13}<\frac{3}{\sqrt{13}}\hat{i}+\frac{2}{\sqrt{13}}\hat{j}> = <9.21E-11\hat{i}+6.14E-11\hat{j}> </math>N/C

<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{Q_2}{||r_2||^2}\hat{r_2}=\frac{1}{4 \pi \epsilon_0}\frac{e}{5}<\frac{-1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j}> = <-1.29E-10\hat{i}+2.58E-10\hat{j}> </math>N/C

Then, simply add the two electric fields together:

<math>\vec{E}=\vec{E_1}+\vec{E_2} = <-3.69E-11\hat{i}+3.19E-10\hat{j}></math>
</div>
</div>

===Difficult - Five Point Charges and Three Dimensions===

[[File:brooksEx2.png]]

If all point charges have a charge of e, what the the net electric field present at point L?
<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
This problem is similar to the previous examples but this one includes the z axis and more points. Since there are 5 points and you have already had some practice, you will only see the procedure for the first one. We will still work through the whole problem. Again, first each vector and magnitude:

<math>\vec{d_5} = 0\hat{i}+0\hat{j}+0\hat{k}-(2\hat{i}-1\hat{j}-1\hat{k})-\Rightarrow\vec{d_5} = -2\hat{i}+1\hat{j}+1\hat{k}</math>

<math>||\vec{d_5}|| = \sqrt{(-2)^2 + 1^2 + 1^2} = 2</math>

Do the same for each of the other point charges and plug them into the electric field formula:

<math>\vec{E_5} = \frac{1}{4 \pi \epsilon_0}\frac{Q_5}{||d_5||^2}\hat{d_5}=\frac{1}{4 \pi \epsilon_0}\frac{e}{4}<\frac{-2}{2}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}> = <-3.6E-10\hat{i}+1.8E-10\hat{j}+1.8E-10\hat{k}></math>N/C

Follow the same steps for the other electric fields and add them all together to get your final answer:

<math>\vec{E_1} = <1.44E-9\hat{i}+0\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_2} = <0\hat{i}-1.44E-9\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_3} = <-1.29E-10\hat{i}+0\hat{j}+2.58E-10\hat{k}></math>N/C

<math>\vec{E_4} = <0\hat{i}+0\hat{j}-1.44E-9\hat{k}></math>N/C

<math>\vec{E} = \vec{E_1}+\vec{E_2}+\vec{E_3}+\vec{E_4}+\vec{E_5}=<9.51E-10\hat{i}-1.26E-9\hat{j}+1.00E-9\hat{k}></math>N/C
</div>
</div>

==Connectedness==
[[File:Superposition Example.PNG|thumb|left|A diagram of a circuit you will be able to analyze further on by using the superposition of currents.]]
The superposition principle is critical as you progress to more complicated topics in Physics. This principle simplifies many topics that you will encounter further ahead, by assuming that their net contribution will be equal to the addition of all the individual ones. You will use superposition with electric fields and forces, as well as magnetic fields and forces. In addition, if in a problem you are only given the final net contribution of any type of field, you can inversely solve it by constructing equations that relate the individual formulas. You can then plug in values, and subtract in order to find the magnitudes of each individual contributor. This will also enable you to better analyze different types of circuits.

However, the superposition principle also aids outside of the world of abstract electromagnetism. It is also useful for static problems, which you might encounter if you are a Civil or Mechanical Engineer, or systems and circuitry, which you might encounter if you are a Biomedical engineer.

==History==
[[File:Danielbernoulli.jpg|thumb|right|Daniel Bernoulli]]
The superposition principle was supposedly first stated by [[Daniel Bernoulli]]. Bernoulli was a famous scientist whose primary work was in fluid mechanics and statistics. If you recognize the name, he is primarily remembered for discovering the Bernoulli Principle. In 1753, he stated that, "The general motion of a vibrating system is given by a superposition of its proper vibrations." This idea was at first rejected by some other popular scientists until it became widely accepted due to the work of Joseph Fourier. Fourier was a famous scientist known for his work on the Fourier Series, which is used in heat transfer and vibrations, as well as his discovery of the Greenhouse Effect. Fournier’s support legitimized Bernoulli’s claims and has made a huge impact on history.

== See also ==

===Next Topics===
Now that you have a solid understanding of the Superposition Principle, you can move on to these topics:
*[[Electric Field]]
**[[Point Charge]]
**[[Electric Dipole]]
*[[Electric Force]]
*[[Magnetic Field]]
*[[Magnetic Force]]
*[[Superposition Principle]]
===Further reading===

Books, Articles or other print media on this topic

[http://www.acoustics.salford.ac.uk/feschools/waves/super.php Superposition of Waves]

[http://whatis.techtarget.com/definition/Schrodingers-cat Schrodingers' Cat from a Superposition Point of View]

===External Links===

[https://www.youtube.com/watch?v=p3Xugztl9Ho Visual Explanation of Superposition in Electric Fields]

[https://www.youtube.com/watch?v=mdulzEfQXDE Electric Fields and Superposition Principle Crash Course Video]

[https://www.youtube.com/watch?v=S1TXN1M9t18 Instructional video on how to calculate the net electric field using the superposition principle]

[https://www.youtube.com/watch?v=lCM5dql_ul0 Additional Video on calculating superposition principle]

[https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1479978&tag=1 Using superposition principle to analyze solar cells]

==References==
[http://physics-help.info/physicsguide/electricity/electric_field.shtml Electric Fields]

[https://en.wikipedia.org/wiki/Superposition_principle Superposition Principle]

[https://en.wikipedia.org/wiki/Daniel_Bernoulli Daniel Bernoulli Biography]

[https://en.wikipedia.org/wiki/Joseph_Fourier Joseph Fourier Biography]
[[Category:Fields]]

File:69832.jpg

2021-11-29T18:05:44Z

Jnation8:

Superposition Principle

2021-11-29T17:53:05Z

Jnation8:

'''Claimed By: Jared Nation'''

== The Big Picture ==
The superposition principle is based on the idea that in a closed system, an object receives a net force equal to the sum of all outside forces acting on it. This is commonly used to calculate the net electric field or magnetic field on an object. More specifically, the Superposition Principle states that the net result of multiple vectors acting on a given place and time is equal to the vector sum of each individual vector. For example, think about two waves colliding. When the waves collide, the height of the combined wave is the sum of the height of the two waves. This principle can be seen throughout nature and theory. T

===The Wave Model===

Superposition is easy to think about in the form of waves. Let's say, you are at the beach watching the ocean. A huge 5 foot wave comes towards the land and is about to crash. However, just as this happens, a smaller 1 foot tall wave forms due to the undercurrent and goes out to the ocean. Before the waves crash, they meet. At this meeting point, the combined waves become 6 feet tall (1 + 5). Why is this? This is because of superposition.

[[File:Wav sup.jpeg|500px|center]]

Let's now think about an orchestra. Before the concert starts, individual instruments practice on stage. Sitting in the audience, you can hear the individual instruments. However, the sound of that one instrument does not nearly compare the the sound of the entire orchestra. When the entire orchestra begins to play, the volume is much louder. This is because of the superposition principle. Each instrument produces sound waves. When an entire orchestra performs at once, the sound waves from each instrument are all summed together. The superposition principle is not only seen in waves. The superposition principle can be seen in forces, electric fields, magnetic fields, etc. However, it is easier to understand when looking at waves that everyone notices.

[[File:Standing_wave_2.gif]]

This gif depicts the superposition principal. The peaks and the throughs (the bottom of the waves) are summed together as the different waves overlap. Just as the waves of the ocean and the sound waves of the orchestra summed together, the waves in this gif sum together.

===A Mathematical Model===

The Superposition Principle is derived from the properties of additivity and homogeneity for linear systems which are defined in terms of a scalar value of a by the following equations:

<math>F(x_1 + x_2) = F(x_1) + F(x_2)\quad Additivity</math>

<math>aF(x) = F(ax)\quad Homogeneity</math>
[[File:Superposition.gif|thumb|Example of the Superposition of Electric Fields using charges with changing locations]]

The principle can be applied to any linear system and can be used to find the net result of functions, vectors, vector fields, etc. For the topic of introductory physics, it will mainly apply to vectors and vector fields such as electric forces and fields.

If given a number of vectors passing through a certain point, the resultant vector is given by simply adding all the the vectors at that point. For example, for a number of uniform electric fields passing though a single point, the resulting electric field at that point is given by:

<math>\vec{E} = \vec{E}_{1} + \vec{E}_{2} +...+ \vec{E}_{n} = \sum_{i=1}^n\vec{E}_{i}</math>

This same concept can be applied to electric forces as well as to magnetic fields and forces. This is more useful when dealing with the effect that multiple point charges have on each other is an area of void of other electric fields. By adding together each of the contributing electric field vectors at a specific point you can figure out the net contribution that all of the present charges have on the overall electric field experienced at said location.

You can calculate the electric field of a single point charge by using the following equation:
<math>\vec{E} = \frac{1}{4 \pi \epsilon_0}\sum_{i=1}^n\frac{q_i}{r_i^2}\hat{r_i}</math>

Where:

<math>\frac{1}{4 \pi \epsilon_0}</math> is a constant that equals <math>9e9</math>.

<math>q_i</math> is the charge of the particle you are studying.

<math>r_i</math> is the magnitude of the distance between the particle and the observation location.

<math>\hat{r_i}</math> is the vector pointing from the particle to the observation location.

Conceptually is is also important to note that an object cannot exert a force on itself, so the sum of all of the forces does not include the object itself.

==Examples==

When attempting to solve these problems be sure to show all steps in your solution and include diagrams whenever possible.

===Simple - Two Point Charges and One Dimension===
[[File:SuperpositionExample.png]]

If <math>Q_1</math> equals 1e-4 C and <math>Q_2</math> equals 1e-5C, what is the net electric field at the midpoint of both charges?

<div class="toccolours mw-collapsible mw-collapsed">

'''''Click for Solution'''''
<div class="mw-collapsible-content">
We know that the distance of each charge to the midpoint is <math>1m</math>. Since this value is the magnitude of the distance it will serve as our <math>r_i</math> value for both of the electric field formulas.

Our <math>\hat{r_1}</math> value for <math>E_1</math> will be <math>1</math> and the <math>\hat{r_2}</math> value for <math>E_2</math> will be <math>-1</math>, if you use the midpoint as your observation point in the x-direction.

Now that we have all the necessary values to use the electric field formula, we can plug them in.

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{q_1}{r_1^2}\hat{r_1} = (9e9)\frac{1e-4C}{1m^2}(1m) = 9e5\frac{N}{C}</math>
<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{q_2}{r_2^2}\hat{r_2} = (9e9)\frac{1e-5C}{1m^2}(-1m) = -9e4\frac{N}{C}</math>

By this point, you have the individual contributions of each charge. To get the net contribution of all the charges, you only need to add them.

<math>\vec{E_1}+\vec{E_2}=9e5\frac{N}{C}-9e4\frac{N}{C}= 8.1e5\frac{N}{C}</math>

This means that the net electric field will point towards the right with a magnitude of <math>8.1e5\frac{N}{C}</math>.
</div>
</div>
=== Medium - Three Point charges and two Dimensions ===
[[File:Superposfile2.JPG]]

----
Using k for the constant term
<math> E = Q1 * rhat/ (r)^2 </math>

<math> F = q3 * E </math>

<math> r = (4d, -3d,0> - (0,3d,)) = (4d,-6d,0) </math>

<math> k * Q3 * Q1 * rhat/ (r)^2 </math>

<math> k* Q^2 * 1/(52d^2) * (2/13 * sqrt(13),-3/13 sqrt(13),0) </math>

=== Medium - Three Point charges and two Dimensions===
[[File:Superpos file.JPG]]

----

What is the magnitude of the net force on the charge -q'?

To begin, we recognize that this problem will be using the superposition principle. It is additionally
important to recognize that objects can't exert forces on themselves. Keeping these in mind, The force exerted on -q' will be a combination of the forces exerted by 3q and negative q.

For this problem, let's consider -q' to be q3, 3q to be q1, and -1 to be q2 and, For the sake of simplicity, we are going to call our constant 'k'

Let's find the forces separately then add them together:

<math> k * (q1*q3)/(r)^2 * rhat </math>

Now that we have the general form of the equation, we plug in variables.

<math> F1 = k * 3q(-q')/(sqrt(3) * l)^2 * (sin(theta),cos(theta),0) </math>
<math>

= k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) </math>

Great! Now we have the first force from q1, now we need to find the force from q2

<math> F2 = k * (q2*q3)/ (r)^2 * rhat </math>

<math> = k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>

Now that we've found both forces, we add them together.

<math> Fnet = F1+F2 </math>
<math> = k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) + k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>
<math> = k * (q*q')/(l^2) * (-2sin(theta),0,0) </math>

Once we were able to recognize our problem, it became as simple as plugging in variables and using algebra to solve the problem.

Additionally, if we were asked "What is the direction of the force on the charge -q'?"

Looking back at our work, we can see that our vertical components cancel out, leaving
<math> -2sin(theta) </math>
Indicating our direction is negative x.

We could get to this conclusion without looking at the math by understanding that electric field falls off in a way that resembles
<math> 1/(r)^2 </math>

For q1, it is sqrt(3) times farther than q2, and 3 times the charge. Which is equivalent to
<math> 3*q/(sqrt(3))^2 </math>
Which is equivalent to q2 in the vertical direction. In the horizontal direction, q2 is slightly closer, meaning that the total net charge should be in the negative x direction.

===Medium - Two Point Charges and Two Dimensions===
[[File:BrooksEx1.png]]

If <math>Q_1</math> and <math>Q_2</math> are positive point charges with a charge of e, what is the net electric field at point P?

<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
To begin this problem, the first step is to find <math>r_1</math> and <math>r_2</math>, the vectors from the charges to point P as well as their magnitudes:

<math>\vec{r_1} = 3\hat{i}+2\hat{j}-(0\hat{i}+0\hat{j})\Rightarrow\vec{r_1} = 3\hat{i}+2\hat{j}</math>

<math>||\vec{r_1}|| = \sqrt{3^2 + 2^2} =\sqrt{13}</math>

<math>\vec{r_2} = 3\hat{i}+2\hat{j}-(4\hat{i}+0\hat{j})\Rightarrow\vec{r_2} = -1\hat{i}+2\hat{j}</math>

<math>||\vec{r_2}|| = \sqrt{-1^2 + 2^2} =\sqrt{5}</math>

Using these in the equation for an electric field from a point charge, you get:

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{Q_1}{||r_1||^2}\hat{r_1}=\frac{1}{4 \pi \epsilon_0}\frac{e}{13}<\frac{3}{\sqrt{13}}\hat{i}+\frac{2}{\sqrt{13}}\hat{j}> = <9.21E-11\hat{i}+6.14E-11\hat{j}> </math>N/C

<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{Q_2}{||r_2||^2}\hat{r_2}=\frac{1}{4 \pi \epsilon_0}\frac{e}{5}<\frac{-1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j}> = <-1.29E-10\hat{i}+2.58E-10\hat{j}> </math>N/C

Then, simply add the two electric fields together:

<math>\vec{E}=\vec{E_1}+\vec{E_2} = <-3.69E-11\hat{i}+3.19E-10\hat{j}></math>
</div>
</div>

===Difficult - Five Point Charges and Three Dimensions===

[[File:brooksEx2.png]]

If all point charges have a charge of e, what the the net electric field present at point L?
<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
This problem is similar to the previous examples but this one includes the z axis and more points. Since there are 5 points and you have already had some practice, you will only see the procedure for the first one. We will still work through the whole problem. Again, first each vector and magnitude:

<math>\vec{d_5} = 0\hat{i}+0\hat{j}+0\hat{k}-(2\hat{i}-1\hat{j}-1\hat{k})-\Rightarrow\vec{d_5} = -2\hat{i}+1\hat{j}+1\hat{k}</math>

<math>||\vec{d_5}|| = \sqrt{(-2)^2 + 1^2 + 1^2} = 2</math>

Do the same for each of the other point charges and plug them into the electric field formula:

<math>\vec{E_5} = \frac{1}{4 \pi \epsilon_0}\frac{Q_5}{||d_5||^2}\hat{d_5}=\frac{1}{4 \pi \epsilon_0}\frac{e}{4}<\frac{-2}{2}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}> = <-3.6E-10\hat{i}+1.8E-10\hat{j}+1.8E-10\hat{k}></math>N/C

Follow the same steps for the other electric fields and add them all together to get your final answer:

<math>\vec{E_1} = <1.44E-9\hat{i}+0\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_2} = <0\hat{i}-1.44E-9\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_3} = <-1.29E-10\hat{i}+0\hat{j}+2.58E-10\hat{k}></math>N/C

<math>\vec{E_4} = <0\hat{i}+0\hat{j}-1.44E-9\hat{k}></math>N/C

<math>\vec{E} = \vec{E_1}+\vec{E_2}+\vec{E_3}+\vec{E_4}+\vec{E_5}=<9.51E-10\hat{i}-1.26E-9\hat{j}+1.00E-9\hat{k}></math>N/C
</div>
</div>

==Connectedness==
[[File:Superposition Example.PNG|thumb|left|A diagram of a circuit you will be able to analyze further on by using the superposition of currents.]]
The superposition principle is critical as you progress to more complicated topics in Physics. This principle simplifies many topics that you will encounter further ahead, by assuming that their net contribution will be equal to the addition of all the individual ones. You will use superposition with electric fields and forces, as well as magnetic fields and forces. In addition, if in a problem you are only given the final net contribution of any type of field, you can inversely solve it by constructing equations that relate the individual formulas. You can then plug in values, and subtract in order to find the magnitudes of each individual contributor. This will also enable you to better analyze different types of circuits.

However, the superposition principle also aids outside of the world of abstract electromagnetism. It is also useful for static problems, which you might encounter if you are a Civil or Mechanical Engineer, or systems and circuitry, which you might encounter if you are a Biomedical engineer.

==History==
[[File:Danielbernoulli.jpg|thumb|right|Daniel Bernoulli]]
The superposition principle was supposedly first stated by [[Daniel Bernoulli]]. Bernoulli was a famous scientist whose primary work was in fluid mechanics and statistics. If you recognize the name, he is primarily remembered for discovering the Bernoulli Principle. In 1753, he stated that, "The general motion of a vibrating system is given by a superposition of its proper vibrations." This idea was at first rejected by some other popular scientists until it became widely accepted due to the work of Joseph Fourier. Fourier was a famous scientist known for his work on the Fourier Series, which is used in heat transfer and vibrations, as well as his discovery of the Greenhouse Effect. Fournier’s support legitimized Bernoulli’s claims and has made a huge impact on history.

== See also ==

===Next Topics===
Now that you have a solid understanding of the Superposition Principle, you can move on to these topics:
*[[Electric Field]]
**[[Point Charge]]
**[[Electric Dipole]]
*[[Electric Force]]
*[[Magnetic Field]]
*[[Magnetic Force]]
*[[Superposition Principle]]
===Further reading===

Books, Articles or other print media on this topic

[http://www.acoustics.salford.ac.uk/feschools/waves/super.php Superposition of Waves]

[http://whatis.techtarget.com/definition/Schrodingers-cat Schrodingers' Cat from a Superposition Point of View]

===External Links===

[https://www.youtube.com/watch?v=p3Xugztl9Ho Visual Explanation of Superposition in Electric Fields]

[https://www.youtube.com/watch?v=mdulzEfQXDE Electric Fields and Superposition Principle Crash Course Video]

[https://www.youtube.com/watch?v=S1TXN1M9t18 Instructional video on how to calculate the net electric field using the superposition principle]

[https://www.youtube.com/watch?v=lCM5dql_ul0 Additional Video on calculating superposition principle]

[https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1479978&tag=1 Using superposition principle to analyze solar cells]

==References==
[http://physics-help.info/physicsguide/electricity/electric_field.shtml Electric Fields]

[https://en.wikipedia.org/wiki/Superposition_principle Superposition Principle]

[https://en.wikipedia.org/wiki/Daniel_Bernoulli Daniel Bernoulli Biography]

[https://en.wikipedia.org/wiki/Joseph_Fourier Joseph Fourier Biography]
[[Category:Fields]]

Superposition Principle

2021-11-29T17:52:03Z

Jnation8: /* The Big Picture */

'''Claimed By: Jared Nation'''

== The Big Picture ==
The superposition principle is based on the idea that in a closed system, an object receives a net force equal to the sum of all outside forces acting on it. This is commonly used to calculate the net electric field or magnetic field on an object. More specifically, the Superposition Principle states that the net result of multiple vectors acting on a given place and time is equal to the vector sum of each individual vector. For example, think about two waves colliding. When the waves collide, the height of the combined wave is the sum of the height of the two waves. This principle can be seen throughout nature and theory. T

===The Wave Model===

Superposition is easy to think about in the form of waves. Let's say, you are at the beach watching the ocean. A huge 5 foot wave comes towards the land and is about to crash. However, just as this happens, a smaller 1 foot tall wave forms due to the undercurrent and goes out to the ocean. Before the waves crash, they meet. At this meeting point, the combined waves become 6 feet tall (1 + 5). Why is this? This is because of superposition.

[[File:Wav sup.jpeg|500px|center]]

Let's now think about an orchestra. Before the concert starts, individual instruments practice on stage. Sitting in the audience, you can hear the individual instruments. However, the sound of that one instrument does not nearly compare the the sound of the entire orchestra. When the entire orchestra begins to play, the volume is much louder. This is because of the superposition principle. Each instrument produces sound waves. When an entire orchestra performs at once, the sound waves from each instrument are all summed together. The superposition principle is not only seen in waves. The superposition principle can be seen in forces, electric fields, magnetic fields, etc. However, it is easier to understand when looking at waves that everyone notices.

[[File:Standing_wave_2.gif]]

This gif depicts the superposition principal. The peaks and the throughs (the bottom of the waves) are summed together as the different waves overlap. Just as the waves of the ocean and the sound waves of the orchestra summed together, the waves in this gif sum together.

Mathematically, this can be represented by with phase ϕ. This equation can be represented in the gif below.

y ( x , t ) = y m sin ( kx - ωt ) + y m sin ( kx - ωt + ϕ ) = 2 y m cos ( ϕ / 2 ) sin ( kx - ωt + ϕ / 2 )

[[File:waves_sup123.gif]]

Another mathematical representation of the superposition of waves can be seen below.

y ( x , t ) = y m sin ( kx - ωt ) + y m sin ( kx + ωt ) = 2 y m sin ( kx ) cos ( ωt )

[[File:waves_two_moving123.gif]]

===A Mathematical Model===

The Superposition Principle is derived from the properties of additivity and homogeneity for linear systems which are defined in terms of a scalar value of a by the following equations:

<math>F(x_1 + x_2) = F(x_1) + F(x_2)\quad Additivity</math>

<math>aF(x) = F(ax)\quad Homogeneity</math>
[[File:Superposition.gif|thumb|Example of the Superposition of Electric Fields using charges with changing locations]]

The principle can be applied to any linear system and can be used to find the net result of functions, vectors, vector fields, etc. For the topic of introductory physics, it will mainly apply to vectors and vector fields such as electric forces and fields.

If given a number of vectors passing through a certain point, the resultant vector is given by simply adding all the the vectors at that point. For example, for a number of uniform electric fields passing though a single point, the resulting electric field at that point is given by:

<math>\vec{E} = \vec{E}_{1} + \vec{E}_{2} +...+ \vec{E}_{n} = \sum_{i=1}^n\vec{E}_{i}</math>

This same concept can be applied to electric forces as well as to magnetic fields and forces. This is more useful when dealing with the effect that multiple point charges have on each other is an area of void of other electric fields. By adding together each of the contributing electric field vectors at a specific point you can figure out the net contribution that all of the present charges have on the overall electric field experienced at said location.

You can calculate the electric field of a single point charge by using the following equation:
<math>\vec{E} = \frac{1}{4 \pi \epsilon_0}\sum_{i=1}^n\frac{q_i}{r_i^2}\hat{r_i}</math>

Where:

<math>\frac{1}{4 \pi \epsilon_0}</math> is a constant that equals <math>9e9</math>.

<math>q_i</math> is the charge of the particle you are studying.

<math>r_i</math> is the magnitude of the distance between the particle and the observation location.

<math>\hat{r_i}</math> is the vector pointing from the particle to the observation location.

Conceptually is is also important to note that an object cannot exert a force on itself, so the sum of all of the forces does not include the object itself.

==Examples==

When attempting to solve these problems be sure to show all steps in your solution and include diagrams whenever possible.

===Simple - Two Point Charges and One Dimension===
[[File:SuperpositionExample.png]]

If <math>Q_1</math> equals 1e-4 C and <math>Q_2</math> equals 1e-5C, what is the net electric field at the midpoint of both charges?

<div class="toccolours mw-collapsible mw-collapsed">

'''''Click for Solution'''''
<div class="mw-collapsible-content">
We know that the distance of each charge to the midpoint is <math>1m</math>. Since this value is the magnitude of the distance it will serve as our <math>r_i</math> value for both of the electric field formulas.

Our <math>\hat{r_1}</math> value for <math>E_1</math> will be <math>1</math> and the <math>\hat{r_2}</math> value for <math>E_2</math> will be <math>-1</math>, if you use the midpoint as your observation point in the x-direction.

Now that we have all the necessary values to use the electric field formula, we can plug them in.

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{q_1}{r_1^2}\hat{r_1} = (9e9)\frac{1e-4C}{1m^2}(1m) = 9e5\frac{N}{C}</math>
<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{q_2}{r_2^2}\hat{r_2} = (9e9)\frac{1e-5C}{1m^2}(-1m) = -9e4\frac{N}{C}</math>

By this point, you have the individual contributions of each charge. To get the net contribution of all the charges, you only need to add them.

<math>\vec{E_1}+\vec{E_2}=9e5\frac{N}{C}-9e4\frac{N}{C}= 8.1e5\frac{N}{C}</math>

This means that the net electric field will point towards the right with a magnitude of <math>8.1e5\frac{N}{C}</math>.
</div>
</div>
=== Medium - Three Point charges and two Dimensions ===
[[File:Superposfile2.JPG]]

----
Using k for the constant term
<math> E = Q1 * rhat/ (r)^2 </math>

<math> F = q3 * E </math>

<math> r = (4d, -3d,0> - (0,3d,)) = (4d,-6d,0) </math>

<math> k * Q3 * Q1 * rhat/ (r)^2 </math>

<math> k* Q^2 * 1/(52d^2) * (2/13 * sqrt(13),-3/13 sqrt(13),0) </math>

=== Medium - Three Point charges and two Dimensions===
[[File:Superpos file.JPG]]

----

What is the magnitude of the net force on the charge -q'?

To begin, we recognize that this problem will be using the superposition principle. It is additionally
important to recognize that objects can't exert forces on themselves. Keeping these in mind, The force exerted on -q' will be a combination of the forces exerted by 3q and negative q.

For this problem, let's consider -q' to be q3, 3q to be q1, and -1 to be q2 and, For the sake of simplicity, we are going to call our constant 'k'

Let's find the forces separately then add them together:

<math> k * (q1*q3)/(r)^2 * rhat </math>

Now that we have the general form of the equation, we plug in variables.

<math> F1 = k * 3q(-q')/(sqrt(3) * l)^2 * (sin(theta),cos(theta),0) </math>
<math>

= k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) </math>

Great! Now we have the first force from q1, now we need to find the force from q2

<math> F2 = k * (q2*q3)/ (r)^2 * rhat </math>

<math> = k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>

Now that we've found both forces, we add them together.

<math> Fnet = F1+F2 </math>
<math> = k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) + k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>
<math> = k * (q*q')/(l^2) * (-2sin(theta),0,0) </math>

Once we were able to recognize our problem, it became as simple as plugging in variables and using algebra to solve the problem.

Additionally, if we were asked "What is the direction of the force on the charge -q'?"

Looking back at our work, we can see that our vertical components cancel out, leaving
<math> -2sin(theta) </math>
Indicating our direction is negative x.

We could get to this conclusion without looking at the math by understanding that electric field falls off in a way that resembles
<math> 1/(r)^2 </math>

For q1, it is sqrt(3) times farther than q2, and 3 times the charge. Which is equivalent to
<math> 3*q/(sqrt(3))^2 </math>
Which is equivalent to q2 in the vertical direction. In the horizontal direction, q2 is slightly closer, meaning that the total net charge should be in the negative x direction.

===Medium - Two Point Charges and Two Dimensions===
[[File:BrooksEx1.png]]

If <math>Q_1</math> and <math>Q_2</math> are positive point charges with a charge of e, what is the net electric field at point P?

<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
To begin this problem, the first step is to find <math>r_1</math> and <math>r_2</math>, the vectors from the charges to point P as well as their magnitudes:

<math>\vec{r_1} = 3\hat{i}+2\hat{j}-(0\hat{i}+0\hat{j})\Rightarrow\vec{r_1} = 3\hat{i}+2\hat{j}</math>

<math>||\vec{r_1}|| = \sqrt{3^2 + 2^2} =\sqrt{13}</math>

<math>\vec{r_2} = 3\hat{i}+2\hat{j}-(4\hat{i}+0\hat{j})\Rightarrow\vec{r_2} = -1\hat{i}+2\hat{j}</math>

<math>||\vec{r_2}|| = \sqrt{-1^2 + 2^2} =\sqrt{5}</math>

Using these in the equation for an electric field from a point charge, you get:

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{Q_1}{||r_1||^2}\hat{r_1}=\frac{1}{4 \pi \epsilon_0}\frac{e}{13}<\frac{3}{\sqrt{13}}\hat{i}+\frac{2}{\sqrt{13}}\hat{j}> = <9.21E-11\hat{i}+6.14E-11\hat{j}> </math>N/C

<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{Q_2}{||r_2||^2}\hat{r_2}=\frac{1}{4 \pi \epsilon_0}\frac{e}{5}<\frac{-1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j}> = <-1.29E-10\hat{i}+2.58E-10\hat{j}> </math>N/C

Then, simply add the two electric fields together:

<math>\vec{E}=\vec{E_1}+\vec{E_2} = <-3.69E-11\hat{i}+3.19E-10\hat{j}></math>
</div>
</div>

===Difficult - Five Point Charges and Three Dimensions===

[[File:brooksEx2.png]]

If all point charges have a charge of e, what the the net electric field present at point L?
<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
This problem is similar to the previous examples but this one includes the z axis and more points. Since there are 5 points and you have already had some practice, you will only see the procedure for the first one. We will still work through the whole problem. Again, first each vector and magnitude:

<math>\vec{d_5} = 0\hat{i}+0\hat{j}+0\hat{k}-(2\hat{i}-1\hat{j}-1\hat{k})-\Rightarrow\vec{d_5} = -2\hat{i}+1\hat{j}+1\hat{k}</math>

<math>||\vec{d_5}|| = \sqrt{(-2)^2 + 1^2 + 1^2} = 2</math>

Do the same for each of the other point charges and plug them into the electric field formula:

<math>\vec{E_5} = \frac{1}{4 \pi \epsilon_0}\frac{Q_5}{||d_5||^2}\hat{d_5}=\frac{1}{4 \pi \epsilon_0}\frac{e}{4}<\frac{-2}{2}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}> = <-3.6E-10\hat{i}+1.8E-10\hat{j}+1.8E-10\hat{k}></math>N/C

Follow the same steps for the other electric fields and add them all together to get your final answer:

<math>\vec{E_1} = <1.44E-9\hat{i}+0\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_2} = <0\hat{i}-1.44E-9\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_3} = <-1.29E-10\hat{i}+0\hat{j}+2.58E-10\hat{k}></math>N/C

<math>\vec{E_4} = <0\hat{i}+0\hat{j}-1.44E-9\hat{k}></math>N/C

<math>\vec{E} = \vec{E_1}+\vec{E_2}+\vec{E_3}+\vec{E_4}+\vec{E_5}=<9.51E-10\hat{i}-1.26E-9\hat{j}+1.00E-9\hat{k}></math>N/C
</div>
</div>

==Connectedness==
[[File:Superposition Example.PNG|thumb|left|A diagram of a circuit you will be able to analyze further on by using the superposition of currents.]]
The superposition principle is critical as you progress to more complicated topics in Physics. This principle simplifies many topics that you will encounter further ahead, by assuming that their net contribution will be equal to the addition of all the individual ones. You will use superposition with electric fields and forces, as well as magnetic fields and forces. In addition, if in a problem you are only given the final net contribution of any type of field, you can inversely solve it by constructing equations that relate the individual formulas. You can then plug in values, and subtract in order to find the magnitudes of each individual contributor. This will also enable you to better analyze different types of circuits.

However, the superposition principle also aids outside of the world of abstract electromagnetism. It is also useful for static problems, which you might encounter if you are a Civil or Mechanical Engineer, or systems and circuitry, which you might encounter if you are a Biomedical engineer.

==History==
[[File:Danielbernoulli.jpg|thumb|right|Daniel Bernoulli]]
The superposition principle was supposedly first stated by [[Daniel Bernoulli]]. Bernoulli was a famous scientist whose primary work was in fluid mechanics and statistics. If you recognize the name, he is primarily remembered for discovering the Bernoulli Principle. In 1753, he stated that, "The general motion of a vibrating system is given by a superposition of its proper vibrations." This idea was at first rejected by some other popular scientists until it became widely accepted due to the work of Joseph Fourier. Fourier was a famous scientist known for his work on the Fourier Series, which is used in heat transfer and vibrations, as well as his discovery of the Greenhouse Effect. Fournier’s support legitimized Bernoulli’s claims and has made a huge impact on history.

== See also ==

===Next Topics===
Now that you have a solid understanding of the Superposition Principle, you can move on to these topics:
*[[Electric Field]]
**[[Point Charge]]
**[[Electric Dipole]]
*[[Electric Force]]
*[[Magnetic Field]]
*[[Magnetic Force]]
*[[Superposition Principle]]
===Further reading===

Books, Articles or other print media on this topic

[http://www.acoustics.salford.ac.uk/feschools/waves/super.php Superposition of Waves]

[http://whatis.techtarget.com/definition/Schrodingers-cat Schrodingers' Cat from a Superposition Point of View]

===External Links===

[https://www.youtube.com/watch?v=p3Xugztl9Ho Visual Explanation of Superposition in Electric Fields]

[https://www.youtube.com/watch?v=mdulzEfQXDE Electric Fields and Superposition Principle Crash Course Video]

[https://www.youtube.com/watch?v=S1TXN1M9t18 Instructional video on how to calculate the net electric field using the superposition principle]

[https://www.youtube.com/watch?v=lCM5dql_ul0 Additional Video on calculating superposition principle]

[https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1479978&tag=1 Using superposition principle to analyze solar cells]

==References==
[http://physics-help.info/physicsguide/electricity/electric_field.shtml Electric Fields]

[https://en.wikipedia.org/wiki/Superposition_principle Superposition Principle]

[https://en.wikipedia.org/wiki/Daniel_Bernoulli Daniel Bernoulli Biography]

[https://en.wikipedia.org/wiki/Joseph_Fourier Joseph Fourier Biography]
[[Category:Fields]]

Superposition Principle

2021-11-29T17:34:33Z

Jnation8:

'''Claimed By: Jared Nation'''

== The Big Picture ==
The superposition principle is based on the idea that in a close system, an object receives a net force equal to the sum of all outside forces acting on it. This is commonly used to calculate the net electric field or magnetic field on an object. More specifically, the Superposition Principle states that the net result of multiple vectors acting on a given place and time is equal to the vector sum of each individual vector. For example, think about two waves colliding. When the waves collide, the height of the combined wave is the sum of the height of the two waves. This principle can be seen throughout nature and theory.

===The Wave Model===

Superposition is easy to think about in the form of waves. Let's say, you are at the beach watching the ocean. A huge 5 foot wave comes towards the land and is about to crash. However, just as this happens, a smaller 1 foot tall wave forms due to the undercurrent and goes out to the ocean. Before the waves crash, they meet. At this meeting point, the combined waves become 6 feet tall (1 + 5). Why is this? This is because of superposition.

[[File:Wav sup.jpeg|500px|center]]

Let's now think about an orchestra. Before the concert starts, individual instruments practice on stage. Sitting in the audience, you can hear the individual instruments. However, the sound of that one instrument does not nearly compare the the sound of the entire orchestra. When the entire orchestra begins to play, the volume is much louder. This is because of the superposition principle. Each instrument produces sound waves. When an entire orchestra performs at once, the sound waves from each instrument are all summed together. The superposition principle is not only seen in waves. The superposition principle can be seen in forces, electric fields, magnetic fields, etc. However, it is easier to understand when looking at waves that everyone notices.

[[File:Standing_wave_2.gif]]

This gif depicts the superposition principal. The peaks and the throughs (the bottom of the waves) are summed together as the different waves overlap. Just as the waves of the ocean and the sound waves of the orchestra summed together, the waves in this gif sum together.

Mathematically, this can be represented by with phase ϕ. This equation can be represented in the gif below.

y ( x , t ) = y m sin ( kx - ωt ) + y m sin ( kx - ωt + ϕ ) = 2 y m cos ( ϕ / 2 ) sin ( kx - ωt + ϕ / 2 )

[[File:waves_sup123.gif]]

Another mathematical representation of the superposition of waves can be seen below.

y ( x , t ) = y m sin ( kx - ωt ) + y m sin ( kx + ωt ) = 2 y m sin ( kx ) cos ( ωt )

[[File:waves_two_moving123.gif]]

===A Mathematical Model===

The Superposition Principle is derived from the properties of additivity and homogeneity for linear systems which are defined in terms of a scalar value of a by the following equations:

<math>F(x_1 + x_2) = F(x_1) + F(x_2)\quad Additivity</math>

<math>aF(x) = F(ax)\quad Homogeneity</math>
[[File:Superposition.gif|thumb|Example of the Superposition of Electric Fields using charges with changing locations]]

The principle can be applied to any linear system and can be used to find the net result of functions, vectors, vector fields, etc. For the topic of introductory physics, it will mainly apply to vectors and vector fields such as electric forces and fields.

If given a number of vectors passing through a certain point, the resultant vector is given by simply adding all the the vectors at that point. For example, for a number of uniform electric fields passing though a single point, the resulting electric field at that point is given by:

<math>\vec{E} = \vec{E}_{1} + \vec{E}_{2} +...+ \vec{E}_{n} = \sum_{i=1}^n\vec{E}_{i}</math>

This same concept can be applied to electric forces as well as to magnetic fields and forces. This is more useful when dealing with the effect that multiple point charges have on each other is an area of void of other electric fields. By adding together each of the contributing electric field vectors at a specific point you can figure out the net contribution that all of the present charges have on the overall electric field experienced at said location.

You can calculate the electric field of a single point charge by using the following equation:
<math>\vec{E} = \frac{1}{4 \pi \epsilon_0}\sum_{i=1}^n\frac{q_i}{r_i^2}\hat{r_i}</math>

Where:

<math>\frac{1}{4 \pi \epsilon_0}</math> is a constant that equals <math>9e9</math>.

<math>q_i</math> is the charge of the particle you are studying.

<math>r_i</math> is the magnitude of the distance between the particle and the observation location.

<math>\hat{r_i}</math> is the vector pointing from the particle to the observation location.

Conceptually is is also important to note that an object cannot exert a force on itself, so the sum of all of the forces does not include the object itself.

==Examples==

When attempting to solve these problems be sure to show all steps in your solution and include diagrams whenever possible.

===Simple - Two Point Charges and One Dimension===
[[File:SuperpositionExample.png]]

If <math>Q_1</math> equals 1e-4 C and <math>Q_2</math> equals 1e-5C, what is the net electric field at the midpoint of both charges?

<div class="toccolours mw-collapsible mw-collapsed">

'''''Click for Solution'''''
<div class="mw-collapsible-content">
We know that the distance of each charge to the midpoint is <math>1m</math>. Since this value is the magnitude of the distance it will serve as our <math>r_i</math> value for both of the electric field formulas.

Our <math>\hat{r_1}</math> value for <math>E_1</math> will be <math>1</math> and the <math>\hat{r_2}</math> value for <math>E_2</math> will be <math>-1</math>, if you use the midpoint as your observation point in the x-direction.

Now that we have all the necessary values to use the electric field formula, we can plug them in.

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{q_1}{r_1^2}\hat{r_1} = (9e9)\frac{1e-4C}{1m^2}(1m) = 9e5\frac{N}{C}</math>
<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{q_2}{r_2^2}\hat{r_2} = (9e9)\frac{1e-5C}{1m^2}(-1m) = -9e4\frac{N}{C}</math>

By this point, you have the individual contributions of each charge. To get the net contribution of all the charges, you only need to add them.

<math>\vec{E_1}+\vec{E_2}=9e5\frac{N}{C}-9e4\frac{N}{C}= 8.1e5\frac{N}{C}</math>

This means that the net electric field will point towards the right with a magnitude of <math>8.1e5\frac{N}{C}</math>.
</div>
</div>
=== Medium - Three Point charges and two Dimensions ===
[[File:Superposfile2.JPG]]

----
Using k for the constant term
<math> E = Q1 * rhat/ (r)^2 </math>

<math> F = q3 * E </math>

<math> r = (4d, -3d,0> - (0,3d,)) = (4d,-6d,0) </math>

<math> k * Q3 * Q1 * rhat/ (r)^2 </math>

<math> k* Q^2 * 1/(52d^2) * (2/13 * sqrt(13),-3/13 sqrt(13),0) </math>

=== Medium - Three Point charges and two Dimensions===
[[File:Superpos file.JPG]]

----

What is the magnitude of the net force on the charge -q'?

To begin, we recognize that this problem will be using the superposition principle. It is additionally
important to recognize that objects can't exert forces on themselves. Keeping these in mind, The force exerted on -q' will be a combination of the forces exerted by 3q and negative q.

For this problem, let's consider -q' to be q3, 3q to be q1, and -1 to be q2 and, For the sake of simplicity, we are going to call our constant 'k'

Let's find the forces separately then add them together:

<math> k * (q1*q3)/(r)^2 * rhat </math>

Now that we have the general form of the equation, we plug in variables.

<math> F1 = k * 3q(-q')/(sqrt(3) * l)^2 * (sin(theta),cos(theta),0) </math>
<math>

= k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) </math>

Great! Now we have the first force from q1, now we need to find the force from q2

<math> F2 = k * (q2*q3)/ (r)^2 * rhat </math>

<math> = k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>

Now that we've found both forces, we add them together.

<math> Fnet = F1+F2 </math>
<math> = k * (-q*q')/ (l^2) * (sin(theta),cos(theta),0) + k * (q*q)/(l^2) * (-sin(theta), cos(theta),0) </math>
<math> = k * (q*q')/(l^2) * (-2sin(theta),0,0) </math>

Once we were able to recognize our problem, it became as simple as plugging in variables and using algebra to solve the problem.

Additionally, if we were asked "What is the direction of the force on the charge -q'?"

Looking back at our work, we can see that our vertical components cancel out, leaving
<math> -2sin(theta) </math>
Indicating our direction is negative x.

We could get to this conclusion without looking at the math by understanding that electric field falls off in a way that resembles
<math> 1/(r)^2 </math>

For q1, it is sqrt(3) times farther than q2, and 3 times the charge. Which is equivalent to
<math> 3*q/(sqrt(3))^2 </math>
Which is equivalent to q2 in the vertical direction. In the horizontal direction, q2 is slightly closer, meaning that the total net charge should be in the negative x direction.

===Medium - Two Point Charges and Two Dimensions===
[[File:BrooksEx1.png]]

If <math>Q_1</math> and <math>Q_2</math> are positive point charges with a charge of e, what is the net electric field at point P?

<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
To begin this problem, the first step is to find <math>r_1</math> and <math>r_2</math>, the vectors from the charges to point P as well as their magnitudes:

<math>\vec{r_1} = 3\hat{i}+2\hat{j}-(0\hat{i}+0\hat{j})\Rightarrow\vec{r_1} = 3\hat{i}+2\hat{j}</math>

<math>||\vec{r_1}|| = \sqrt{3^2 + 2^2} =\sqrt{13}</math>

<math>\vec{r_2} = 3\hat{i}+2\hat{j}-(4\hat{i}+0\hat{j})\Rightarrow\vec{r_2} = -1\hat{i}+2\hat{j}</math>

<math>||\vec{r_2}|| = \sqrt{-1^2 + 2^2} =\sqrt{5}</math>

Using these in the equation for an electric field from a point charge, you get:

<math>\vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{Q_1}{||r_1||^2}\hat{r_1}=\frac{1}{4 \pi \epsilon_0}\frac{e}{13}<\frac{3}{\sqrt{13}}\hat{i}+\frac{2}{\sqrt{13}}\hat{j}> = <9.21E-11\hat{i}+6.14E-11\hat{j}> </math>N/C

<math>\vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{Q_2}{||r_2||^2}\hat{r_2}=\frac{1}{4 \pi \epsilon_0}\frac{e}{5}<\frac{-1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j}> = <-1.29E-10\hat{i}+2.58E-10\hat{j}> </math>N/C

Then, simply add the two electric fields together:

<math>\vec{E}=\vec{E_1}+\vec{E_2} = <-3.69E-11\hat{i}+3.19E-10\hat{j}></math>
</div>
</div>

===Difficult - Five Point Charges and Three Dimensions===

[[File:brooksEx2.png]]

If all point charges have a charge of e, what the the net electric field present at point L?
<div class="toccolours mw-collapsible mw-collapsed">
'''''Click for Solution'''''
<div class="mw-collapsible-content">
This problem is similar to the previous examples but this one includes the z axis and more points. Since there are 5 points and you have already had some practice, you will only see the procedure for the first one. We will still work through the whole problem. Again, first each vector and magnitude:

<math>\vec{d_5} = 0\hat{i}+0\hat{j}+0\hat{k}-(2\hat{i}-1\hat{j}-1\hat{k})-\Rightarrow\vec{d_5} = -2\hat{i}+1\hat{j}+1\hat{k}</math>

<math>||\vec{d_5}|| = \sqrt{(-2)^2 + 1^2 + 1^2} = 2</math>

Do the same for each of the other point charges and plug them into the electric field formula:

<math>\vec{E_5} = \frac{1}{4 \pi \epsilon_0}\frac{Q_5}{||d_5||^2}\hat{d_5}=\frac{1}{4 \pi \epsilon_0}\frac{e}{4}<\frac{-2}{2}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}> = <-3.6E-10\hat{i}+1.8E-10\hat{j}+1.8E-10\hat{k}></math>N/C

Follow the same steps for the other electric fields and add them all together to get your final answer:

<math>\vec{E_1} = <1.44E-9\hat{i}+0\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_2} = <0\hat{i}-1.44E-9\hat{j}+0\hat{k}></math>N/C

<math>\vec{E_3} = <-1.29E-10\hat{i}+0\hat{j}+2.58E-10\hat{k}></math>N/C

<math>\vec{E_4} = <0\hat{i}+0\hat{j}-1.44E-9\hat{k}></math>N/C

<math>\vec{E} = \vec{E_1}+\vec{E_2}+\vec{E_3}+\vec{E_4}+\vec{E_5}=<9.51E-10\hat{i}-1.26E-9\hat{j}+1.00E-9\hat{k}></math>N/C
</div>
</div>

==Connectedness==
[[File:Superposition Example.PNG|thumb|left|A diagram of a circuit you will be able to analyze further on by using the superposition of currents.]]
The superposition principle is critical as you progress to more complicated topics in Physics. This principle simplifies many topics that you will encounter further ahead, by assuming that their net contribution will be equal to the addition of all the individual ones. You will use superposition with electric fields and forces, as well as magnetic fields and forces. In addition, if in a problem you are only given the final net contribution of any type of field, you can inversely solve it by constructing equations that relate the individual formulas. You can then plug in values, and subtract in order to find the magnitudes of each individual contributor. This will also enable you to better analyze different types of circuits.

However, the superposition principle also aids outside of the world of abstract electromagnetism. It is also useful for static problems, which you might encounter if you are a Civil or Mechanical Engineer, or systems and circuitry, which you might encounter if you are a Biomedical engineer.

==History==
[[File:Danielbernoulli.jpg|thumb|right|Daniel Bernoulli]]
The superposition principle was supposedly first stated by [[Daniel Bernoulli]]. Bernoulli was a famous scientist whose primary work was in fluid mechanics and statistics. If you recognize the name, he is primarily remembered for discovering the Bernoulli Principle. In 1753, he stated that, "The general motion of a vibrating system is given by a superposition of its proper vibrations." This idea was at first rejected by some other popular scientists until it became widely accepted due to the work of Joseph Fourier. Fourier was a famous scientist known for his work on the Fourier Series, which is used in heat transfer and vibrations, as well as his discovery of the Greenhouse Effect. Fournier’s support legitimized Bernoulli’s claims and has made a huge impact on history.

== See also ==

===Next Topics===
Now that you have a solid understanding of the Superposition Principle, you can move on to these topics:
*[[Electric Field]]
**[[Point Charge]]
**[[Electric Dipole]]
*[[Electric Force]]
*[[Magnetic Field]]
*[[Magnetic Force]]
*[[Superposition Principle]]
===Further reading===

Books, Articles or other print media on this topic

[http://www.acoustics.salford.ac.uk/feschools/waves/super.php Superposition of Waves]

[http://whatis.techtarget.com/definition/Schrodingers-cat Schrodingers' Cat from a Superposition Point of View]

===External Links===

[https://www.youtube.com/watch?v=p3Xugztl9Ho Visual Explanation of Superposition in Electric Fields]

[https://www.youtube.com/watch?v=mdulzEfQXDE Electric Fields and Superposition Principle Crash Course Video]

[https://www.youtube.com/watch?v=S1TXN1M9t18 Instructional video on how to calculate the net electric field using the superposition principle]

[https://www.youtube.com/watch?v=lCM5dql_ul0 Additional Video on calculating superposition principle]

[https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1479978&tag=1 Using superposition principle to analyze solar cells]

==References==
[http://physics-help.info/physicsguide/electricity/electric_field.shtml Electric Fields]

[https://en.wikipedia.org/wiki/Superposition_principle Superposition Principle]

[https://en.wikipedia.org/wiki/Daniel_Bernoulli Daniel Bernoulli Biography]

[https://en.wikipedia.org/wiki/Joseph_Fourier Joseph Fourier Biography]
[[Category:Fields]]

Electric Field

2021-11-29T17:33:02Z

Jnation8:

==The Main Idea==

Here, the concept of an '''Electric Field''' produced by a point charge is explored qualitatively and quantitatively through models, examples, and a simulation. An '''Electric Field''' is a useful concept to describe how any charged particle or collection of charged particles (positive / negative) would affect surrounding charge.

The '''Electric Field''' of a point charge is spherically symmetric, (same at all points of equal radius from the source. Hence, it is useful to speak of the electric field at a certain radius (not at a certain <math>(x,y,z)</math> position), which is explained in [[electric Field#A Mathematical Model| the mathematical model]].

Keep in mind, the electric field is a vector quantity, meaning it has a magnitude and direction. The SI units are N/C.

===A Mathematical Model===
The '''Electric Field''' vector <math>\bigl( \mathbf{E}_{s} \bigl)</math> of a point source charge <math>\bigl( Q_{s} \bigl)</math> gives the magnitude and direction of the Electrostatic Force vector <math>\bigl( \mathbf{F}_{s} \bigl)</math> exerted on a unit charge (<math>1</math> Coulomb) by <math>Q_{s}</math>, as a function of position <math>\bigl( \mathbf{r} = (x,y,z) \bigl)</math>. More generally however, the Electrostatic Force vector exerted on any point charge <math>\bigl( q \bigl)</math> by a point source charge <math>\bigl( Q_{s} \bigl)</math> is related to the source charge's '''Electric Field''' vector by:

::<math>\mathbf{F}_{s} ( \mathbf{r} ) = q \mathbf{E}_{s} ( \mathbf{r} )</math>

This definition requires an understanding of the Electrostatic Force (Coulomb's Law), and its mathematical description. If you are not familiar with this yet, read over the [[Electric Force]] page and come back.

Since the Electric Force is defined as:

::<math>\mathbf{F}( \mathbf{r} ) = \frac{1}{4\pi\epsilon_{o}}\frac{q_{1} q_{2}}{r^{2}} \hat{\mathbf{r}}</math>, where
:::<math>\epsilon_{o}</math> is the permittivity of free space with a value of <math>8.854 \times 10^{-12} \frac{\text{C}^2}{\text{N} \cdot \text{m}^2}</math>
:::<math>q_{1}</math> and <math>q_{2}</math> are two different point charges
:::<math>r</math> is the distance between the two point charges, which can also be written as <math>|\mathbf{r}|</math>, the magnitude of the vector connecting the two charges' positions
:::<math>\hat{\mathbf{r}}</math> is the unit vector pointing from charge one to charge two, or from charge two to charge one, depending on whether the force on charge two or charge one is sought for.

The '''Electric Field''' of a source charge <math>Q_{s}</math> is:

::<math>
\begin{align}
\mathbf{E}_{s} ( \mathbf{r}) & = \frac{\mathbf{F}_{s} ( \mathbf{r} )}{q} \\
& = \frac{1}{4\pi\epsilon_{o}}\frac{Q_{s}}{r^{2}}\hat{\mathbf{r}}
\end{align}
</math>

As evidently based from the mathematical formula of an electric field for a point charge (which can also be applied to a spherically uniformly charged body), the electric field has an '''inverse-square relationship''' with the radius regarding the observation location. Essentially, as distance (radius) increases from the point charge's source, it's electric field at that observation location weakens by a corresponding factor of <math>{r^{-2}}</math>

Radially, the magnitude of a point charge's '''Electric Field''' looks something like this:

[[File:MagnitudeofEField.jpg|center|700px|thumb|<math>2 \times 10^{-15} \ \text{C}</math> charge's '''Electric Field''' magnitude as a function of radius.]]

A point charge's '''Electric Field''' is also related to its Electric Potential. If you are unfamiliar with the idea of electric potential, then review these pages ([[Electric Field and Electric Potential]] and [[Electric Potential]]) and come back.

A charge's '''Electric Field''' and Electric Potential <math>V</math> are related by:

::<math>V_{ab} = -\int_{\mathbf{b}}^{\mathbf{a}} \mathbf{E} \cdot d\mathbf{L}</math>, where
:::<math>V_{ab}</math> is the potential difference between points <math>\mathbf{a}</math> and <math>\mathbf{b}</math>
:::<math>\mathbf{E}</math> is the '''Electric Field'''
:::<math>d\mathbf{L}</math> is an infinitesimal length along the path between <math>\mathbf{a}</math> and <math>\mathbf{b}</math>

This relation is less useful for us unless we use a straight line approximation, such that:

::<math>
\begin{align}
V_{ab} & = -\mathbf{E} \cdot \Delta \mathbf{L} \\
& = - \bigl( E_{x}, E_{y}, E_{z} \bigl) \cdot \bigl( \Delta L_{x}, \Delta L_{y}, \Delta L_{z} \bigl) \\
& = - \bigl( E_{x}\Delta L_{x} + E_{y}\Delta L_{y} + E_{z}\Delta L_{z} \bigl) \\
\end{align}
</math>

This leads to:

::<math>\mathbf{E} (x,y,z) = - \biggl( \frac{\Delta V_{x}}{\Delta L_{x}}, \frac{\Delta V_{y}}{\Delta L_{y}}, \frac{\Delta V_{z}}{\Delta L_{z}} \biggl)</math>

By convention, the '''Electric Field''' due to a positive point charge always points away from itself, and the '''Electric Field''' of a negative point charge always points towards itself as shown below:
[[File:Posandnegefield.png|center]]

Opposite charges will attract each other, and like charges will repel each other, as shown below:
[[File:Multiplechargeefield.png|center]]

Lastly, the Principle of Superposition is directly applicable to finding the '''Electric Field''' due to multiple point source charges, using the a vector sum:

::<math>
\begin{align}
\mathbf{E}_{sum} (\mathbf{r}) & = \mathbf{E}_{1} + \mathbf{E}_{2} + \mathbf{E}_{3} + \cdots + \mathbf{E}_{N} \\
& = \sum_{1}^{N} \mathbf{E}_{n} \\
& = \sum_{1}^{N} \frac{1}{4 \pi \epsilon_{o}} \frac{Q_{s_{n}}}{r_{n}^{2}} \hat{\mathbf{r}}_n
\end{align}
</math>
::*When using this, be careful to take note that the '''Electric Field''' of a negative charge points in the opposite direction as a positive charge.

*'''Critical Formulas:'''
**<math>\mathbf{E} ( \mathbf{r}) = \frac{\mathbf{F} ( \mathbf{r} )}{q}</math>
**<math>\mathbf{E} ( \mathbf{r}) = \frac{1}{4\pi\epsilon_{o}}\frac{Q}{r^{2}}\hat{\mathbf{r}}</math>
**<math>\mathbf{E} (x,y,z) = - \biggl( \frac{\Delta V_{x}}{\Delta L_{x}}, \frac{\Delta V_{y}}{\Delta L_{y}}, \frac{\Delta V_{z}}{\Delta L_{z}} \biggl)</math>
**<math>\mathbf{E}_{sum} (\mathbf{r}) = \sum_{1}^{N} \frac{1}{4 \pi \epsilon_{o}} \frac{Q_{s_{n}}}{r_{n}^{2}} \hat{\mathbf{r}}_n</math>

[[File:NormalEField.png|right|250px|thumb|Normal view of simulated electric field]]

The '''Electric Field''' of a dipole perpendicular to axis is:
::<math>
\begin{align}
\rho = qs\rho (hat)
\end{align}
</math>

*p is the unit vector from the - to the + of the dipole

*q is the magnitude of the charge of one side of the dipole

*s is the separation distance between the charges that make up the dipole

When observation location X is perpendicular to the center of the dipole (i.e. vector r runs through the middle of the dipole at s/2), then you can use a special formula to calculate the E-field of the dipole at that location.

::<math>
\begin{align}
E_{dipole\perp} = \frac{1}{4 \pi \epsilon_{o}} \frac{q}{|r|^2 + (\frac{S}{2})^2} (\hat{\mathbf{r}}_{+} - \hat{\mathbf{r}}_{-})
\end{align}
</math>

::<math>
\begin{align}
E_{dipole\perp} = \frac{1}{4 \pi \epsilon_{o}}
\frac{-p}{(|r|^2 + (\frac{S}{2})^2)^\frac{3}{2}}
\end{align}
</math>

Special Case:
When r >> s (when vector between observation location and dipole is much greater than the separation distance of the dipole)

In this case, s becomes negligible, and the equation becomes

::<math>
\begin{align}
E_{dipole\perp} = \frac{1}{4 \pi \epsilon_{o}}
\frac{qs}{r^3}
\end{align}
</math>

===A Computational Model===

###--Create Electric Field Lines of a Positive Charge at the Origin--###
#==============================================================#
#---Import statements for VPython---#
from __future__ import division
from visual import *
#---Import function used to find combinations---#
from itertools import combinations
#==============================================================#
#---Create scene---#
scene.center = vector(0,0,0) #-Position of source charge-#
scene.height = 800 #-Set height of frame of scene-#
scene.width = 800 #-Set width of frame of scene-#
scene.range = 4 #-Set range of scene-#
scene.userzoom = 1 #-Allow user to zoom in/out: CTRL & move in/out on trackpad-#
scene.userspin = 1 #-Allow user to rotate camera angle: SHIFT & OPTION & move around on track pad-#
#==============================================================#
#---Specify point charge attributes---#
sourceCharge = 3*10**(-11) #-Coulombs of charge-#
sourcePos = vector(0,0,0) #-Position of source charge-#
###--Modeling source point charge as a sphere with radius 0.1 meters--###
sourceObj = sphere(pos = sourcePos, radius = 0.1, color = color.cyan)
#==============================================================#
#---Set range (0 to 3) and possible inputs for the coordinates (0.5 step)---#
###--Many of the same number included to allow for combinations such as (1,1,1).
#The itertools.combinations function will only use each element of the...
#list once, starting from the beginning.
#Repeating each coordinate many times with intermixing, grants...
[[File:CenteredAndDistantEField.png|right|250px|thumb|Distant view of simulated electric field]]
#all combinations of points, with repeats however.
#Later, a for loop will be used to eliminate repeats.
#This can be optimized later if need be.---------------###
posXYZ = [0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3]
#==============================================================#
[[File:RotatedAndZoomedInEField.png|right|250px|thumb|Rotated and zoomed in view of simulated electric field]]
#---Create combinations of points (x,y,z) for later use---#
###--prelimPoints will be a list of tuples of tuples--##
#ie: [((,,),(,,),(,,),(,,)) , ((,,),(,,)) ,..., ((,,),(,,))]
prelimPoints = [tuple(combinations(posXYZ, 3))]
###--Pull the points out of the grouping tuples and add them to a...
#new list alphaPoints------------------------###
alphaPoints = []
for groupingTuple in prelimPoints:
for XYZ in groupingTuple:
if XYZ not in alphaPoints: #-Check for repeat (x,y,z)-#
alphaPoints.append(XYZ)
##--The negative of this tuple may not be in the combinations:
#check to see-------------##
first = -XYZ[0]
second = -XYZ[1]
third = -XYZ[2]
negXYZ = (first, second, third)
if negXYZ not in alphaPoints:
alphaPoints.append(negXYZ)
##--Swap x and z coordinates for futher combination checking--##
first = XYZ[2]
second = XYZ[1]
third = XYZ[0]
reverseXYZ = (first, second, third)
if reverseXYZ not in alphaPoints:
alphaPoints.append(reverseXYZ)
##--The negative of the x and z coordinate swap may not be in...
#the combinations: check to see---------##
first = -XYZ[2]
second = -XYZ[1]
third = -XYZ[0]
reverseXYZneg = (first, second, third)
if reverseXYZneg not in alphaPoints:
alphaPoints.append(reverseXYZneg)
##--Make x [3], y [0], and z [1] to check for more combinations--##
first = XYZ[1]
second = XYZ[2]
third = XYZ[0]
shiftedXYZ = (first, second, third)
if shiftedXYZ not in alphaPoints:
alphaPoints.append(shiftedXYZ)
##--The negative of the shifted XYZ may not be in the combinations:
#check to see---------------##
first = -XYZ[1]
second = -XYZ[2]
third = -XYZ[0]
shiftedXYZneg = (first, second, third)
if shiftedXYZneg not in alphaPoints:
alphaPoints.append(shiftedXYZneg)
###--------This should be enough recombining---------###
#================================================================#
[[File:SideAngleAndTopViewEField.png|right|250px|thumb|Rotated top view of simulated electric field]]
#---Create a new list of tuples that contain the points, magnitude,...
#and direction (betaPoints)-----------#
#ie: [((x,y,z), mag((x,y,z)), norm((x,y,z))),...]
betaPoints = []
for XYZ in alphaPoints:
Mag = mag(XYZ)
Dir = norm(XYZ)
betaPoints.append((XYZ, Mag, Dir))
#================================================================#
#---Sort the tuples based on their magnitudes from least to greatest...
#using sorted().
#key = lamda x: x[1] tells the sorted function to sort the tuples...
#based on their second component...their magnitudes--------#
charliePoints = sorted(betaPoints, key = lambda x: x[1])
#================================================================#
#---Calculate parts of electric field equation:
#E = 1/(4*pi*epsilon0) * Q/(magnitude)**2
epsilonO = 8.854*(10**(-12)) #-N*(m/C)**2-#
k = 1/(4*pi*(epsilonO)) #-N*(m/C)**2-#
chargeContri = k*sourceCharge #-N*(m**2/C)-#
#================================================================#
#---Loop through points and find mag of electric field:
#add it to a new list with the existing tuple info-------#
deltaPoints = []
for XYZ in charliePoints:
try: ###-Avoid divide by 0 error in (x,y,z) = (0,0,0)-###
magEfield = chargeContri*(1/(XYZ[1])**2)
except:
magEfield = 0
tupEfield = (XYZ[0], XYZ[1], XYZ[2], magEfield)
deltaPoints.append(tupEfield)
#================================================================#
[[File:SIdeAngleAndSideViewEField.png|right|250px|thumb|Side angle of simulated electric field]]
#---Loop through points and create an arrow at that point proportional in...
#length to the magnitude of the electric field there.
#Also, the arrow points in the direction of the electric field there.
#Color coding is based on 0.25 meter increments:
#stronger field = redder; weaker field = blue
for XYZ in deltaPoints:
if XYZ[1] <= 0.25:
lengthP = XYZ[3]*0.5
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.000, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 0.5:
lengthP = XYZ[3]*0.7
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.200, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 1:
lengthP = XYZ[3]*0.9
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.300, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 1.25:
lengthP = XYZ[3]*1.1
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.400, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 1.5:
lengthP = XYZ[3]*1.3
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.500, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 1.75:
lengthP = XYZ[3]*1.5
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.600, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2:
lengthP = XYZ[3]*1.7
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.700, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2.25:
lengthP = XYZ[3]*1.9
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.800, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2.5:
lengthP = XYZ[3]*2.1
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.900, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2.75:
lengthP = XYZ[3]*2.3
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 1.000, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
else:
lengthP = XYZ[3]*2.5
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = color.blue,
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)

* This link [https://phet.colorado.edu/en/simulation/charges-and-fields Charges and Fields] provides a PhET simulation of '''Electric Fields'''. Play with it!

* Or if you prefer something with more action, explore this [https://phet.colorado.edu/sims/electric-hockey/electric-hockey Hockey Game] to gain a deeper visual understanding of electric fields and their effects on charges

==Examples==
===Simple===
*''Question'':
::In the following figure, the red circles represent positive point charges, and the blue circles represent negative point charges. If the yellow arrows are meant to represent the '''Electric Field''' due to each point charge, '''''which field(s) and charge(s) are correctly matched?''''' (Only take into account direction)

[[File:ElectricFieldSimpleExample.png|600px|center]]

*''Solution'':
::Since '''Electric Field''' lines always point away from a positive point charge, Option (C.) cannot be correct. Likewise, '''Electric Field''' lines always point towards a negative charge. Therefore, Option (A.) is also incorrect.
::Option (B.) shows a positive charge with an '''Electric Field''' pointing radially outwards. This is correct. Option (D.) shows a negative charge with an '''Electric Field''' pointing radially inwards. This is also correct.
:::'''Answer:''' Options (B.) & (D.)

===Middling===
*''Question'':
:: Four point charges <math>\big(q_{1}, q_{2}, q_{3}, \text{and} \ q_{4} \big)</math>, are each located at a distance <math>d</math> along either the <math>x</math> or <math>y</math> axes, as shown in the figure below.
:*'''A.)''' '''''What is the net Electric Field at the origin?'''''
:*'''B.)''' '''''If <math>\ |q_{3}| = |q_{1}| \ \text{and} \ |q_{4}| = |q_{2}|</math> what does the Electric Field at the origin reduce to?'''''
[[File:ElectricFieldMiddlingExample.png|600px|center]]

*''Solution'':
:*'''A.)''' To find the net '''Electric Field''' at the origin, we must first find the '''Electric Field''' due to each charge at the origin.
::*Starting with <math>q_{1}</math>, its general '''Electric Field''' can be described as:
::::<math>\mathbf{E}_{1} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{1}|}{r_{1}^2} \hat{\mathbf{r}}_{1}</math>
::::<math>r_{1}</math> is measured relative to the location of <math>q_{1}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{1}</math>, which is along the y-axis. Since it is a positive charge, its '''Electric Field''' at the origin will point "down" the y-axis (away from the charge):
::::<math>\mathbf{E}_{1} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{1}|}{d^2} (-\mathbf{j})</math> where <math>\mathbf{j}</math> is the unit vector in the y-direction.

::*For <math>q_{2}</math> we have:
::::<math>\mathbf{E}_{2} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{2}|}{r_{2}^2} \hat{\mathbf{r}}_{2}</math>
::::<math>r_{2}</math> is measured relative to the location of <math>q_{2}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{2}</math>, which is along the x-axis. Since it is a positive charge, its '''Electric Field''' at the origin will point to the left (away from the charge):
::::<math>\mathbf{E}_{2} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{2}|}{d^2} (-\mathbf{i})</math> where <math>\mathbf{i}</math> is the unit vector in the x-direction.

::*For <math>q_{3}</math> we have:
::::<math>\mathbf{E}_{3} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{3}|}{r_{3}^2} \hat{\mathbf{r}}_{3}</math>
::::<math>r_{3}</math> is measured relative to the location of <math>q_{3}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{3}</math>, which is along the y-axis. Since it is a negative charge, its '''Electric Field''' at the origin will point "down" the y-axis (towards the charge):
::::<math>\mathbf{E}_{3} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{3}|}{d^2} (-\mathbf{j})</math> where <math>\mathbf{j}</math> is the same unit vector in the y-direction from earlier.

::*For <math>q_{4}</math> the electric field is:
::::<math>\mathbf{E}_{4} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{4}|}{r_{4}^2} \hat{\mathbf{r}}_{4}</math>
::::<math>r_{4}</math> is measured relative to the location of <math>q_{4}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{4}</math>, which is along the x-axis. Since it is a positive charge, its '''Electric Field''' at the origin will point to the right (away from the charge):
::::<math>\mathbf{E}_{4} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{4}|}{d^2} (\mathbf{i})</math> where <math>\mathbf{i}</math> is the same unit vector in the x-direction from earlier.

::Now that we have the four '''Electric Fields''' present at the origin, we can use the Principle of Superposition to find the '''net''' '''Electric Field''' at the origin:

:::<math>
\begin{align}
\mathbf{E}_{net} &= \mathbf{E}_{1} + \mathbf{E}_{2} + \mathbf{E}_{3} + \mathbf{E}_{4} \\
&= \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{1}|}{d^2} (-\mathbf{j}) + \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{2}|}{d^2} (-\mathbf{i}) + \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{3}|}{d^2} (-\mathbf{j}) + \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{4}|}{d^2} (\mathbf{i}) \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ -|q_{1}| \mathbf{j} -|q_{2}| \mathbf{i} -|q_{3}| \mathbf{j} + |q_{4}| \mathbf{i} \Big] \\
\mathbf{E}_{net} &= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{4}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{3}| \big)\mathbf{j} \Big]
\end{align}
</math>

[[File:ElectricFieldMiddlingExampleAnswer.png|400px|right|thumb|Part '''(B)''' answer]]

:*'''B.)''' We will simply plug in the specified values into our answer from '''(A)''':
:::<math>
\begin{align}
\mathbf{E}_{net} &= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{4}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{3}| \big)\mathbf{j} \Big] \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{2}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{1}| \big)\mathbf{j} \Big] \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ 0 \mathbf{i} - 2|q_{1}| \mathbf{j} \Big] \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ - 2|q_{1}| \mathbf{j} \Big] \\
\mathbf{E}_{net} &= - \frac{1}{2 \pi \epsilon_{0} d^{2}} |q_{1}| \mathbf{j} \\
\end{align}
</math>

:::'''Answer:'''
:::*'''A.)''' <math>\mathbf{E}_{net} = \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{4}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{3}| \big)\mathbf{j} \Big]</math>
:::*'''B.)''' <math>\mathbf{E}_{net} = - \frac{1}{2 \pi \epsilon_{0} d^{2}} |q_{1}| \mathbf{j}</math>

===Difficult===
*''Question'':
::A ring of evenly distributed charge of radius <math>a</math> is centered on the origin in the xy-plane. The ring has a total charge <math>Q</math>. '''''Show that the Electric Field due to this ring is 0 at the origin.'''''
[[File:ElectricFieldDifficultExample.png|600px|center]]

*''Solution'':
::The '''Electric Field''' due to a point charge is given by:
:::<math>\mathbf{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{|Q|}{| \mathbf{r} - \mathbf{r}^{'} |^{2}} \frac{\mathbf{r} - \mathbf{r}^{'}}{| \mathbf{r} - \mathbf{r}^{'} |}</math>
:::This equation is equivalent to the formula presented in the [[Electric Field#A Mathematical Model | Mathematical Model]]. The reason it looks so different is due to a few assumptions in the mathematical model that we have stopped using:
:::# The source charge is located at the origin (our ring of charge is around the origin)
:::# The distance between the source charge and the observing location is simply expressed as a distance <math>r</math> (like in the [[Electric Field#Middling| Middling Example]]). Now, instead we will represent the distance as the magnitude of the difference in position between the source and observer <math>\big( | \mathbf{r} - \mathbf{r}^{'} | \big)</math>.
:::# Subsequently, our unit vector in the direction of the field <math>\big( \hat{\mathbf{r}} \big)</math> is not simply expressed as a typical unit vector (like in the middling example). It has now become the vector joining the source and observer divided by the magnitude of this same vector <math>\bigg( \frac{\mathbf{r} - \mathbf{r}^{'}}{| \mathbf{r} - \mathbf{r}^{'} |} \bigg) </math>.

::Another complication this problem presents is:
::::Where is the source charge?
:::To answer this, notice that the ring has an evenly distributed TOTAL charge <math>Q</math> and a radius <math>a</math>. Also, notice that the "source" position is constantly changing as you go around the ring. This issue makes it much more convenient to speak of the line charge DENSITY at a point along the ring instead of the TOTAL charge. This will allow us to treat the ring as many, many little source charges. The line charge density is simply the charge on the line divided by the length of that line (circumference), since the charge is evenly distributed about the ring:

::::<math>\rho_{L} = \frac{Q}{2 \pi a}</math>

:::This allows us to represent a differential amount of source charge as a product of the line charge density and a differential length:

::::<math>dQ = \rho_{L} dL</math>

:::The next question is: What is a differential length around the ring?
:::The differential length is a differential arc length <math>(s = r \theta)</math> around the circle dependent on the change in angle:

::::<math>dL = a d\theta</math>

:::Therefore:
::::<math>
\begin{align}
dQ &= \frac{Q}{2 \pi a} a d\theta \\
&= \frac{Q}{2 \pi} d\theta \\
\end{align}
</math>

:::Now we can sum each of these differential source charge's contribution to the '''Electric Field''' at the origin using an integral:
::::<math>\mathbf{E} = \int \frac{1}{4 \pi \epsilon_{0}} \frac{\frac{Q}{2 \pi} d\theta}{| \mathbf{r} - \mathbf{r}^{'} |^{2}} \frac{\mathbf{r} - \mathbf{r}^{'}}{| \mathbf{r} - \mathbf{r}^{'} |}</math>

:::The only things left to find are the generic source position (a vector that can describe the position of each differential source charge along the ring) and the observer location. The observer location is given to us; the origin:
::::<math>\mathbf{r} = 0\mathbf{i} +0\mathbf{j} + 0\mathbf{k}</math>

:::The source position is easiest to describe as a radius from the origin (polar coordinates):
::::<math>\mathbf{r}^{'} = a \hat{ \mathbf{a}}_{r}</math> where <math>\hat{\mathbf{a}}_{r}</math> is a unit vector in the radial direction

:::Therefore:
::::<math>
\begin{align}
\mathbf{r} - \mathbf{r}^{'} &= \big( 0\mathbf{i} +0\mathbf{j} + 0\mathbf{k} \big) - \big( a\hat{ \mathbf{a}}_{r} \big) \\
&= -a\hat{ \mathbf{a}}_{r} \\

|\mathbf{r} - \mathbf{r}^{'}| &= \sqrt{(-a)^{2}} \\
&= a \\
\end{align}
</math>

:::Plugging these into the '''Electric Field''' integral gives:
::::<math>
\begin{align}
\mathbf{E} &= \int \frac{1}{4 \pi \epsilon_{0}} \frac{\frac{Q}{2 \pi} d\theta}{a^2} \frac{-a \hat{ \mathbf{a}}_{r}}{a} \\
&= - \int \frac{1}{8 {\pi}^{2} \epsilon_{0}} \frac{Q}{a^2} \hat{ \mathbf{a}}_{r} d\theta \\
&= - \frac{Q}{a^{2} 8 {\pi}^{2} \epsilon_{0}} \int \hat{ \mathbf{a}}_{r} d\theta \\
\end{align}
</math>

::*<math>\theta</math> is the angle from the x-axis.
::*To integrate over the entire ring, we set the bounds of <math>\theta</math> as <math>[0, 2 \pi)</math>.
::*Also, as of right now, the integral would not evaluate to 0. This is because <math>\hat{ \mathbf{a}}_{r}</math> has a hidden dependence on <math>\theta</math>:
::::<math>\hat{ \mathbf{a}}_{r} = \text{cos}( \theta) \mathbf{i} + \text{sin}( \theta) \mathbf{j}</math>

:::Plugging this information in gives:
::::<math>
\begin{alignat}{3}
\mathbf{E} &= - \frac{Q}{a^{2} 8 {\pi}^{2} \epsilon_{0}} \int_{0}^{2 \pi} \big( \text{cos}( \theta) \mathbf{i} + \text{sin}( \theta) \mathbf{j} \big) d\theta \\
\int_{0}^{2 \pi} \text{cos}( \theta) \mathbf{i} \ d\theta &= 0 \\
\int_{0}^{2 \pi} \text{sin}( \theta) \mathbf{j} \ d\theta &= 0 \\
\end{alignat}
</math>

:::Therefore:
::::<math>\mathbf{E} = 0</math> at the origin.

==Connectedness==
The real world applications of electric fields are endless. Here are some:
[[File:electricmotor.jpg|400px|right]]
*'''Electric Motors:'''<br>
:Electric motors convert Electrical Energy into Mechanical Energy through '''Electric Fields'''. Whenever electric motors are turned on, '''Electric Fields''' are generated. This is because in order to turn an electric motor, an '''Electric Field''' must first be generated, which then generates a Magnetic Field, thus making the motor spin. Electric motors are used in cars, elevators, fans, refrigerators, and many more applications.

*'''Computers:'''<br>
:Computers use circuits, electric fans, and transistors to work. All of these use '''Electric Fields''' to push charge through a circuit, spin fans, and allow logic to be implemented in electronics.

*'''Painting:'''<br>
:'''Electric Fields''' are also used in some paintings. The '''Electric Field''' generates charges on the surface of the material being painted on, and an opposite charge is generated on the paint. Paint that touches the material sticks, and excess paint falls off to go back into the system.

*'''Cancer Treatment:'''<br>
:Recently, weak '''Electric Fields''' have been used to kill cancer cells. This treatment works best for brain and breast cancers, and it has no effect on normal cells. In lab and animal tests, this treatment killed cancer cells of every type tested; however, this is still a developing treatment.

*'''Military and Defense:'''<br>
:'''Electric Fields''' are commonly used in various weapons platforms. Weapons used to rely primarily on explosives; however, electric weapons use stored electrical energy to attack targets. There are two general types: directed-energy weapons (DEWs) and electromagnetic (EM) weapons. DEWs include lasers, radio frequency weapons, and more. EM weapons include rail guns, coil guns, etc. For example, rail guns use EM force to launch high velocity projectiles at a target. They work by using very high electrical currents to induce magnetic fields that accelerate a projectile to extremely high speeds (up to Mach 6).

==History==
'''Electric Fields''' are created by Electric charges. The original discovery of the Electric charge is not explicitly known, but in 1675 the esteemed chemist Robert Boyle, known for Boyle's Law, discovered the attraction and repulsion of certain particles in a vacuum. Almost 100 years later in the 18th century, the American Benjamin Franklin first coined the phrases 'positive' and 'negative' (later developed into proton and electron) for these particles with attractive and repulsive properties. Finally, in the 19th century Michael Faraday utilized his Electrolysis process to discover the discrete nature of Electric charge.

==See also==
The ability to understand '''Electric Fields''' helps set the basis for the introduction to [[Electric Force]] (as we discussed <math> \mathbf{F} = q\mathbf{E}</math> ). The introduction of Electric Force will attach the specific charge of the particles with the '''Electric Field''' that they produce, resulting in the Electric Force. Electric Force will lay the ground work for understanding the force that particles have in different systems and environments, and eventually lead to the introduction of [[Magnetic Force]].
The understanding of '''Electric Fields''' is a doorway into many various fields, only some of which will be covered in Physics 2212. The fundamental understanding of '''Electric Fields''' will prove to be very important further along when Magnetic Fields are introduced, as they share many qualities. The understanding of Electric and Magnetic Fields will be used throughout the semester to learn about various Electromagnetic concepts, and ultimately to understanding and apply Maxwell's Equations.
Please see related topics:

===Further reading===
*[[Electric Potential]]<br>
*[[Electric Force]]<br>
*[[Lorentz Force]]<br>
*[[Electric Polarization]]<br>
*[[Charged Ring]]<br>

===External links===
*[https://www.youtube.com/watch?v=EPIhhbwbCNc&list=PLX2gX-ftPVXUcMGbk1A7UbNtgadPsK5BD&index=9 A Youtube Playlist That Does A Great Job Going Step By Step And Reviewing Topics]

*[http://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Lines Further Review On Electric Field Lines.]

*[https://phet.colorado.edu/en/simulation/charges-and-fields Get A Better Understanding Of Fields Through Hands On Manipulation In PhET. This Can Be Very Helpful For Getting An Intuitive Understanding Of Fields.]

*[https://en.wikipedia.org/wiki/Electric_field Wikipedia Electric Field]

*[https://www.khanacademy.org/science/physics/electric-charge-electric-force-and-voltage/electric-field/v/electrostatics-part-2 Electric Field]

==References==
*[https://openstax.org/details/books/university-physics-volume-2 OpenStax Volume on Electricity and Magnetism]<br>
*Hayt & Buck 9th Edition Engineering Electromagnetics<br>
*Matter and Interactions<br>

Electric Field

2021-11-29T17:28:55Z

Jnation8:

'''Jared Nation (FALL 2021)'''
==The Main Idea==

Here, the concept of an '''Electric Field''' produced by a point charge is explored qualitatively and quantitatively through models, examples, and a simulation. An '''Electric Field''' is a useful concept to describe how any charged particle or collection of charged particles (positive / negative) would affect surrounding charge.

The '''Electric Field''' of a point charge is spherically symmetric, (same at all points of equal radius from the source. Hence, it is useful to speak of the electric field at a certain radius (not at a certain <math>(x,y,z)</math> position), which is explained in [[electric Field#A Mathematical Model| the mathematical model]].

Keep in mind, the electric field is a vector quantity, meaning it has a magnitude and direction. The SI units are N/C.

===A Mathematical Model===
The '''Electric Field''' vector <math>\bigl( \mathbf{E}_{s} \bigl)</math> of a point source charge <math>\bigl( Q_{s} \bigl)</math> gives the magnitude and direction of the Electrostatic Force vector <math>\bigl( \mathbf{F}_{s} \bigl)</math> exerted on a unit charge (<math>1</math> Coulomb) by <math>Q_{s}</math>, as a function of position <math>\bigl( \mathbf{r} = (x,y,z) \bigl)</math>. More generally however, the Electrostatic Force vector exerted on any point charge <math>\bigl( q \bigl)</math> by a point source charge <math>\bigl( Q_{s} \bigl)</math> is related to the source charge's '''Electric Field''' vector by:

::<math>\mathbf{F}_{s} ( \mathbf{r} ) = q \mathbf{E}_{s} ( \mathbf{r} )</math>

This definition requires an understanding of the Electrostatic Force (Coulomb's Law), and its mathematical description. If you are not familiar with this yet, read over the [[Electric Force]] page and come back.

Since the Electric Force is defined as:

::<math>\mathbf{F}( \mathbf{r} ) = \frac{1}{4\pi\epsilon_{o}}\frac{q_{1} q_{2}}{r^{2}} \hat{\mathbf{r}}</math>, where
:::<math>\epsilon_{o}</math> is the permittivity of free space with a value of <math>8.854 \times 10^{-12} \frac{\text{C}^2}{\text{N} \cdot \text{m}^2}</math>
:::<math>q_{1}</math> and <math>q_{2}</math> are two different point charges
:::<math>r</math> is the distance between the two point charges, which can also be written as <math>|\mathbf{r}|</math>, the magnitude of the vector connecting the two charges' positions
:::<math>\hat{\mathbf{r}}</math> is the unit vector pointing from charge one to charge two, or from charge two to charge one, depending on whether the force on charge two or charge one is sought for.

The '''Electric Field''' of a source charge <math>Q_{s}</math> is:

::<math>
\begin{align}
\mathbf{E}_{s} ( \mathbf{r}) & = \frac{\mathbf{F}_{s} ( \mathbf{r} )}{q} \\
& = \frac{1}{4\pi\epsilon_{o}}\frac{Q_{s}}{r^{2}}\hat{\mathbf{r}}
\end{align}
</math>

As evidently based from the mathematical formula of an electric field for a point charge (which can also be applied to a spherically uniformly charged body), the electric field has an '''inverse-square relationship''' with the radius regarding the observation location. Essentially, as distance (radius) increases from the point charge's source, it's electric field at that observation location weakens by a corresponding factor of <math>{r^{-2}}</math>

Radially, the magnitude of a point charge's '''Electric Field''' looks something like this:

[[File:MagnitudeofEField.jpg|center|700px|thumb|<math>2 \times 10^{-15} \ \text{C}</math> charge's '''Electric Field''' magnitude as a function of radius.]]

A point charge's '''Electric Field''' is also related to its Electric Potential. If you are unfamiliar with the idea of electric potential, then review these pages ([[Electric Field and Electric Potential]] and [[Electric Potential]]) and come back.

A charge's '''Electric Field''' and Electric Potential <math>V</math> are related by:

::<math>V_{ab} = -\int_{\mathbf{b}}^{\mathbf{a}} \mathbf{E} \cdot d\mathbf{L}</math>, where
:::<math>V_{ab}</math> is the potential difference between points <math>\mathbf{a}</math> and <math>\mathbf{b}</math>
:::<math>\mathbf{E}</math> is the '''Electric Field'''
:::<math>d\mathbf{L}</math> is an infinitesimal length along the path between <math>\mathbf{a}</math> and <math>\mathbf{b}</math>

This relation is less useful for us unless we use a straight line approximation, such that:

::<math>
\begin{align}
V_{ab} & = -\mathbf{E} \cdot \Delta \mathbf{L} \\
& = - \bigl( E_{x}, E_{y}, E_{z} \bigl) \cdot \bigl( \Delta L_{x}, \Delta L_{y}, \Delta L_{z} \bigl) \\
& = - \bigl( E_{x}\Delta L_{x} + E_{y}\Delta L_{y} + E_{z}\Delta L_{z} \bigl) \\
\end{align}
</math>

This leads to:

::<math>\mathbf{E} (x,y,z) = - \biggl( \frac{\Delta V_{x}}{\Delta L_{x}}, \frac{\Delta V_{y}}{\Delta L_{y}}, \frac{\Delta V_{z}}{\Delta L_{z}} \biggl)</math>

By convention, the '''Electric Field''' due to a positive point charge always points away from itself, and the '''Electric Field''' of a negative point charge always points towards itself as shown below:
[[File:Posandnegefield.png|center]]

Opposite charges will attract each other, and like charges will repel each other, as shown below:
[[File:Multiplechargeefield.png|center]]

Lastly, the Principle of Superposition is directly applicable to finding the '''Electric Field''' due to multiple point source charges, using the a vector sum:

::<math>
\begin{align}
\mathbf{E}_{sum} (\mathbf{r}) & = \mathbf{E}_{1} + \mathbf{E}_{2} + \mathbf{E}_{3} + \cdots + \mathbf{E}_{N} \\
& = \sum_{1}^{N} \mathbf{E}_{n} \\
& = \sum_{1}^{N} \frac{1}{4 \pi \epsilon_{o}} \frac{Q_{s_{n}}}{r_{n}^{2}} \hat{\mathbf{r}}_n
\end{align}
</math>
::*When using this, be careful to take note that the '''Electric Field''' of a negative charge points in the opposite direction as a positive charge.

*'''Critical Formulas:'''
**<math>\mathbf{E} ( \mathbf{r}) = \frac{\mathbf{F} ( \mathbf{r} )}{q}</math>
**<math>\mathbf{E} ( \mathbf{r}) = \frac{1}{4\pi\epsilon_{o}}\frac{Q}{r^{2}}\hat{\mathbf{r}}</math>
**<math>\mathbf{E} (x,y,z) = - \biggl( \frac{\Delta V_{x}}{\Delta L_{x}}, \frac{\Delta V_{y}}{\Delta L_{y}}, \frac{\Delta V_{z}}{\Delta L_{z}} \biggl)</math>
**<math>\mathbf{E}_{sum} (\mathbf{r}) = \sum_{1}^{N} \frac{1}{4 \pi \epsilon_{o}} \frac{Q_{s_{n}}}{r_{n}^{2}} \hat{\mathbf{r}}_n</math>

[[File:NormalEField.png|right|250px|thumb|Normal view of simulated electric field]]

The '''Electric Field''' of a dipole perpendicular to axis is:
::<math>
\begin{align}
\rho = qs\rho (hat)
\end{align}
</math>

*p is the unit vector from the - to the + of the dipole

*q is the magnitude of the charge of one side of the dipole

*s is the separation distance between the charges that make up the dipole

When observation location X is perpendicular to the center of the dipole (i.e. vector r runs through the middle of the dipole at s/2), then you can use a special formula to calculate the E-field of the dipole at that location.

::<math>
\begin{align}
E_{dipole\perp} = \frac{1}{4 \pi \epsilon_{o}} \frac{q}{|r|^2 + (\frac{S}{2})^2} (\hat{\mathbf{r}}_{+} - \hat{\mathbf{r}}_{-})
\end{align}
</math>

::<math>
\begin{align}
E_{dipole\perp} = \frac{1}{4 \pi \epsilon_{o}}
\frac{-p}{(|r|^2 + (\frac{S}{2})^2)^\frac{3}{2}}
\end{align}
</math>

Special Case:
When r >> s (when vector between observation location and dipole is much greater than the separation distance of the dipole)

In this case, s becomes negligible, and the equation becomes

::<math>
\begin{align}
E_{dipole\perp} = \frac{1}{4 \pi \epsilon_{o}}
\frac{qs}{r^3}
\end{align}
</math>

===A Computational Model===

###--Create Electric Field Lines of a Positive Charge at the Origin--###
#==============================================================#
#---Import statements for VPython---#
from __future__ import division
from visual import *
#---Import function used to find combinations---#
from itertools import combinations
#==============================================================#
#---Create scene---#
scene.center = vector(0,0,0) #-Position of source charge-#
scene.height = 800 #-Set height of frame of scene-#
scene.width = 800 #-Set width of frame of scene-#
scene.range = 4 #-Set range of scene-#
scene.userzoom = 1 #-Allow user to zoom in/out: CTRL & move in/out on trackpad-#
scene.userspin = 1 #-Allow user to rotate camera angle: SHIFT & OPTION & move around on track pad-#
#==============================================================#
#---Specify point charge attributes---#
sourceCharge = 3*10**(-11) #-Coulombs of charge-#
sourcePos = vector(0,0,0) #-Position of source charge-#
###--Modeling source point charge as a sphere with radius 0.1 meters--###
sourceObj = sphere(pos = sourcePos, radius = 0.1, color = color.cyan)
#==============================================================#
#---Set range (0 to 3) and possible inputs for the coordinates (0.5 step)---#
###--Many of the same number included to allow for combinations such as (1,1,1).
#The itertools.combinations function will only use each element of the...
#list once, starting from the beginning.
#Repeating each coordinate many times with intermixing, grants...
[[File:CenteredAndDistantEField.png|right|250px|thumb|Distant view of simulated electric field]]
#all combinations of points, with repeats however.
#Later, a for loop will be used to eliminate repeats.
#This can be optimized later if need be.---------------###
posXYZ = [0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3]
#==============================================================#
[[File:RotatedAndZoomedInEField.png|right|250px|thumb|Rotated and zoomed in view of simulated electric field]]
#---Create combinations of points (x,y,z) for later use---#
###--prelimPoints will be a list of tuples of tuples--##
#ie: [((,,),(,,),(,,),(,,)) , ((,,),(,,)) ,..., ((,,),(,,))]
prelimPoints = [tuple(combinations(posXYZ, 3))]
###--Pull the points out of the grouping tuples and add them to a...
#new list alphaPoints------------------------###
alphaPoints = []
for groupingTuple in prelimPoints:
for XYZ in groupingTuple:
if XYZ not in alphaPoints: #-Check for repeat (x,y,z)-#
alphaPoints.append(XYZ)
##--The negative of this tuple may not be in the combinations:
#check to see-------------##
first = -XYZ[0]
second = -XYZ[1]
third = -XYZ[2]
negXYZ = (first, second, third)
if negXYZ not in alphaPoints:
alphaPoints.append(negXYZ)
##--Swap x and z coordinates for futher combination checking--##
first = XYZ[2]
second = XYZ[1]
third = XYZ[0]
reverseXYZ = (first, second, third)
if reverseXYZ not in alphaPoints:
alphaPoints.append(reverseXYZ)
##--The negative of the x and z coordinate swap may not be in...
#the combinations: check to see---------##
first = -XYZ[2]
second = -XYZ[1]
third = -XYZ[0]
reverseXYZneg = (first, second, third)
if reverseXYZneg not in alphaPoints:
alphaPoints.append(reverseXYZneg)
##--Make x [3], y [0], and z [1] to check for more combinations--##
first = XYZ[1]
second = XYZ[2]
third = XYZ[0]
shiftedXYZ = (first, second, third)
if shiftedXYZ not in alphaPoints:
alphaPoints.append(shiftedXYZ)
##--The negative of the shifted XYZ may not be in the combinations:
#check to see---------------##
first = -XYZ[1]
second = -XYZ[2]
third = -XYZ[0]
shiftedXYZneg = (first, second, third)
if shiftedXYZneg not in alphaPoints:
alphaPoints.append(shiftedXYZneg)
###--------This should be enough recombining---------###
#================================================================#
[[File:SideAngleAndTopViewEField.png|right|250px|thumb|Rotated top view of simulated electric field]]
#---Create a new list of tuples that contain the points, magnitude,...
#and direction (betaPoints)-----------#
#ie: [((x,y,z), mag((x,y,z)), norm((x,y,z))),...]
betaPoints = []
for XYZ in alphaPoints:
Mag = mag(XYZ)
Dir = norm(XYZ)
betaPoints.append((XYZ, Mag, Dir))
#================================================================#
#---Sort the tuples based on their magnitudes from least to greatest...
#using sorted().
#key = lamda x: x[1] tells the sorted function to sort the tuples...
#based on their second component...their magnitudes--------#
charliePoints = sorted(betaPoints, key = lambda x: x[1])
#================================================================#
#---Calculate parts of electric field equation:
#E = 1/(4*pi*epsilon0) * Q/(magnitude)**2
epsilonO = 8.854*(10**(-12)) #-N*(m/C)**2-#
k = 1/(4*pi*(epsilonO)) #-N*(m/C)**2-#
chargeContri = k*sourceCharge #-N*(m**2/C)-#
#================================================================#
#---Loop through points and find mag of electric field:
#add it to a new list with the existing tuple info-------#
deltaPoints = []
for XYZ in charliePoints:
try: ###-Avoid divide by 0 error in (x,y,z) = (0,0,0)-###
magEfield = chargeContri*(1/(XYZ[1])**2)
except:
magEfield = 0
tupEfield = (XYZ[0], XYZ[1], XYZ[2], magEfield)
deltaPoints.append(tupEfield)
#================================================================#
[[File:SIdeAngleAndSideViewEField.png|right|250px|thumb|Side angle of simulated electric field]]
#---Loop through points and create an arrow at that point proportional in...
#length to the magnitude of the electric field there.
#Also, the arrow points in the direction of the electric field there.
#Color coding is based on 0.25 meter increments:
#stronger field = redder; weaker field = blue
for XYZ in deltaPoints:
if XYZ[1] <= 0.25:
lengthP = XYZ[3]*0.5
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.000, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 0.5:
lengthP = XYZ[3]*0.7
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.200, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 1:
lengthP = XYZ[3]*0.9
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.300, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 1.25:
lengthP = XYZ[3]*1.1
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.400, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 1.5:
lengthP = XYZ[3]*1.3
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.500, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 1.75:
lengthP = XYZ[3]*1.5
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.600, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2:
lengthP = XYZ[3]*1.7
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.700, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2.25:
lengthP = XYZ[3]*1.9
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.800, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2.5:
lengthP = XYZ[3]*2.1
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.900, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2.75:
lengthP = XYZ[3]*2.3
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 1.000, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
else:
lengthP = XYZ[3]*2.5
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = color.blue,
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)

* This link [https://phet.colorado.edu/en/simulation/charges-and-fields Charges and Fields] provides a PhET simulation of '''Electric Fields'''. Play with it!

* Or if you prefer something with more action, explore this [https://phet.colorado.edu/sims/electric-hockey/electric-hockey Hockey Game] to gain a deeper visual understanding of electric fields and their effects on charges

==Examples==
===Simple===
*''Question'':
::In the following figure, the red circles represent positive point charges, and the blue circles represent negative point charges. If the yellow arrows are meant to represent the '''Electric Field''' due to each point charge, '''''which field(s) and charge(s) are correctly matched?''''' (Only take into account direction)

[[File:ElectricFieldSimpleExample.png|600px|center]]

*''Solution'':
::Since '''Electric Field''' lines always point away from a positive point charge, Option (C.) cannot be correct. Likewise, '''Electric Field''' lines always point towards a negative charge. Therefore, Option (A.) is also incorrect.
::Option (B.) shows a positive charge with an '''Electric Field''' pointing radially outwards. This is correct. Option (D.) shows a negative charge with an '''Electric Field''' pointing radially inwards. This is also correct.
:::'''Answer:''' Options (B.) & (D.)

===Middling===
*''Question'':
:: Four point charges <math>\big(q_{1}, q_{2}, q_{3}, \text{and} \ q_{4} \big)</math>, are each located at a distance <math>d</math> along either the <math>x</math> or <math>y</math> axes, as shown in the figure below.
:*'''A.)''' '''''What is the net Electric Field at the origin?'''''
:*'''B.)''' '''''If <math>\ |q_{3}| = |q_{1}| \ \text{and} \ |q_{4}| = |q_{2}|</math> what does the Electric Field at the origin reduce to?'''''
[[File:ElectricFieldMiddlingExample.png|600px|center]]

*''Solution'':
:*'''A.)''' To find the net '''Electric Field''' at the origin, we must first find the '''Electric Field''' due to each charge at the origin.
::*Starting with <math>q_{1}</math>, its general '''Electric Field''' can be described as:
::::<math>\mathbf{E}_{1} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{1}|}{r_{1}^2} \hat{\mathbf{r}}_{1}</math>
::::<math>r_{1}</math> is measured relative to the location of <math>q_{1}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{1}</math>, which is along the y-axis. Since it is a positive charge, its '''Electric Field''' at the origin will point "down" the y-axis (away from the charge):
::::<math>\mathbf{E}_{1} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{1}|}{d^2} (-\mathbf{j})</math> where <math>\mathbf{j}</math> is the unit vector in the y-direction.

::*For <math>q_{2}</math> we have:
::::<math>\mathbf{E}_{2} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{2}|}{r_{2}^2} \hat{\mathbf{r}}_{2}</math>
::::<math>r_{2}</math> is measured relative to the location of <math>q_{2}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{2}</math>, which is along the x-axis. Since it is a positive charge, its '''Electric Field''' at the origin will point to the left (away from the charge):
::::<math>\mathbf{E}_{2} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{2}|}{d^2} (-\mathbf{i})</math> where <math>\mathbf{i}</math> is the unit vector in the x-direction.

::*For <math>q_{3}</math> we have:
::::<math>\mathbf{E}_{3} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{3}|}{r_{3}^2} \hat{\mathbf{r}}_{3}</math>
::::<math>r_{3}</math> is measured relative to the location of <math>q_{3}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{3}</math>, which is along the y-axis. Since it is a negative charge, its '''Electric Field''' at the origin will point "down" the y-axis (towards the charge):
::::<math>\mathbf{E}_{3} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{3}|}{d^2} (-\mathbf{j})</math> where <math>\mathbf{j}</math> is the same unit vector in the y-direction from earlier.

::*For <math>q_{4}</math> the electric field is:
::::<math>\mathbf{E}_{4} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{4}|}{r_{4}^2} \hat{\mathbf{r}}_{4}</math>
::::<math>r_{4}</math> is measured relative to the location of <math>q_{4}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{4}</math>, which is along the x-axis. Since it is a positive charge, its '''Electric Field''' at the origin will point to the right (away from the charge):
::::<math>\mathbf{E}_{4} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{4}|}{d^2} (\mathbf{i})</math> where <math>\mathbf{i}</math> is the same unit vector in the x-direction from earlier.

::Now that we have the four '''Electric Fields''' present at the origin, we can use the Principle of Superposition to find the '''net''' '''Electric Field''' at the origin:

:::<math>
\begin{align}
\mathbf{E}_{net} &= \mathbf{E}_{1} + \mathbf{E}_{2} + \mathbf{E}_{3} + \mathbf{E}_{4} \\
&= \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{1}|}{d^2} (-\mathbf{j}) + \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{2}|}{d^2} (-\mathbf{i}) + \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{3}|}{d^2} (-\mathbf{j}) + \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{4}|}{d^2} (\mathbf{i}) \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ -|q_{1}| \mathbf{j} -|q_{2}| \mathbf{i} -|q_{3}| \mathbf{j} + |q_{4}| \mathbf{i} \Big] \\
\mathbf{E}_{net} &= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{4}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{3}| \big)\mathbf{j} \Big]
\end{align}
</math>

[[File:ElectricFieldMiddlingExampleAnswer.png|400px|right|thumb|Part '''(B)''' answer]]

:*'''B.)''' We will simply plug in the specified values into our answer from '''(A)''':
:::<math>
\begin{align}
\mathbf{E}_{net} &= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{4}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{3}| \big)\mathbf{j} \Big] \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{2}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{1}| \big)\mathbf{j} \Big] \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ 0 \mathbf{i} - 2|q_{1}| \mathbf{j} \Big] \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ - 2|q_{1}| \mathbf{j} \Big] \\
\mathbf{E}_{net} &= - \frac{1}{2 \pi \epsilon_{0} d^{2}} |q_{1}| \mathbf{j} \\
\end{align}
</math>

:::'''Answer:'''
:::*'''A.)''' <math>\mathbf{E}_{net} = \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{4}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{3}| \big)\mathbf{j} \Big]</math>
:::*'''B.)''' <math>\mathbf{E}_{net} = - \frac{1}{2 \pi \epsilon_{0} d^{2}} |q_{1}| \mathbf{j}</math>

===Difficult===
*''Question'':
::A ring of evenly distributed charge of radius <math>a</math> is centered on the origin in the xy-plane. The ring has a total charge <math>Q</math>. '''''Show that the Electric Field due to this ring is 0 at the origin.'''''
[[File:ElectricFieldDifficultExample.png|600px|center]]

*''Solution'':
::The '''Electric Field''' due to a point charge is given by:
:::<math>\mathbf{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{|Q|}{| \mathbf{r} - \mathbf{r}^{'} |^{2}} \frac{\mathbf{r} - \mathbf{r}^{'}}{| \mathbf{r} - \mathbf{r}^{'} |}</math>
:::This equation is equivalent to the formula presented in the [[Electric Field#A Mathematical Model | Mathematical Model]]. The reason it looks so different is due to a few assumptions in the mathematical model that we have stopped using:
:::# The source charge is located at the origin (our ring of charge is around the origin)
:::# The distance between the source charge and the observing location is simply expressed as a distance <math>r</math> (like in the [[Electric Field#Middling| Middling Example]]). Now, instead we will represent the distance as the magnitude of the difference in position between the source and observer <math>\big( | \mathbf{r} - \mathbf{r}^{'} | \big)</math>.
:::# Subsequently, our unit vector in the direction of the field <math>\big( \hat{\mathbf{r}} \big)</math> is not simply expressed as a typical unit vector (like in the middling example). It has now become the vector joining the source and observer divided by the magnitude of this same vector <math>\bigg( \frac{\mathbf{r} - \mathbf{r}^{'}}{| \mathbf{r} - \mathbf{r}^{'} |} \bigg) </math>.

::Another complication this problem presents is:
::::Where is the source charge?
:::To answer this, notice that the ring has an evenly distributed TOTAL charge <math>Q</math> and a radius <math>a</math>. Also, notice that the "source" position is constantly changing as you go around the ring. This issue makes it much more convenient to speak of the line charge DENSITY at a point along the ring instead of the TOTAL charge. This will allow us to treat the ring as many, many little source charges. The line charge density is simply the charge on the line divided by the length of that line (circumference), since the charge is evenly distributed about the ring:

::::<math>\rho_{L} = \frac{Q}{2 \pi a}</math>

:::This allows us to represent a differential amount of source charge as a product of the line charge density and a differential length:

::::<math>dQ = \rho_{L} dL</math>

:::The next question is: What is a differential length around the ring?
:::The differential length is a differential arc length <math>(s = r \theta)</math> around the circle dependent on the change in angle:

::::<math>dL = a d\theta</math>

:::Therefore:
::::<math>
\begin{align}
dQ &= \frac{Q}{2 \pi a} a d\theta \\
&= \frac{Q}{2 \pi} d\theta \\
\end{align}
</math>

:::Now we can sum each of these differential source charge's contribution to the '''Electric Field''' at the origin using an integral:
::::<math>\mathbf{E} = \int \frac{1}{4 \pi \epsilon_{0}} \frac{\frac{Q}{2 \pi} d\theta}{| \mathbf{r} - \mathbf{r}^{'} |^{2}} \frac{\mathbf{r} - \mathbf{r}^{'}}{| \mathbf{r} - \mathbf{r}^{'} |}</math>

:::The only things left to find are the generic source position (a vector that can describe the position of each differential source charge along the ring) and the observer location. The observer location is given to us; the origin:
::::<math>\mathbf{r} = 0\mathbf{i} +0\mathbf{j} + 0\mathbf{k}</math>

:::The source position is easiest to describe as a radius from the origin (polar coordinates):
::::<math>\mathbf{r}^{'} = a \hat{ \mathbf{a}}_{r}</math> where <math>\hat{\mathbf{a}}_{r}</math> is a unit vector in the radial direction

:::Therefore:
::::<math>
\begin{align}
\mathbf{r} - \mathbf{r}^{'} &= \big( 0\mathbf{i} +0\mathbf{j} + 0\mathbf{k} \big) - \big( a\hat{ \mathbf{a}}_{r} \big) \\
&= -a\hat{ \mathbf{a}}_{r} \\

|\mathbf{r} - \mathbf{r}^{'}| &= \sqrt{(-a)^{2}} \\
&= a \\
\end{align}
</math>

:::Plugging these into the '''Electric Field''' integral gives:
::::<math>
\begin{align}
\mathbf{E} &= \int \frac{1}{4 \pi \epsilon_{0}} \frac{\frac{Q}{2 \pi} d\theta}{a^2} \frac{-a \hat{ \mathbf{a}}_{r}}{a} \\
&= - \int \frac{1}{8 {\pi}^{2} \epsilon_{0}} \frac{Q}{a^2} \hat{ \mathbf{a}}_{r} d\theta \\
&= - \frac{Q}{a^{2} 8 {\pi}^{2} \epsilon_{0}} \int \hat{ \mathbf{a}}_{r} d\theta \\
\end{align}
</math>

::*<math>\theta</math> is the angle from the x-axis.
::*To integrate over the entire ring, we set the bounds of <math>\theta</math> as <math>[0, 2 \pi)</math>.
::*Also, as of right now, the integral would not evaluate to 0. This is because <math>\hat{ \mathbf{a}}_{r}</math> has a hidden dependence on <math>\theta</math>:
::::<math>\hat{ \mathbf{a}}_{r} = \text{cos}( \theta) \mathbf{i} + \text{sin}( \theta) \mathbf{j}</math>

:::Plugging this information in gives:
::::<math>
\begin{alignat}{3}
\mathbf{E} &= - \frac{Q}{a^{2} 8 {\pi}^{2} \epsilon_{0}} \int_{0}^{2 \pi} \big( \text{cos}( \theta) \mathbf{i} + \text{sin}( \theta) \mathbf{j} \big) d\theta \\
\int_{0}^{2 \pi} \text{cos}( \theta) \mathbf{i} \ d\theta &= 0 \\
\int_{0}^{2 \pi} \text{sin}( \theta) \mathbf{j} \ d\theta &= 0 \\
\end{alignat}
</math>

:::Therefore:
::::<math>\mathbf{E} = 0</math> at the origin.

==Connectedness==
The real world applications of electric fields are endless. Here are some:
[[File:electricmotor.jpg|400px|right]]
*'''Electric Motors:'''<br>
:Electric motors convert Electrical Energy into Mechanical Energy through '''Electric Fields'''. Whenever electric motors are turned on, '''Electric Fields''' are generated. This is because in order to turn an electric motor, an '''Electric Field''' must first be generated, which then generates a Magnetic Field, thus making the motor spin. Electric motors are used in cars, elevators, fans, refrigerators, and many more applications.

*'''Computers:'''<br>
:Computers use circuits, electric fans, and transistors to work. All of these use '''Electric Fields''' to push charge through a circuit, spin fans, and allow logic to be implemented in electronics.

*'''Painting:'''<br>
:'''Electric Fields''' are also used in some paintings. The '''Electric Field''' generates charges on the surface of the material being painted on, and an opposite charge is generated on the paint. Paint that touches the material sticks, and excess paint falls off to go back into the system.

*'''Cancer Treatment:'''<br>
:Recently, weak '''Electric Fields''' have been used to kill cancer cells. This treatment works best for brain and breast cancers, and it has no effect on normal cells. In lab and animal tests, this treatment killed cancer cells of every type tested; however, this is still a developing treatment.

*'''Military and Defense:'''<br>
:'''Electric Fields''' are commonly used in various weapons platforms. Weapons used to rely primarily on explosives; however, electric weapons use stored electrical energy to attack targets. There are two general types: directed-energy weapons (DEWs) and electromagnetic (EM) weapons. DEWs include lasers, radio frequency weapons, and more. EM weapons include rail guns, coil guns, etc. For example, rail guns use EM force to launch high velocity projectiles at a target. They work by using very high electrical currents to induce magnetic fields that accelerate a projectile to extremely high speeds (up to Mach 6).

==History==
'''Electric Fields''' are created by Electric charges. The original discovery of the Electric charge is not explicitly known, but in 1675 the esteemed chemist Robert Boyle, known for Boyle's Law, discovered the attraction and repulsion of certain particles in a vacuum. Almost 100 years later in the 18th century, the American Benjamin Franklin first coined the phrases 'positive' and 'negative' (later developed into proton and electron) for these particles with attractive and repulsive properties. Finally, in the 19th century Michael Faraday utilized his Electrolysis process to discover the discrete nature of Electric charge.

==See also==
The ability to understand '''Electric Fields''' helps set the basis for the introduction to [[Electric Force]] (as we discussed <math> \mathbf{F} = q\mathbf{E}</math> ). The introduction of Electric Force will attach the specific charge of the particles with the '''Electric Field''' that they produce, resulting in the Electric Force. Electric Force will lay the ground work for understanding the force that particles have in different systems and environments, and eventually lead to the introduction of [[Magnetic Force]].
The understanding of '''Electric Fields''' is a doorway into many various fields, only some of which will be covered in Physics 2212. The fundamental understanding of '''Electric Fields''' will prove to be very important further along when Magnetic Fields are introduced, as they share many qualities. The understanding of Electric and Magnetic Fields will be used throughout the semester to learn about various Electromagnetic concepts, and ultimately to understanding and apply Maxwell's Equations.
Please see related topics:

===Further reading===
*[[Electric Potential]]<br>
*[[Electric Force]]<br>
*[[Lorentz Force]]<br>
*[[Electric Polarization]]<br>
*[[Charged Ring]]<br>

===External links===
*[https://www.youtube.com/watch?v=EPIhhbwbCNc&list=PLX2gX-ftPVXUcMGbk1A7UbNtgadPsK5BD&index=9 A Youtube Playlist That Does A Great Job Going Step By Step And Reviewing Topics]

*[http://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Lines Further Review On Electric Field Lines.]

*[https://phet.colorado.edu/en/simulation/charges-and-fields Get A Better Understanding Of Fields Through Hands On Manipulation In PhET. This Can Be Very Helpful For Getting An Intuitive Understanding Of Fields.]

*[https://en.wikipedia.org/wiki/Electric_field Wikipedia Electric Field]

*[https://www.khanacademy.org/science/physics/electric-charge-electric-force-and-voltage/electric-field/v/electrostatics-part-2 Electric Field]

==References==
*[https://openstax.org/details/books/university-physics-volume-2 OpenStax Volume on Electricity and Magnetism]<br>
*Hayt & Buck 9th Edition Engineering Electromagnetics<br>
*Matter and Interactions<br>

Electric Field

2021-11-29T17:26:40Z

Jnation8:

==The Main Idea==

'''Jared Nation (FALL 2021)'''

Here, the concept of an '''Electric Field''' produced by a point charge is explored qualitatively and quantitatively through models, examples, and a simulation. An '''Electric Field''' is a useful concept to describe how any charged particle or collection of charged particles (positive / negative) would affect surrounding charge.

The '''Electric Field''' of a point charge is spherically symmetric, (same at all points of equal radius from the source. Hence, it is useful to speak of the electric field at a certain radius (not at a certain <math>(x,y,z)</math> position), which is explained in [[electric Field#A Mathematical Model| the mathematical model]].

Keep in mind, the electric field is a vector quantity, meaning it has a magnitude and direction. The SI units are N/C.

===A Mathematical Model===
The '''Electric Field''' vector <math>\bigl( \mathbf{E}_{s} \bigl)</math> of a point source charge <math>\bigl( Q_{s} \bigl)</math> gives the magnitude and direction of the Electrostatic Force vector <math>\bigl( \mathbf{F}_{s} \bigl)</math> exerted on a unit charge (<math>1</math> Coulomb) by <math>Q_{s}</math>, as a function of position <math>\bigl( \mathbf{r} = (x,y,z) \bigl)</math>. More generally however, the Electrostatic Force vector exerted on any point charge <math>\bigl( q \bigl)</math> by a point source charge <math>\bigl( Q_{s} \bigl)</math> is related to the source charge's '''Electric Field''' vector by:

::<math>\mathbf{F}_{s} ( \mathbf{r} ) = q \mathbf{E}_{s} ( \mathbf{r} )</math>

This definition requires an understanding of the Electrostatic Force (Coulomb's Law), and its mathematical description. If you are not familiar with this yet, read over the [[Electric Force]] page and come back.

Since the Electric Force is defined as:

::<math>\mathbf{F}( \mathbf{r} ) = \frac{1}{4\pi\epsilon_{o}}\frac{q_{1} q_{2}}{r^{2}} \hat{\mathbf{r}}</math>, where
:::<math>\epsilon_{o}</math> is the permittivity of free space with a value of <math>8.854 \times 10^{-12} \frac{\text{C}^2}{\text{N} \cdot \text{m}^2}</math>
:::<math>q_{1}</math> and <math>q_{2}</math> are two different point charges
:::<math>r</math> is the distance between the two point charges, which can also be written as <math>|\mathbf{r}|</math>, the magnitude of the vector connecting the two charges' positions
:::<math>\hat{\mathbf{r}}</math> is the unit vector pointing from charge one to charge two, or from charge two to charge one, depending on whether the force on charge two or charge one is sought for.

The '''Electric Field''' of a source charge <math>Q_{s}</math> is:

::<math>
\begin{align}
\mathbf{E}_{s} ( \mathbf{r}) & = \frac{\mathbf{F}_{s} ( \mathbf{r} )}{q} \\
& = \frac{1}{4\pi\epsilon_{o}}\frac{Q_{s}}{r^{2}}\hat{\mathbf{r}}
\end{align}
</math>

As evidently based from the mathematical formula of an electric field for a point charge (which can also be applied to a spherically uniformly charged body), the electric field has an '''inverse-square relationship''' with the radius regarding the observation location. Essentially, as distance (radius) increases from the point charge's source, it's electric field at that observation location weakens by a corresponding factor of <math>{r^{-2}}</math>

Radially, the magnitude of a point charge's '''Electric Field''' looks something like this:

[[File:MagnitudeofEField.jpg|center|700px|thumb|<math>2 \times 10^{-15} \ \text{C}</math> charge's '''Electric Field''' magnitude as a function of radius.]]

A point charge's '''Electric Field''' is also related to its Electric Potential. If you are unfamiliar with the idea of electric potential, then review these pages ([[Electric Field and Electric Potential]] and [[Electric Potential]]) and come back.

A charge's '''Electric Field''' and Electric Potential <math>V</math> are related by:

::<math>V_{ab} = -\int_{\mathbf{b}}^{\mathbf{a}} \mathbf{E} \cdot d\mathbf{L}</math>, where
:::<math>V_{ab}</math> is the potential difference between points <math>\mathbf{a}</math> and <math>\mathbf{b}</math>
:::<math>\mathbf{E}</math> is the '''Electric Field'''
:::<math>d\mathbf{L}</math> is an infinitesimal length along the path between <math>\mathbf{a}</math> and <math>\mathbf{b}</math>

This relation is less useful for us unless we use a straight line approximation, such that:

::<math>
\begin{align}
V_{ab} & = -\mathbf{E} \cdot \Delta \mathbf{L} \\
& = - \bigl( E_{x}, E_{y}, E_{z} \bigl) \cdot \bigl( \Delta L_{x}, \Delta L_{y}, \Delta L_{z} \bigl) \\
& = - \bigl( E_{x}\Delta L_{x} + E_{y}\Delta L_{y} + E_{z}\Delta L_{z} \bigl) \\
\end{align}
</math>

This leads to:

::<math>\mathbf{E} (x,y,z) = - \biggl( \frac{\Delta V_{x}}{\Delta L_{x}}, \frac{\Delta V_{y}}{\Delta L_{y}}, \frac{\Delta V_{z}}{\Delta L_{z}} \biggl)</math>

By convention, the '''Electric Field''' due to a positive point charge always points away from itself, and the '''Electric Field''' of a negative point charge always points towards itself as shown below:
[[File:Posandnegefield.png|center]]

Opposite charges will attract each other, and like charges will repel each other, as shown below:
[[File:Multiplechargeefield.png|center]]

Lastly, the Principle of Superposition is directly applicable to finding the '''Electric Field''' due to multiple point source charges, using the a vector sum:

::<math>
\begin{align}
\mathbf{E}_{sum} (\mathbf{r}) & = \mathbf{E}_{1} + \mathbf{E}_{2} + \mathbf{E}_{3} + \cdots + \mathbf{E}_{N} \\
& = \sum_{1}^{N} \mathbf{E}_{n} \\
& = \sum_{1}^{N} \frac{1}{4 \pi \epsilon_{o}} \frac{Q_{s_{n}}}{r_{n}^{2}} \hat{\mathbf{r}}_n
\end{align}
</math>
::*When using this, be careful to take note that the '''Electric Field''' of a negative charge points in the opposite direction as a positive charge.

*'''Critical Formulas:'''
**<math>\mathbf{E} ( \mathbf{r}) = \frac{\mathbf{F} ( \mathbf{r} )}{q}</math>
**<math>\mathbf{E} ( \mathbf{r}) = \frac{1}{4\pi\epsilon_{o}}\frac{Q}{r^{2}}\hat{\mathbf{r}}</math>
**<math>\mathbf{E} (x,y,z) = - \biggl( \frac{\Delta V_{x}}{\Delta L_{x}}, \frac{\Delta V_{y}}{\Delta L_{y}}, \frac{\Delta V_{z}}{\Delta L_{z}} \biggl)</math>
**<math>\mathbf{E}_{sum} (\mathbf{r}) = \sum_{1}^{N} \frac{1}{4 \pi \epsilon_{o}} \frac{Q_{s_{n}}}{r_{n}^{2}} \hat{\mathbf{r}}_n</math>

[[File:NormalEField.png|right|250px|thumb|Normal view of simulated electric field]]

The '''Electric Field''' of a dipole perpendicular to axis is:
::<math>
\begin{align}
\rho = qs\rho (hat)
\end{align}
</math>

*p is the unit vector from the - to the + of the dipole

*q is the magnitude of the charge of one side of the dipole

*s is the separation distance between the charges that make up the dipole

When observation location X is perpendicular to the center of the dipole (i.e. vector r runs through the middle of the dipole at s/2), then you can use a special formula to calculate the E-field of the dipole at that location.

::<math>
\begin{align}
E_{dipole\perp} = \frac{1}{4 \pi \epsilon_{o}} \frac{q}{|r|^2 + (\frac{S}{2})^2} (\hat{\mathbf{r}}_{+} - \hat{\mathbf{r}}_{-})
\end{align}
</math>

::<math>
\begin{align}
E_{dipole\perp} = \frac{1}{4 \pi \epsilon_{o}}
\frac{-p}{(|r|^2 + (\frac{S}{2})^2)^\frac{3}{2}}
\end{align}
</math>

Special Case:
When r >> s (when vector between observation location and dipole is much greater than the separation distance of the dipole)

In this case, s becomes negligible, and the equation becomes

::<math>
\begin{align}
E_{dipole\perp} = \frac{1}{4 \pi \epsilon_{o}}
\frac{qs}{r^3}
\end{align}
</math>

===A Computational Model===

###--Create Electric Field Lines of a Positive Charge at the Origin--###
#==============================================================#
#---Import statements for VPython---#
from __future__ import division
from visual import *
#---Import function used to find combinations---#
from itertools import combinations
#==============================================================#
#---Create scene---#
scene.center = vector(0,0,0) #-Position of source charge-#
scene.height = 800 #-Set height of frame of scene-#
scene.width = 800 #-Set width of frame of scene-#
scene.range = 4 #-Set range of scene-#
scene.userzoom = 1 #-Allow user to zoom in/out: CTRL & move in/out on trackpad-#
scene.userspin = 1 #-Allow user to rotate camera angle: SHIFT & OPTION & move around on track pad-#
#==============================================================#
#---Specify point charge attributes---#
sourceCharge = 3*10**(-11) #-Coulombs of charge-#
sourcePos = vector(0,0,0) #-Position of source charge-#
###--Modeling source point charge as a sphere with radius 0.1 meters--###
sourceObj = sphere(pos = sourcePos, radius = 0.1, color = color.cyan)
#==============================================================#
#---Set range (0 to 3) and possible inputs for the coordinates (0.5 step)---#
###--Many of the same number included to allow for combinations such as (1,1,1).
#The itertools.combinations function will only use each element of the...
#list once, starting from the beginning.
#Repeating each coordinate many times with intermixing, grants...
[[File:CenteredAndDistantEField.png|right|250px|thumb|Distant view of simulated electric field]]
#all combinations of points, with repeats however.
#Later, a for loop will be used to eliminate repeats.
#This can be optimized later if need be.---------------###
posXYZ = [0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3,
0, -0.5, 1, -1.5, 2, -2.5, 3,
0, 0.5, -1, 1.5, -2, 2.5, -3]
#==============================================================#
[[File:RotatedAndZoomedInEField.png|right|250px|thumb|Rotated and zoomed in view of simulated electric field]]
#---Create combinations of points (x,y,z) for later use---#
###--prelimPoints will be a list of tuples of tuples--##
#ie: [((,,),(,,),(,,),(,,)) , ((,,),(,,)) ,..., ((,,),(,,))]
prelimPoints = [tuple(combinations(posXYZ, 3))]
###--Pull the points out of the grouping tuples and add them to a...
#new list alphaPoints------------------------###
alphaPoints = []
for groupingTuple in prelimPoints:
for XYZ in groupingTuple:
if XYZ not in alphaPoints: #-Check for repeat (x,y,z)-#
alphaPoints.append(XYZ)
##--The negative of this tuple may not be in the combinations:
#check to see-------------##
first = -XYZ[0]
second = -XYZ[1]
third = -XYZ[2]
negXYZ = (first, second, third)
if negXYZ not in alphaPoints:
alphaPoints.append(negXYZ)
##--Swap x and z coordinates for futher combination checking--##
first = XYZ[2]
second = XYZ[1]
third = XYZ[0]
reverseXYZ = (first, second, third)
if reverseXYZ not in alphaPoints:
alphaPoints.append(reverseXYZ)
##--The negative of the x and z coordinate swap may not be in...
#the combinations: check to see---------##
first = -XYZ[2]
second = -XYZ[1]
third = -XYZ[0]
reverseXYZneg = (first, second, third)
if reverseXYZneg not in alphaPoints:
alphaPoints.append(reverseXYZneg)
##--Make x [3], y [0], and z [1] to check for more combinations--##
first = XYZ[1]
second = XYZ[2]
third = XYZ[0]
shiftedXYZ = (first, second, third)
if shiftedXYZ not in alphaPoints:
alphaPoints.append(shiftedXYZ)
##--The negative of the shifted XYZ may not be in the combinations:
#check to see---------------##
first = -XYZ[1]
second = -XYZ[2]
third = -XYZ[0]
shiftedXYZneg = (first, second, third)
if shiftedXYZneg not in alphaPoints:
alphaPoints.append(shiftedXYZneg)
###--------This should be enough recombining---------###
#================================================================#
[[File:SideAngleAndTopViewEField.png|right|250px|thumb|Rotated top view of simulated electric field]]
#---Create a new list of tuples that contain the points, magnitude,...
#and direction (betaPoints)-----------#
#ie: [((x,y,z), mag((x,y,z)), norm((x,y,z))),...]
betaPoints = []
for XYZ in alphaPoints:
Mag = mag(XYZ)
Dir = norm(XYZ)
betaPoints.append((XYZ, Mag, Dir))
#================================================================#
#---Sort the tuples based on their magnitudes from least to greatest...
#using sorted().
#key = lamda x: x[1] tells the sorted function to sort the tuples...
#based on their second component...their magnitudes--------#
charliePoints = sorted(betaPoints, key = lambda x: x[1])
#================================================================#
#---Calculate parts of electric field equation:
#E = 1/(4*pi*epsilon0) * Q/(magnitude)**2
epsilonO = 8.854*(10**(-12)) #-N*(m/C)**2-#
k = 1/(4*pi*(epsilonO)) #-N*(m/C)**2-#
chargeContri = k*sourceCharge #-N*(m**2/C)-#
#================================================================#
#---Loop through points and find mag of electric field:
#add it to a new list with the existing tuple info-------#
deltaPoints = []
for XYZ in charliePoints:
try: ###-Avoid divide by 0 error in (x,y,z) = (0,0,0)-###
magEfield = chargeContri*(1/(XYZ[1])**2)
except:
magEfield = 0
tupEfield = (XYZ[0], XYZ[1], XYZ[2], magEfield)
deltaPoints.append(tupEfield)
#================================================================#
[[File:SIdeAngleAndSideViewEField.png|right|250px|thumb|Side angle of simulated electric field]]
#---Loop through points and create an arrow at that point proportional in...
#length to the magnitude of the electric field there.
#Also, the arrow points in the direction of the electric field there.
#Color coding is based on 0.25 meter increments:
#stronger field = redder; weaker field = blue
for XYZ in deltaPoints:
if XYZ[1] <= 0.25:
lengthP = XYZ[3]*0.5
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.000, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 0.5:
lengthP = XYZ[3]*0.7
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.200, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 1:
lengthP = XYZ[3]*0.9
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.300, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 1.25:
lengthP = XYZ[3]*1.1
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.400, 0.000),
length = lengthP,
headwidth = lengthP*0.2,
headlength = lengthP*0.25)
elif XYZ[1] <= 1.5:
lengthP = XYZ[3]*1.3
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.500, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 1.75:
lengthP = XYZ[3]*1.5
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.600, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2:
lengthP = XYZ[3]*1.7
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.700, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2.25:
lengthP = XYZ[3]*1.9
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.800, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2.5:
lengthP = XYZ[3]*2.1
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 0.900, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
elif XYZ[1] <= 2.75:
lengthP = XYZ[3]*2.3
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = vector(1.000, 1.000, 0.000),
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)
else:
lengthP = XYZ[3]*2.5
arroW = arrow(pos=vector(XYZ[0]), axis=XYZ[2],
color = color.blue,
length = lengthP,
headwidth = lengthP*1,
headlength = lengthP*1)

* This link [https://phet.colorado.edu/en/simulation/charges-and-fields Charges and Fields] provides a PhET simulation of '''Electric Fields'''. Play with it!

* Or if you prefer something with more action, explore this [https://phet.colorado.edu/sims/electric-hockey/electric-hockey Hockey Game] to gain a deeper visual understanding of electric fields and their effects on charges

==Examples==
===Simple===
*''Question'':
::In the following figure, the red circles represent positive point charges, and the blue circles represent negative point charges. If the yellow arrows are meant to represent the '''Electric Field''' due to each point charge, '''''which field(s) and charge(s) are correctly matched?''''' (Only take into account direction)

[[File:ElectricFieldSimpleExample.png|600px|center]]

*''Solution'':
::Since '''Electric Field''' lines always point away from a positive point charge, Option (C.) cannot be correct. Likewise, '''Electric Field''' lines always point towards a negative charge. Therefore, Option (A.) is also incorrect.
::Option (B.) shows a positive charge with an '''Electric Field''' pointing radially outwards. This is correct. Option (D.) shows a negative charge with an '''Electric Field''' pointing radially inwards. This is also correct.
:::'''Answer:''' Options (B.) & (D.)

===Middling===
*''Question'':
:: Four point charges <math>\big(q_{1}, q_{2}, q_{3}, \text{and} \ q_{4} \big)</math>, are each located at a distance <math>d</math> along either the <math>x</math> or <math>y</math> axes, as shown in the figure below.
:*'''A.)''' '''''What is the net Electric Field at the origin?'''''
:*'''B.)''' '''''If <math>\ |q_{3}| = |q_{1}| \ \text{and} \ |q_{4}| = |q_{2}|</math> what does the Electric Field at the origin reduce to?'''''
[[File:ElectricFieldMiddlingExample.png|600px|center]]

*''Solution'':
:*'''A.)''' To find the net '''Electric Field''' at the origin, we must first find the '''Electric Field''' due to each charge at the origin.
::*Starting with <math>q_{1}</math>, its general '''Electric Field''' can be described as:
::::<math>\mathbf{E}_{1} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{1}|}{r_{1}^2} \hat{\mathbf{r}}_{1}</math>
::::<math>r_{1}</math> is measured relative to the location of <math>q_{1}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{1}</math>, which is along the y-axis. Since it is a positive charge, its '''Electric Field''' at the origin will point "down" the y-axis (away from the charge):
::::<math>\mathbf{E}_{1} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{1}|}{d^2} (-\mathbf{j})</math> where <math>\mathbf{j}</math> is the unit vector in the y-direction.

::*For <math>q_{2}</math> we have:
::::<math>\mathbf{E}_{2} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{2}|}{r_{2}^2} \hat{\mathbf{r}}_{2}</math>
::::<math>r_{2}</math> is measured relative to the location of <math>q_{2}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{2}</math>, which is along the x-axis. Since it is a positive charge, its '''Electric Field''' at the origin will point to the left (away from the charge):
::::<math>\mathbf{E}_{2} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{2}|}{d^2} (-\mathbf{i})</math> where <math>\mathbf{i}</math> is the unit vector in the x-direction.

::*For <math>q_{3}</math> we have:
::::<math>\mathbf{E}_{3} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{3}|}{r_{3}^2} \hat{\mathbf{r}}_{3}</math>
::::<math>r_{3}</math> is measured relative to the location of <math>q_{3}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{3}</math>, which is along the y-axis. Since it is a negative charge, its '''Electric Field''' at the origin will point "down" the y-axis (towards the charge):
::::<math>\mathbf{E}_{3} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{3}|}{d^2} (-\mathbf{j})</math> where <math>\mathbf{j}</math> is the same unit vector in the y-direction from earlier.

::*For <math>q_{4}</math> the electric field is:
::::<math>\mathbf{E}_{4} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{4}|}{r_{4}^2} \hat{\mathbf{r}}_{4}</math>
::::<math>r_{4}</math> is measured relative to the location of <math>q_{4}</math>.
::::The origin is a distance <math>d</math> away from <math>q_{4}</math>, which is along the x-axis. Since it is a positive charge, its '''Electric Field''' at the origin will point to the right (away from the charge):
::::<math>\mathbf{E}_{4} = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{4}|}{d^2} (\mathbf{i})</math> where <math>\mathbf{i}</math> is the same unit vector in the x-direction from earlier.

::Now that we have the four '''Electric Fields''' present at the origin, we can use the Principle of Superposition to find the '''net''' '''Electric Field''' at the origin:

:::<math>
\begin{align}
\mathbf{E}_{net} &= \mathbf{E}_{1} + \mathbf{E}_{2} + \mathbf{E}_{3} + \mathbf{E}_{4} \\
&= \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{1}|}{d^2} (-\mathbf{j}) + \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{2}|}{d^2} (-\mathbf{i}) + \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{3}|}{d^2} (-\mathbf{j}) + \frac{1}{4 \pi \epsilon_{0}} \frac{|q_{4}|}{d^2} (\mathbf{i}) \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ -|q_{1}| \mathbf{j} -|q_{2}| \mathbf{i} -|q_{3}| \mathbf{j} + |q_{4}| \mathbf{i} \Big] \\
\mathbf{E}_{net} &= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{4}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{3}| \big)\mathbf{j} \Big]
\end{align}
</math>

[[File:ElectricFieldMiddlingExampleAnswer.png|400px|right|thumb|Part '''(B)''' answer]]

:*'''B.)''' We will simply plug in the specified values into our answer from '''(A)''':
:::<math>
\begin{align}
\mathbf{E}_{net} &= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{4}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{3}| \big)\mathbf{j} \Big] \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{2}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{1}| \big)\mathbf{j} \Big] \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ 0 \mathbf{i} - 2|q_{1}| \mathbf{j} \Big] \\
&= \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ - 2|q_{1}| \mathbf{j} \Big] \\
\mathbf{E}_{net} &= - \frac{1}{2 \pi \epsilon_{0} d^{2}} |q_{1}| \mathbf{j} \\
\end{align}
</math>

:::'''Answer:'''
:::*'''A.)''' <math>\mathbf{E}_{net} = \frac{1}{4 \pi \epsilon_{0} d^{2}} \Big[ \big( |q_{4}| - |q_{2}| \big)\mathbf{i} - \big( |q_{1}| + |q_{3}| \big)\mathbf{j} \Big]</math>
:::*'''B.)''' <math>\mathbf{E}_{net} = - \frac{1}{2 \pi \epsilon_{0} d^{2}} |q_{1}| \mathbf{j}</math>

===Difficult===
*''Question'':
::A ring of evenly distributed charge of radius <math>a</math> is centered on the origin in the xy-plane. The ring has a total charge <math>Q</math>. '''''Show that the Electric Field due to this ring is 0 at the origin.'''''
[[File:ElectricFieldDifficultExample.png|600px|center]]

*''Solution'':
::The '''Electric Field''' due to a point charge is given by:
:::<math>\mathbf{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{|Q|}{| \mathbf{r} - \mathbf{r}^{'} |^{2}} \frac{\mathbf{r} - \mathbf{r}^{'}}{| \mathbf{r} - \mathbf{r}^{'} |}</math>
:::This equation is equivalent to the formula presented in the [[Electric Field#A Mathematical Model | Mathematical Model]]. The reason it looks so different is due to a few assumptions in the mathematical model that we have stopped using:
:::# The source charge is located at the origin (our ring of charge is around the origin)
:::# The distance between the source charge and the observing location is simply expressed as a distance <math>r</math> (like in the [[Electric Field#Middling| Middling Example]]). Now, instead we will represent the distance as the magnitude of the difference in position between the source and observer <math>\big( | \mathbf{r} - \mathbf{r}^{'} | \big)</math>.
:::# Subsequently, our unit vector in the direction of the field <math>\big( \hat{\mathbf{r}} \big)</math> is not simply expressed as a typical unit vector (like in the middling example). It has now become the vector joining the source and observer divided by the magnitude of this same vector <math>\bigg( \frac{\mathbf{r} - \mathbf{r}^{'}}{| \mathbf{r} - \mathbf{r}^{'} |} \bigg) </math>.

::Another complication this problem presents is:
::::Where is the source charge?
:::To answer this, notice that the ring has an evenly distributed TOTAL charge <math>Q</math> and a radius <math>a</math>. Also, notice that the "source" position is constantly changing as you go around the ring. This issue makes it much more convenient to speak of the line charge DENSITY at a point along the ring instead of the TOTAL charge. This will allow us to treat the ring as many, many little source charges. The line charge density is simply the charge on the line divided by the length of that line (circumference), since the charge is evenly distributed about the ring:

::::<math>\rho_{L} = \frac{Q}{2 \pi a}</math>

:::This allows us to represent a differential amount of source charge as a product of the line charge density and a differential length:

::::<math>dQ = \rho_{L} dL</math>

:::The next question is: What is a differential length around the ring?
:::The differential length is a differential arc length <math>(s = r \theta)</math> around the circle dependent on the change in angle:

::::<math>dL = a d\theta</math>

:::Therefore:
::::<math>
\begin{align}
dQ &= \frac{Q}{2 \pi a} a d\theta \\
&= \frac{Q}{2 \pi} d\theta \\
\end{align}
</math>

:::Now we can sum each of these differential source charge's contribution to the '''Electric Field''' at the origin using an integral:
::::<math>\mathbf{E} = \int \frac{1}{4 \pi \epsilon_{0}} \frac{\frac{Q}{2 \pi} d\theta}{| \mathbf{r} - \mathbf{r}^{'} |^{2}} \frac{\mathbf{r} - \mathbf{r}^{'}}{| \mathbf{r} - \mathbf{r}^{'} |}</math>

:::The only things left to find are the generic source position (a vector that can describe the position of each differential source charge along the ring) and the observer location. The observer location is given to us; the origin:
::::<math>\mathbf{r} = 0\mathbf{i} +0\mathbf{j} + 0\mathbf{k}</math>

:::The source position is easiest to describe as a radius from the origin (polar coordinates):
::::<math>\mathbf{r}^{'} = a \hat{ \mathbf{a}}_{r}</math> where <math>\hat{\mathbf{a}}_{r}</math> is a unit vector in the radial direction

:::Therefore:
::::<math>
\begin{align}
\mathbf{r} - \mathbf{r}^{'} &= \big( 0\mathbf{i} +0\mathbf{j} + 0\mathbf{k} \big) - \big( a\hat{ \mathbf{a}}_{r} \big) \\
&= -a\hat{ \mathbf{a}}_{r} \\

|\mathbf{r} - \mathbf{r}^{'}| &= \sqrt{(-a)^{2}} \\
&= a \\
\end{align}
</math>

:::Plugging these into the '''Electric Field''' integral gives:
::::<math>
\begin{align}
\mathbf{E} &= \int \frac{1}{4 \pi \epsilon_{0}} \frac{\frac{Q}{2 \pi} d\theta}{a^2} \frac{-a \hat{ \mathbf{a}}_{r}}{a} \\
&= - \int \frac{1}{8 {\pi}^{2} \epsilon_{0}} \frac{Q}{a^2} \hat{ \mathbf{a}}_{r} d\theta \\
&= - \frac{Q}{a^{2} 8 {\pi}^{2} \epsilon_{0}} \int \hat{ \mathbf{a}}_{r} d\theta \\
\end{align}
</math>

::*<math>\theta</math> is the angle from the x-axis.
::*To integrate over the entire ring, we set the bounds of <math>\theta</math> as <math>[0, 2 \pi)</math>.
::*Also, as of right now, the integral would not evaluate to 0. This is because <math>\hat{ \mathbf{a}}_{r}</math> has a hidden dependence on <math>\theta</math>:
::::<math>\hat{ \mathbf{a}}_{r} = \text{cos}( \theta) \mathbf{i} + \text{sin}( \theta) \mathbf{j}</math>

:::Plugging this information in gives:
::::<math>
\begin{alignat}{3}
\mathbf{E} &= - \frac{Q}{a^{2} 8 {\pi}^{2} \epsilon_{0}} \int_{0}^{2 \pi} \big( \text{cos}( \theta) \mathbf{i} + \text{sin}( \theta) \mathbf{j} \big) d\theta \\
\int_{0}^{2 \pi} \text{cos}( \theta) \mathbf{i} \ d\theta &= 0 \\
\int_{0}^{2 \pi} \text{sin}( \theta) \mathbf{j} \ d\theta &= 0 \\
\end{alignat}
</math>

:::Therefore:
::::<math>\mathbf{E} = 0</math> at the origin.

==Connectedness==
The real world applications of electric fields are endless. Here are some:
[[File:electricmotor.jpg|400px|right]]
*'''Electric Motors:'''<br>
:Electric motors convert Electrical Energy into Mechanical Energy through '''Electric Fields'''. Whenever electric motors are turned on, '''Electric Fields''' are generated. This is because in order to turn an electric motor, an '''Electric Field''' must first be generated, which then generates a Magnetic Field, thus making the motor spin. Electric motors are used in cars, elevators, fans, refrigerators, and many more applications.

*'''Computers:'''<br>
:Computers use circuits, electric fans, and transistors to work. All of these use '''Electric Fields''' to push charge through a circuit, spin fans, and allow logic to be implemented in electronics.

*'''Painting:'''<br>
:'''Electric Fields''' are also used in some paintings. The '''Electric Field''' generates charges on the surface of the material being painted on, and an opposite charge is generated on the paint. Paint that touches the material sticks, and excess paint falls off to go back into the system.

*'''Cancer Treatment:'''<br>
:Recently, weak '''Electric Fields''' have been used to kill cancer cells. This treatment works best for brain and breast cancers, and it has no effect on normal cells. In lab and animal tests, this treatment killed cancer cells of every type tested; however, this is still a developing treatment.

*'''Military and Defense:'''<br>
:'''Electric Fields''' are commonly used in various weapons platforms. Weapons used to rely primarily on explosives; however, electric weapons use stored electrical energy to attack targets. There are two general types: directed-energy weapons (DEWs) and electromagnetic (EM) weapons. DEWs include lasers, radio frequency weapons, and more. EM weapons include rail guns, coil guns, etc. For example, rail guns use EM force to launch high velocity projectiles at a target. They work by using very high electrical currents to induce magnetic fields that accelerate a projectile to extremely high speeds (up to Mach 6).

==History==
'''Electric Fields''' are created by Electric charges. The original discovery of the Electric charge is not explicitly known, but in 1675 the esteemed chemist Robert Boyle, known for Boyle's Law, discovered the attraction and repulsion of certain particles in a vacuum. Almost 100 years later in the 18th century, the American Benjamin Franklin first coined the phrases 'positive' and 'negative' (later developed into proton and electron) for these particles with attractive and repulsive properties. Finally, in the 19th century Michael Faraday utilized his Electrolysis process to discover the discrete nature of Electric charge.

==See also==
The ability to understand '''Electric Fields''' helps set the basis for the introduction to [[Electric Force]] (as we discussed <math> \mathbf{F} = q\mathbf{E}</math> ). The introduction of Electric Force will attach the specific charge of the particles with the '''Electric Field''' that they produce, resulting in the Electric Force. Electric Force will lay the ground work for understanding the force that particles have in different systems and environments, and eventually lead to the introduction of [[Magnetic Force]].
The understanding of '''Electric Fields''' is a doorway into many various fields, only some of which will be covered in Physics 2212. The fundamental understanding of '''Electric Fields''' will prove to be very important further along when Magnetic Fields are introduced, as they share many qualities. The understanding of Electric and Magnetic Fields will be used throughout the semester to learn about various Electromagnetic concepts, and ultimately to understanding and apply Maxwell's Equations.
Please see related topics:

===Further reading===
*[[Electric Potential]]<br>
*[[Electric Force]]<br>
*[[Lorentz Force]]<br>
*[[Electric Polarization]]<br>
*[[Charged Ring]]<br>

===External links===
*[https://www.youtube.com/watch?v=EPIhhbwbCNc&list=PLX2gX-ftPVXUcMGbk1A7UbNtgadPsK5BD&index=9 A Youtube Playlist That Does A Great Job Going Step By Step And Reviewing Topics]

*[http://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Lines Further Review On Electric Field Lines.]

*[https://phet.colorado.edu/en/simulation/charges-and-fields Get A Better Understanding Of Fields Through Hands On Manipulation In PhET. This Can Be Very Helpful For Getting An Intuitive Understanding Of Fields.]

*[https://en.wikipedia.org/wiki/Electric_field Wikipedia Electric Field]

*[https://www.khanacademy.org/science/physics/electric-charge-electric-force-and-voltage/electric-field/v/electrostatics-part-2 Electric Field]

==References==
*[https://openstax.org/details/books/university-physics-volume-2 OpenStax Volume on Electricity and Magnetism]<br>
*Hayt & Buck 9th Edition Engineering Electromagnetics<br>
*Matter and Interactions<br>

Escape Velocity

2021-11-29T17:22:49Z

Jnation8:

[[File:Voyager Path czech version.jpg|400px|thumb|right| A diagram showing the paths of Voyager 1 and 2.]]

Escape velocity is defined as the minimum velocity required for an object to escape the gravitational force of a large object. The sum of an object's kinetic energy and its gravitational potential energy is equal to zero. The gravitational potential energy is negative due to the fact that kinetic energy is always positive. The velocity of the object will be zero at infinite distance from the center of gravity. There is no net force on an object as it escapes and zero acceleration is perceived.

==The Main Idea==

The formula for escape velocity at a given distance from a body is calculated by the formula
:<math>v_e = \sqrt{\frac{2GM}{r}},</math>

where <math>G</math> is the universal [[gravitational constant]] (<math>G = 6.7\times 10^{-11} \text{N}\cdot\text{m}^2 / \text{kg}^2</math>), <math>M</math> is the mass of the large body to be escaped, and <math>r</math> the distance from the [[center of mass]] of the mass <math>M</math> to the object. This equation assumes there are no other forces acting on either body. As a side note, the escape velocity stated here could really be called escape speed due to the fact that the quantity is independent of direction. Notice that the equation does not include the mass of the orbiting body.

===A Mathematical Model===

When we model escape velocity, we consider the situation when an object's velocity takes it to a point an infinite distance away. The lowest possible escape velocity has a final speed of zero, and any speed higher results in a nonzero final speed. To derive the formula for the escape velocity, the energy principle is used, and we assume that the only two objects in our system are the orbiting body and the planet.

In our system, the energy principle states that

:<math>(K + U_g)_i = (K + U_g)_f \,</math>

When finding the minimum escape velocity, <math>K-f = 0 </math> because we take the final velocity to be zero, and <math>U_{gf} = 0 </math> because its final distance is expressed as infinity, therefore

:<math>(K + U_g)_i = \frac{1}{2}mv_e^2 + \frac{-GMm}{r} = 0</math>

Solving for <math>v_e</math> yields:

:<math>v_e = \sqrt{\frac{2GM}{r}}</math>

Note that because both the kinetic and potential energy terms contain a common factor <math>m</math>, the final escape velocity is independent of the mass of the orbiting body.

'''Bound versus unbound systems'''

When an object is orbiting a massive body, it can be in one of two states: bound and unbound. If the object is in a bound state, we see an elliptical trajectory, in which the orbiting body never escapes the gravitational influence of the more massive body. In an unbound state, however, we observe a parabolic or hyperbolic trajectory, in which the object is able to escape the gravitational influence of the orbiting body and escape to infinity. The diagram to the left shows an unbound system, in which the sum of the kinetic and potential energy of the orbiting body is greater then 0. As distance goes to infinity in this system, gravitational potential energy approaches zero, but the object retains a positive kinetic energy, and therefore a positive velocity. The image on the right shows a bound system, in which the sum of the kinetic and potential energy is negative. In this system, kinetic energy reaches zero at a specific maximum distance, at which point the object begins to fall back towards the massive body, never to escape.

[[File:Ediagram_MSPaint_UnboundSystemWithExtraEnergy.png|500px|thumb|left|The energy diagram of an unbound system, in which the object has excess kinetic energy. The vertical axis represents energy, while the horizontal axis represents distance. At exactly the escape velocity, the sum of <math>K</math> and <math>U</math> is exactly 0.]]
[[File:EDiagram_MSPaint_BoundSystem.png|500px|thumb|right|The energy diagram of a bound system, in which the object has insufficient kinetic energy to escape. The vertical axis represents energy, while the horizontal axis represents distance. ]]

<br clear=all>

===A Computational Model===

Here is simulation that can be used to experiment with different factors that affect escape velocity:https://phet.colorado.edu/en/simulation/gravity-and-orbits.

==Examples==

===Simple===

'''Question'''

Compute the escape velocity for Earth if its mass is <math>5.98 \times 10^{24} \text{kg}</math> and its radius is <math>6.37 \times 10^{6} \text{m}</math>.

'''Solution'''

<math>
v_e = \sqrt{\frac{2GM}{r}} \\
= \sqrt{\frac{2(6.7\times 10^{-11} \text{N}\cdot\text{m}^2 / \text{kg}^2)(5.98\times 10^{24} \text{kg})}{(6.37\times 10^6 \text{m})}}\\
= 1.12 \times 10^4 \text{m/s}
</math>

</div>

===Middling===

'''Question'''

If the escape velocity of a planet with the same mass as Earth is greater than the escape velocity of Earth, is the planet larger or smaller than Earth?

'''Solution'''

Given the formula: <math> v_e = \sqrt{\frac{2GM}{r}} </math>

And rearranging for radius: <math> r = \frac{2GM}{v^2_e} </math>

We can see that since the denominator of the faction is larger with a constant numerator, then the radius must be smaller. Therefore, the planet is smaller than Earth.

===Difficult===

'''Question'''

The radius of Jupiter is <math>71.5\times 10^6 \text{m}</math>, and its mass is <math>1900\times 10^{24} \text{kg}</math>. What is the escape speed of an object launched straight up from just above the atmosphere of Jupiter?

'''Solution'''

<div style="">
System = Jupiter + object
<math>

\Delta E = 0\\
v_i = ?\\
v_f = 0 \text{m/s}\\
r_i = 71.5\times 10^6 \text{m}\\
r_f = \infty\\
m = m_{Object}\\
M = m_{Jupiter} = 1900\times 10^{24} \text{kg} \\

</math>

Starting from the Energy Principle:

<math>

\Delta E = W + Q\\
\Delta E = 0 + 0 = 0\\
\Delta K + \Delta U = 0\\
\frac{1}{2}m(v_f^2-v_i^2) + (\frac{-GMm}{r_f} - \frac{-GMm}{r_i}) = 0\\
\frac{1}{2}m(0-v_i^2) + (0 - \frac{-GMm}{r_i}) = 0\\
-\frac{1}{2}mv_i^2 + \frac{GMm}{r_i} = 0\\
\frac{GMm}{r_i} = \frac{1}{2}mv_i^2\\
\frac{GM}{r_i} = \frac{1}{2}v_i^2\\
v_i = \sqrt{\frac{2GM}{r_i}}\\
= \sqrt{\frac{2(6.7\times 10^{-11} \text{N}\cdot\text{m}^2 / \text{kg}^2)(1900\times 10^{24} \text{kg})}{(71.5\times 10^6 \text{m})}}\\
= 5.97 \times 10^4 \text{m/s}
</math>

==Connectedness==

In real life, escape velocity is not as easy to calculate as in the above examples. The primary complicating factor is the fact that there are more than two bodies in the universe, so in systems with multiple massive attracting bodies, escape velocity can become more complicated. One example is the escape velocity of an object from Earth: an object that achieves the escape velocity of Earth could theoretically escape Earth's influence, but it would remain in orbit around the sun unless its speed were much greater. The concept of escape velocity is widely used in orbital mechanics and rocketry and is critical for the planning of space missions.

==History==
Escape velocity stems from the concept of gravity, which was pioneered by Sir Issac Newton. Escape velocity became more important as people looked towards putting objects and people in space. In 1957, the Soviets were the first to launch a satellite into space using the idea of escape velocity to put the object in orbit.

==See Also==

===Further Reading===

https://www.nasa.gov/audience/foreducators/k-4/features/F_Escape_Velocity.html

http://www.qrg.northwestern.edu/projects/vss/docs/space-environment/2-whats-escape-velocity.html

===External links===
http://www.scientificamerican.com/article/bring-science-home-reaction-time/

https://www.youtube.com/watch?v=7w56rwAtUZU

==References==
“A Brief History of Space Exploration.” The Aerospace Corporation, The Aerospace Corporation, 1 June 2018, aerospace.org/article/brief-history-space-exploration.

"Escape Velocity | Physics." Encyclopedia Britannica Online. Encyclopedia Britannica, n.d. Web. 05 Dec. 2015.
[http://www.britannica.com/science/escape-velocity]

Giancoli, Douglas C. "Physics for Scientists and Engineers with Modern Physics." Google Books. Google, n.d. Web. 05 Dec. 2015.
[https://books.google.com/books?id=xz-UEdtRmzkC&pg=PA199&dq=escape+velocity+gravitational+potential+energy&hl=en&sa=X&ved=0CC0Q6AEwA2oVChMI8_PO4_PBxwIVBJmICh3T6gGl#v=onepage&q=escape%20velocity%20gravitational%20potential%20energy&f=false]

“Gravity And Orbits.” PhET, University of Colorado Boulder, 5 Aug. 2019, phet.colorado.edu/en/simulation/gravity-and-orbits.

"Escape Velocity." Wikipedia. Wikimedia Foundation, n.d. Web. 05 Dec. 2015.
[https://en.wikipedia.org/wiki/Escape_velocity]

Velocity, Escape, and ©200. ESCAPE VELOCITY EXAMPLES (n.d.): n. pag. 13 June 2003. Web. 5 Dec. 2015.
[http://www.beaconlearningcenter.com/documents/1483_01.pdf]

[[Category:Energy]]

Escape Velocity

2021-11-29T17:19:19Z

Jnation8:

[[File:Voyager Path czech version.jpg|400px|thumb|right| A diagram showing the paths of Voyager 1 and 2.]]
'''Claimed By: Jared Nation Fall 2021'''

Escape velocity is defined as the minimum velocity required for an object to escape the gravitational force of a large object. The sum of an object's kinetic energy and its gravitational potential energy is equal to zero. The gravitational potential energy is negative due to the fact that kinetic energy is always positive. The velocity of the object will be zero at infinite distance from the center of gravity. There is no net force on an object as it escapes and zero acceleration is perceived.

==The Main Idea==

The formula for escape velocity at a given distance from a body is calculated by the formula
:<math>v_e = \sqrt{\frac{2GM}{r}},</math>

where <math>G</math> is the universal [[gravitational constant]] (<math>G = 6.7\times 10^{-11} \text{N}\cdot\text{m}^2 / \text{kg}^2</math>), <math>M</math> is the mass of the large body to be escaped, and <math>r</math> the distance from the [[center of mass]] of the mass <math>M</math> to the object. This equation assumes there are no other forces acting on either body. As a side note, the escape velocity stated here could really be called escape speed due to the fact that the quantity is independent of direction. Notice that the equation does not include the mass of the orbiting body.

===A Mathematical Model===

When we model escape velocity, we consider the situation when an object's velocity takes it to a point an infinite distance away. The lowest possible escape velocity has a final speed of zero, and any speed higher results in a nonzero final speed. To derive the formula for the escape velocity, the energy principle is used, and we assume that the only two objects in our system are the orbiting body and the planet.

In our system, the energy principle states that

:<math>(K + U_g)_i = (K + U_g)_f \,</math>

When finding the minimum escape velocity, <math>K-f = 0 </math> because we take the final velocity to be zero, and <math>U_{gf} = 0 </math> because its final distance is expressed as infinity, therefore

:<math>(K + U_g)_i = \frac{1}{2}mv_e^2 + \frac{-GMm}{r} = 0</math>

Solving for <math>v_e</math> yields:

:<math>v_e = \sqrt{\frac{2GM}{r}}</math>

Note that because both the kinetic and potential energy terms contain a common factor <math>m</math>, the final escape velocity is independent of the mass of the orbiting body.

'''Bound versus unbound systems'''

When an object is orbiting a massive body, it can be in one of two states: bound and unbound. If the object is in a bound state, we see an elliptical trajectory, in which the orbiting body never escapes the gravitational influence of the more massive body. In an unbound state, however, we observe a parabolic or hyperbolic trajectory, in which the object is able to escape the gravitational influence of the orbiting body and escape to infinity. The diagram to the left shows an unbound system, in which the sum of the kinetic and potential energy of the orbiting body is greater then 0. As distance goes to infinity in this system, gravitational potential energy approaches zero, but the object retains a positive kinetic energy, and therefore a positive velocity. The image on the right shows a bound system, in which the sum of the kinetic and potential energy is negative. In this system, kinetic energy reaches zero at a specific maximum distance, at which point the object begins to fall back towards the massive body, never to escape.

[[File:Ediagram_MSPaint_UnboundSystemWithExtraEnergy.png|500px|thumb|left|The energy diagram of an unbound system, in which the object has excess kinetic energy. The vertical axis represents energy, while the horizontal axis represents distance. At exactly the escape velocity, the sum of <math>K</math> and <math>U</math> is exactly 0.]]
[[File:EDiagram_MSPaint_BoundSystem.png|500px|thumb|right|The energy diagram of a bound system, in which the object has insufficient kinetic energy to escape. The vertical axis represents energy, while the horizontal axis represents distance. ]]

<br clear=all>

===A Computational Model===

Here is simulation that can be used to experiment with different factors that affect escape velocity:https://phet.colorado.edu/en/simulation/gravity-and-orbits.

==Examples==

===Simple===

'''Question'''

Compute the escape velocity for Earth if its mass is <math>5.98 \times 10^{24} \text{kg}</math> and its radius is <math>6.37 \times 10^{6} \text{m}</math>.

'''Solution'''

<math>
v_e = \sqrt{\frac{2GM}{r}} \\
= \sqrt{\frac{2(6.7\times 10^{-11} \text{N}\cdot\text{m}^2 / \text{kg}^2)(5.98\times 10^{24} \text{kg})}{(6.37\times 10^6 \text{m})}}\\
= 1.12 \times 10^4 \text{m/s}
</math>

</div>

===Middling===

'''Question'''

If the escape velocity of a planet with the same mass as Earth is greater than the escape velocity of Earth, is the planet larger or smaller than Earth?

'''Solution'''

Given the formula: <math> v_e = \sqrt{\frac{2GM}{r}} </math>

And rearranging for radius: <math> r = \frac{2GM}{v^2_e} </math>

We can see that since the denominator of the faction is larger with a constant numerator, then the radius must be smaller. Therefore, the planet is smaller than Earth.

===Difficult===

'''Question'''

The radius of Jupiter is <math>71.5\times 10^6 \text{m}</math>, and its mass is <math>1900\times 10^{24} \text{kg}</math>. What is the escape speed of an object launched straight up from just above the atmosphere of Jupiter?

'''Solution'''

<div style="">
System = Jupiter + object
<math>

\Delta E = 0\\
v_i = ?\\
v_f = 0 \text{m/s}\\
r_i = 71.5\times 10^6 \text{m}\\
r_f = \infty\\
m = m_{Object}\\
M = m_{Jupiter} = 1900\times 10^{24} \text{kg} \\

</math>

Starting from the Energy Principle:

<math>

\Delta E = W + Q\\
\Delta E = 0 + 0 = 0\\
\Delta K + \Delta U = 0\\
\frac{1}{2}m(v_f^2-v_i^2) + (\frac{-GMm}{r_f} - \frac{-GMm}{r_i}) = 0\\
\frac{1}{2}m(0-v_i^2) + (0 - \frac{-GMm}{r_i}) = 0\\
-\frac{1}{2}mv_i^2 + \frac{GMm}{r_i} = 0\\
\frac{GMm}{r_i} = \frac{1}{2}mv_i^2\\
\frac{GM}{r_i} = \frac{1}{2}v_i^2\\
v_i = \sqrt{\frac{2GM}{r_i}}\\
= \sqrt{\frac{2(6.7\times 10^{-11} \text{N}\cdot\text{m}^2 / \text{kg}^2)(1900\times 10^{24} \text{kg})}{(71.5\times 10^6 \text{m})}}\\
= 5.97 \times 10^4 \text{m/s}
</math>

==Connectedness==

In real life, escape velocity is not as easy to calculate as in the above examples. The primary complicating factor is the fact that there are more than two bodies in the universe, so in systems with multiple massive attracting bodies, escape velocity can become more complicated. One example is the escape velocity of an object from Earth: an object that achieves the escape velocity of Earth could theoretically escape Earth's influence, but it would remain in orbit around the sun unless its speed were much greater. The concept of escape velocity is widely used in orbital mechanics and rocketry and is critical for the planning of space missions.

==History==
Escape velocity stems from the concept of gravity, which was pioneered by Sir Issac Newton. Escape velocity became more important as people looked towards putting objects and people in space. In 1957, the Soviets were the first to launch a satellite into space using the idea of escape velocity to put the object in orbit.

==See Also==

===Further Reading===

https://www.nasa.gov/audience/foreducators/k-4/features/F_Escape_Velocity.html

http://www.qrg.northwestern.edu/projects/vss/docs/space-environment/2-whats-escape-velocity.html

===External links===
http://www.scientificamerican.com/article/bring-science-home-reaction-time/

https://www.youtube.com/watch?v=7w56rwAtUZU

==References==
“A Brief History of Space Exploration.” The Aerospace Corporation, The Aerospace Corporation, 1 June 2018, aerospace.org/article/brief-history-space-exploration.

"Escape Velocity | Physics." Encyclopedia Britannica Online. Encyclopedia Britannica, n.d. Web. 05 Dec. 2015.
[http://www.britannica.com/science/escape-velocity]

Giancoli, Douglas C. "Physics for Scientists and Engineers with Modern Physics." Google Books. Google, n.d. Web. 05 Dec. 2015.
[https://books.google.com/books?id=xz-UEdtRmzkC&pg=PA199&dq=escape+velocity+gravitational+potential+energy&hl=en&sa=X&ved=0CC0Q6AEwA2oVChMI8_PO4_PBxwIVBJmICh3T6gGl#v=onepage&q=escape%20velocity%20gravitational%20potential%20energy&f=false]

“Gravity And Orbits.” PhET, University of Colorado Boulder, 5 Aug. 2019, phet.colorado.edu/en/simulation/gravity-and-orbits.

"Escape Velocity." Wikipedia. Wikimedia Foundation, n.d. Web. 05 Dec. 2015.
[https://en.wikipedia.org/wiki/Escape_velocity]

Velocity, Escape, and ©200. ESCAPE VELOCITY EXAMPLES (n.d.): n. pag. 13 June 2003. Web. 5 Dec. 2015.
[http://www.beaconlearningcenter.com/documents/1483_01.pdf]

[[Category:Energy]]

SI Units

2021-11-29T17:10:25Z

Jnation8:

This page describes the International System of Units and lists its units.

==The Main Idea==

The International System of Units is a set of units allowing for the quantification of each physical dimension. It was invented in France as an extension of the metric system under the name "Système Internationale d'Unités," which is why it is abbreviated to "SI". SI Units are practically universally used for scientific applications, and are used in most countries in everyday life as well, although the United States, which uses customary units, is a notable exception. Most of quantities in this course will be given in SI units.

The International System of Units has seven base units with specific definitions and countless derived units which are created by multiplying and dividing the base units. For example, the meter (m) is the base unit for the dimension distance and the second (s) is the base unit for the dimension time. The meter per second (m/s) measures the dimension speed and is a derived unit because it can be obtained by dividing the meter by the second.

Because measurements must often be made on many different scales, the International System of Units also defines a variety of [[#Prefixes|prefixes]] for use with its units. The prefixes scale the values of each unit by powers of 10.

The International System of Units is adaptable and can change over time to meet the evolving needs of scientists. Over the last several centuries, the number of base units has grown from 3 to 7 as a result of the discovery of new dimensions that could not be expressed in terms of existing base units. The number of derived units recognized as meaningful has also grown substantially and will likely continue to as the human understanding of the physical world improves. Additionally, the seven base units have undergone redefinition at several points in time, most recently on May 20 2019. The purpose of each redefinition is not to change the units' values (in fact, their values are changed as little as possible), but to make their definitions more universally constant and reproducible.

===The Seven Base Units===

The values of the seven base units are historically motivated, but most of their current definitions have been changed from the original. Today, each of the 7 base unit is defined by fixing a specific natural universal constant at a certain numerical value.

====Second (s)====

The second measures time.

Today, the second is defined by fixing the value of the cesium hyperfine transition frequency <math>\Delta v_{Cs}</math> at exactly 9192631770 hz (transitions per second). This defines the second as the 9192631770 times the period of the transition between the two levels of the ground state of cesium-133 atom. Atomic clocks make use of this phenomenon by counting the cesium transitions as a mechanism for keeping time! This definition was instated in 1967.

The second was initially defined as a fraction (1/86400) of a day of 24 hours of 60 minutes of 60 seconds. This definition is poor primarily because the duration of the day changes slightly over time. In 1956 the second was defined as 1/31556925.9747 of the tropical year. (The tropical year is the time it takes the sun to return to the same position in the season cycle, which is about 20 minutes shorter than the causal year.) However, this is still not as elegant as the current definition because it is based on our particular solar system, which may not always exist or be available for measuring.

====Meter (m)====

The meter measures distance.

Today, the meter is defined by fixing the value of the speed of light in a vacuum <math>c</math> at exactly 299,792,458 m/s. Since the second is already defined, this defines the meter as the distance traveled by light in a vacuum in 1/29979248<sup>th</sup> of a second.

The meter was originally defined in 1973 as 1/10,000,000 of the meridian through Paris between the North Pole and the Equator. In 1960 it was redefined as 1650763.73 wavelengths of the radiation in a vacuum corresponding to the transition between 2p10 and 5d5 transition levels of the krypton-86 atom.

====Kilogram (kg)====

The kilogram measures mass.

Today, the kilogram is defined by fixing the value of Planck's constant <math>h</math> at exactly 6.62607015×10−34 J s. Recall that the joule is not a base unit, but a derived unit equal to the kg m<sup>2</sup>/s<sup>2</sup>. Since the meter and the second are already defined, this leaves only one possible value for the kilogram. This definition was instated as part of the SI unit redefinition of May 20, 2019. [https://www.youtube.com/watch?v=c_e1wITe_ig&t=521s This video] by the science YouTube channel Veritasium describes the redefinition of base units of May 20, 2019, with an emphasis on the new definition of the kilogram.

In 1973, the SI unit for mass was called the grave. It was defined as being the mass of 1 liter of pure water at its freezing temperature. The grave was renamed to the kilogram, and later, in 1889, it was redefined as the mass of a specific sample of a platinum alloy. The sample is called "Le Grand K" and is kept in a humidity-controlled vault just outside of Paris, France. This definition was inadequate because it was not universally reproducible, so they considered a few alternate definitions, and, in 2019, redefined it to the current definition. [https://www.youtube.com/watch?v=ZMByI4s-D-Y This video from 2013], also by Veritasium, explains the shortcomings of Le Grand K and describes a proposed alternate definition that was being considered at that time (although it did not end up being the chosen definition).

====Ampere (A)====

The ampere measures electrical current.

Today the ampere is defined by fixing the value of the elementary charge <math>e</math> at exactly 1.602176634e-19 C. This defines the Coulomb, and since an ampere is 1 Coulomb per second (and the second is already defined), it also defines the Ampere.

In 1881, the ampere was defined as a tenth of the current required to create a magnetic field of one Oersted at the center of a 1cm arc of wire with curve radius 1cm. In 1946, the ampere was defined as the current required to create a force of 2e-7 Newtons per meter of length between two straight parallel conductors of infinite length placed 1m apart in a vacuum. [https://www.youtube.com/watch?v=YmCrFPC1qi4 This video] by the physics YouTube channel Kaustubhan describes the 1946 definition of the Ampere.

====Kelvin (K)====

The Kelvin measures temperature.

Today, the Kelvin is defined by fixing the value of the Boltzmann constant <math>k</math> at exactly 1.380649 x 10–23 J/K. Since the joule is a derived unit derived from the second, the meter, and the kilogram, each of which has already been defined, this leaves only one possible value for the Kelvin.

====Mole (mol)====

The mole measures the amount of a substance present by number of particles.

Today, the mole is defined by fixing the value of the Avogadro constant <math>N_A</math> at exactly 6.022140857 × 10<sup>23</sup> mol<sup>-1</sup>. This defines the mol to be 6.022140857 × 10<sup>23</sup> mol particles.

The mol was originally defined in 1967 the mole as the number of carbon-12 atoms necessary to comprise a 12 gram sample.

====Candela (cd)====

The candela measures luminous intensity.

Today, the candela is defined by fixing the value of the luminous efficacy constant <math>K_{cd}</math> at exactly 683 lumens per watt. This defines the candela as the luminous instensity of a source that emits radiation of a frequency 5.4e14 Hz in 1/683 of a steradian. (A steradian is the 3D analog for the radian; it is a part of the sphere's surface area equalling the radius squared in area.) This is about the luminous intensity of a candle.

===Derived Units===

There are many useful quantities in physics that cannot be measured in a single base unit. These must be expressed in derived units, which are obtained by multiplying and dividing the base units. For example, the meter (m) is the base unit for the dimension distance and the second (s) is the base unit for the dimension time. The meter per second (m/s) measures the dimension speed and is a derived unit because it can be obtained by dividing the meter by the second. Speed does not get its own base unit because it does not need one; speed is fundamentally a relationship between distance and time (specifically, distance traveled and the time it takes to travel that distance). Other examples of derived SI units are the kg/m<sup>3</sup>, which measures density, and the kgm/s, which measures momentum.

Some derived units are used so frequently in physics that they are given their own names. Below are several such derived units:

<table border="1">
<tr>
<th>Unit</th><th>Symbol</th><th>In Terms of Base Units</th><th>Dimension</th>
</tr>
<tr>
<td>hertz</td><td>Hz</td><td><math>\mathrm{s}^{-1}</math></td><td>frequency</td>
</tr>
<tr>
<td>newton</td><td>N</td><td><math>\frac{\mathrm{kg}\mathrm{m}}{\mathrm{s}^2}</math></td><td>force</td>
</tr>
<tr>
<td>pascal</td><td>Pa</td><td><math>\frac{\mathrm{kg}}{\mathrm{m}\mathrm{s}^2}</math></td><td>pressure</td>
</tr>
<tr>
<td>joule</td><td>J</td><td><math>\frac{\mathrm{kg}\mathrm{m}^2}{\mathrm{s}^2}</math></td><td>energy</td>
</tr>
<tr>
<td>watt</td><td>W</td><td><math>\frac{\mathrm{kg}\mathrm{m}^2}{\mathrm{s}^3}</math></td><td>power</td>
</tr>
<tr>
<td>coulomb</td><td>C</td><td><math>\mathrm{As}</math></td><td>charge</td>
</tr>
<tr>
<td>volt</td><td>V</td><td><math>\frac{\mathrm{kg}\mathrm{m}^2}{\mathrm{A}\mathrm{s}^3}</math></td><td>electric potential</td>
</tr>
</table>

This list is not comprehensive, and there are many unnamed derived units that are commonly used as well.

===Prefixes===

Because measurements must often be made on many different scales, the International System of Units also defines a variety of prefixes for use with its units. The prefixes scale the values of each unit by powers of 10. For example, the kilometer is 1000 times as long as the meter because of the prefix "kilo." The use of prefixes allows even large and small measurements to be reported with reasonable and easy to read numbers. For example, it is easier to express the atomic radius of the hydrogen atom as 53 picometers than 5.3 x 10<sup>-11</sup> meters.

Each prefix has a symbol, which can be combined with the symbol of a unit. For example, the symbol for pico is "p", so "pm" is the symbol for picometer.

The following table shows the common prefixes used with SI units.

<table border="1">
<tr>
<th>Prefix</th><th>Symbol</th><th>Multiplier</th>
</tr>
<tr>
<td>peta</td><td>P</td><td>10<sup>15</sup></td>
</tr>
<tr>
<td>tera</td><td>T</td><td>10<sup>12</sup></td>
</tr>
<tr>
<td>giga</td><td>G</td><td>10<sup>9</sup></td>
</tr>
<tr>
<td>mega</td><td>M</td><td>10<sup>6</sup></td>
</tr>
<tr>
<td>kilo</td><td>k</td><td>1000 (10<sup>3</sup>)</td>
</tr>
<tr>
<td>hecto</td><td>h</td><td>100 (10<sup>2</sup>)</td>
</tr>
<tr>
<td>deka</td><td>da</td><td>10 (10<sup>1</sup>)</td>
</tr>
<tr>
<td>deci</td><td>d</td><td>.1 (10<sup>-1</sup>)</td>
</tr>
<tr>
<td>centi</td><td>c</td><td>.01 (10<sup>-2</sup>)</td>
</tr>
<tr>
<td>milli</td><td>m</td><td>.001 (10<sup>-3</sup>)</td>
</tr>
<tr>
<td>micro</td><td><math>\mu</math></td><td>10<sup>-6</sup></td>
</tr>
<tr>
<td>nano</td><td>n</td><td>10<sup>-9</sup></td>
</tr>
<tr>
<td>pico</td><td>p</td><td>10<sup>-12</sup></td>
</tr>
<tr>
<td>femto</td><td>f</td><td>10<sup>-15</sup></td>
</tr>
</table>

There are several popular mnemonic devices for remembering some of the prefixes. A common example is "King Henry Died by Drinking Chocolate Milk." The first letter of each word represents a prefix, starting at kilo and ending at mili. The "b" in "by" stands for "base unit," which in this context means a unit with no prefixes.

===Dimensional Analysis===

Dimensional analysis is the analysis of the units used in expressions and equations. Dimensional analysis revolves around the fact that when two quantities, each measured with a specific unit, are multiplied or divided, their units are multiplied or divided as well. For example, consider car traveling in a straight line at a constant speed of 20m/s. Suppose you want to know how far it travels in 30 seconds. To calculate the answer, simply multiply 20m/s by 30s, which is 600m. Notice that the meter per second times the second equals the meter.

A result of the above rule is the fact that both sides of an equation must have the same units. This makes sense because two quantities in different dimensions can't be compared. To illustrate this fact, consider the ideal gas law:

<math>PV=nRT</math>

where <math>P</math> is the pressure of the gas, <math>V</math> is its volume, <math>n</math> is the number moles of the gas, <math>R</math> is the ideal gas constant, and <math>T</math> is the temperature of the gas.

For the ideal gas law to be numerically true, the ideal gas constant <math>R</math> must be in the correct units. The SI units for pressure, volume, and temperature respectively are the kilopascal (kPa), the cubic meter (m<sup>3</sup>), and Kelvin (K). If these units are used, <math>R</math> must be given in m<sup>3</sup> kPa / mol K. That way, both sides of the equation are in m<sup>3</sup> kPa. R can also be given in other units, as long as they are a unit of volume times a unit of pressure divided by moles and a unit of temperature. Its value depends on the units used. In fact, all natural constants can have different values depending on the units in which they are given. In this class, they are typically given in their SI units.

All equations in physics, like <math>PV=nRT</math>, are true for all units, as long as their two sides agree in units, which requires the correct versions of any natural constants. Generally, SI units are preferred.

The fact that both sides of an equation must have the same units can be useful for determining the units of a quantity, or, if the units of all quantities are known, for verifying that a derived equation makes sense and is possible.

===Unit Conversion===

Measured quantities can be converted from one unit to another as long as the relationship between the two units is known. To do so, the quantity should be multiplied by the ratio of the two units' values, with the quantity's current unit in the denominator and the target unit in the numerator. This ratio is called a conversion factor. For example, suppose you want to convert 7cm to inches, given that an inch is equivalent to 2.54cm. 7cm should be multiplied by the ratio of the inch to the centimeter, which is <math>\frac{1\mathrm{in}}{2.54\mathrm{cm}}</math>:

<math>7\mathrm{cm} * \frac{1\mathrm{in}}{2.54\mathrm{cm}} = 2.76\mathrm{in}</math>.

It is also possible to use this method to convert between units that measure different dimensions, as long as they are proportional to each other and the relationship between them is known. For example, it is possible to convert from the volume of a sample of a substance to its mass if its density is known:

<math>.125\mathrm{m}^3\mathrm{ steel} * \frac{8050\mathrm{kg steel}}{1 \mathrm{m}^3\mathrm{ steel}} = 1006.25\mathrm{kg steel}</math>

Sometimes, it is necessary to perform many conversions in sequence. For example, consider a person who wants to convert from meters to feet, but the only relationship they know between SI and customary length units is that an inch is equivalent to 2.54cm. They would first have to convert meters to centimeters, then centimeters to inches, then inches to feet. In such a situation, it is often helpful to use a notation such as the one below to keep track of the conversions:

[[File:SIUnitsFenceConversion.png]]

Each column represents the conversion from one unit to another using a different conversion factor. Each unit in the top (except the target unit) should cancel with a different unit in the bottom.

It is also possible to convert between complex units with both a numerator and a denominator. The following chart converts from meters per second to miles per hour. Some columns are for converting from meters to miles and others are for converting from seconds to hours.

[[File:SIUnitsFenceConversion2.png]]

Finally, it is worth noting that when converting between higher powers of units, such as cubic inches to cubic cm, their conversion factors are raised to that power also. That is, one inch is equivalent to 2.54 cm, so one cubic inch is equivalent to 2.54<sup>3</sup> = 16.39 cubic cm.

==Examples==

===Simple===

The law of universal gravitation states that the magnitude of the gravitational force between two objects of masses <math>m_1</math> and <math>m_2</math> is given by

<math>|\vec{F}_{grav}| = \frac{Gm_1m_2}{r^2}</math>,

where <math>G</math> is the universal gravitational force constant and <math>r</math> is the distance between the objects' centers of mass.

Assuming that <math>|\vec{F}_{grav}|</math>, <math>m_1</math>, <math>m_2</math>, and <math>r</math> are given in their respective standard SI units, what units should G have?

Solution:

The left side of the equation is in N because that is the SI unit for force. The right side of the equation must therefore also be in N. Ignoring the G, the right side of the equation has units of kg<sup>2</sup>/m<sup>2</sup>. G must introduce the unit N in the numerator while canceling the kg<sup>2</sup>/m<sup>2</sup>, so G has units of N*m<sup>2</sup>/kg<sup>2</sup>. This is equivalent to m<sup>3</sup>/kgs<sup>2</sup> because 1N = 1kgm/s<sup>2</sup>. These two answers can also be written respectively as N*m<sup>2</sup>*kg<sup>-2</sup> and m<sup>3</sup>*kg<sup>-1</sup>*s<sup>-2</sup>. Any of the 4 answers is correct.

===Middling===

A rectangular prism made out of aluminum (density: 2,710kg/m<sup>3</sup>) has sides of length 1in, 2in, and 6in. 1 inch is equivalent to 2.54cm. What is the weight of the sample near the surface of the earth?

Solution:

The rectangular prism has a volume of 1in x 2in x 6in = 12in<sup>3</sup>. Let us use the unit conversion notation discussed earlier in the page to convert this volume to force:

[[File:SIUnitsAluminumBlockSoln.png]]

Note that although g is normally thought of as having as units m/s<sup>2</sup>, we used it above as having units N/kg. These units make more sense when it serves as a constant of proportionality between weight (measured in N) and mass (measured in kg). These units are equivalent to m/s<sup>2</sup>.

===Difficult===

A garage door has a width of 2.5m, a width of 2.2m, and a thickness of 4cm. The average density of the materials that comprise it is 273 kg/m<sup>3</sup>.

The garage works like most home garages; it can slide up and down along a rail to open and close respectively, and the rail has a 90<math>^\circ</math> turn (of small radius) in it so that when the garage door is closed, it is vertical, but when it is open, it hangs horizontally below the ceiling. When the garage is partially opened, the horizontal part of it is supported by the rail while the vertical part hangs freely. Because the garage door is intended to move at a constant speed, it is fitted with a spring designed to cancel the weight of the vertical part of the garage door regardless of the door's position. This makes it easier for the cable system to move the garage door. The spring obeys [[Hooke's Law]]. It is at its equilibrium length when the garage door is open and entirely supported by the rail, and as the garage door closes and more of its weight becomes unsupported, the spring extends to apply a force equal to the unsupported weight in the opposite direction.

[[File:SIUnitsGarageDoor.png]]

What should be the spring constant k of the spring?

Solution:

Let d be the length of the vertical part of the garage door, in m. The volume of the vertical part of the garage door is 2.5 x .04 x d m<sup>3</sup> = .1d m<sup>3</sup>. Its mass is .1d m<sup>3</sup> x 273 kg/m<sup>3</sup> = 27.3d kg. Its weight is 27.3d kg x 9.8 N/kg = 268d N. The displacement of the spring from its equilibrium length is also d due to the geometry of the garage door system. According to Hooke's Law, the spring must exert a force whose magnitude is given by f = kd, where k is the spring constant of the spring. The magnitude of the force of the spring must equal the weight of the vertical part of the garage door, which is 268d Newtons, so k must equal 268 N/m.

==Connectedness==

Physics can be thought of as the study of quantifying natural phenomena, which could not be done without some system of units allowing for measurements and calculations to be made. The SI unit system is a good candidate for a number of reasons: its units can apply to quantities on many different scales, new and existing units can be defined and redefined as necessary, and each unit is defined in such a way that they can be replicated with high precision by anyone at any place and time, provided they have the resources to perform certain experiments. The SI system is connected to all of physics (and, indeed, all of science) because practically the entire scientific community has agreed to use it for all measurements and calculations. A basic understanding of the SI system is necessary for any type of technical industry (engineering, architecture, etc.) in most countries outside the united states, where the SI system is used for such industries (the United States uses the customary system for all industries except pure science). An in-depth understanding of the SI system is necessary for certain industries with very small tolerances, such as the creation of precision machine tools.

==History==

The Metric System was created around the time of the French Revolution. On June 22, 1799, two platinum standards defining the meter and the kilogram were deposited in the Archives de la Republic in Paris. This can be seen as the first step in the development of the present International System of Units. The values of the meter and the kg were based on the physical world; the meter was based on the dimensions of the earth and the kilogram was based on the density of water. These two units formed the foundation of what was then called the metric system. The metric system was abandoned and then readopted by France, and slowly gained popularity in the scientific communities of other countries. [[James Maxwell]] proposed the addition of the second as a third base unit for time, which was quickly instated, and Giovanni Giorgi, an Italian physicist and electrical engineer, advocated for the addition of a fourth base unit to describe electromagnetic phenomena. The ampere was defined in 1935 for this purpose. By the 20th century, the metric system was commonplace across scientific communities across the world.

The SI unit system was established in 1960 by the 11th General Conference on Weights and Measures (CGPM), a part of the International Bureau of Weights and Measures (BIPM). The CGPM is an international authority on units recognized by most scientists which aims to standardized the units used around the world. The SI system adopted the four base units of the metric system and added three additional base units: the Kelvin, the candela, and the mole. Since then, the CGPM has continued to occasionally redefine the base SI units.

==See Also==

===Further reading===

SI Units for Clinical Measurement 1st Edition by Donald S. Young

===External Links===

https://www.youtube.com/watch?v=c_e1wITe_ig&t=521s (Video describing the redefinition of SI base units of May 20 2019)

http://wps.prenhall.com/wps/media/objects/165/169061/blb9ch0104.html (Pearson educational site)

http://physics.nist.gov/ (National institute of standards and Technology)

https://www.bipm.org/en/measurement-units/ (International Bureau of Weights and Measures site)

==References==

Matter & Interactions, Vol. I: Modern Mechanics, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

Tutorial & Drill Problems for General Chemistry (and Intro) By Walter S. Hamilton, Ph.D.

Base units of the SI, fundamental constants and modern quantum physics By Christian J Bordé, Published 15 September 2005

http://physics.nist.gov/.

https://www.bipm.org/en/measurement-units/

[[Category: The Scientific Method]]

SI Units

2021-11-29T17:08:41Z

Jnation8:

This page describes the International System of Units and lists its units.

'''Claimed By: Jared Nation'''
==The Main Idea==

The International System of Units is a set of units allowing for the quantification of each physical dimension. It was invented in France as an extension of the metric system under the name "Système Internationale d'Unités," which is why it is abbreviated to "SI". SI Units are practically universally used for scientific applications, and are used in most countries in everyday life as well, although the United States, which uses customary units, is a notable exception. Most of quantities in this course will be given in SI units.

The International System of Units has seven base units with specific definitions and countless derived units which are created by multiplying and dividing the base units. For example, the meter (m) is the base unit for the dimension distance and the second (s) is the base unit for the dimension time. The meter per second (m/s) measures the dimension speed and is a derived unit because it can be obtained by dividing the meter by the second.

Because measurements must often be made on many different scales, the International System of Units also defines a variety of [[#Prefixes|prefixes]] for use with its units. The prefixes scale the values of each unit by powers of 10.

The International System of Units is adaptable and can change over time to meet the evolving needs of scientists. Over the last several centuries, the number of base units has grown from 3 to 7 as a result of the discovery of new dimensions that could not be expressed in terms of existing base units. The number of derived units recognized as meaningful has also grown substantially and will likely continue to as the human understanding of the physical world improves. Additionally, the seven base units have undergone redefinition at several points in time, most recently on May 20 2019. The purpose of each redefinition is not to change the units' values (in fact, their values are changed as little as possible), but to make their definitions more universally constant and reproducible.

===The Seven Base Units===

The values of the seven base units are historically motivated, but most of their current definitions have been changed from the original. Today, each of the 7 base unit is defined by fixing a specific natural universal constant at a certain numerical value.

====Second (s)====

The second measures time.

Today, the second is defined by fixing the value of the cesium hyperfine transition frequency <math>\Delta v_{Cs}</math> at exactly 9192631770 hz (transitions per second). This defines the second as the 9192631770 times the period of the transition between the two levels of the ground state of cesium-133 atom. Atomic clocks make use of this phenomenon by counting the cesium transitions as a mechanism for keeping time! This definition was instated in 1967.

The second was initially defined as a fraction (1/86400) of a day of 24 hours of 60 minutes of 60 seconds. This definition is poor primarily because the duration of the day changes slightly over time. In 1956 the second was defined as 1/31556925.9747 of the tropical year. (The tropical year is the time it takes the sun to return to the same position in the season cycle, which is about 20 minutes shorter than the causal year.) However, this is still not as elegant as the current definition because it is based on our particular solar system, which may not always exist or be available for measuring.

====Meter (m)====

The meter measures distance.

Today, the meter is defined by fixing the value of the speed of light in a vacuum <math>c</math> at exactly 299,792,458 m/s. Since the second is already defined, this defines the meter as the distance traveled by light in a vacuum in 1/29979248<sup>th</sup> of a second.

The meter was originally defined in 1973 as 1/10,000,000 of the meridian through Paris between the North Pole and the Equator. In 1960 it was redefined as 1650763.73 wavelengths of the radiation in a vacuum corresponding to the transition between 2p10 and 5d5 transition levels of the krypton-86 atom.

====Kilogram (kg)====

The kilogram measures mass.

Today, the kilogram is defined by fixing the value of Planck's constant <math>h</math> at exactly 6.62607015×10−34 J s. Recall that the joule is not a base unit, but a derived unit equal to the kg m<sup>2</sup>/s<sup>2</sup>. Since the meter and the second are already defined, this leaves only one possible value for the kilogram. This definition was instated as part of the SI unit redefinition of May 20, 2019. [https://www.youtube.com/watch?v=c_e1wITe_ig&t=521s This video] by the science YouTube channel Veritasium describes the redefinition of base units of May 20, 2019, with an emphasis on the new definition of the kilogram.

In 1973, the SI unit for mass was called the grave. It was defined as being the mass of 1 liter of pure water at its freezing temperature. The grave was renamed to the kilogram, and later, in 1889, it was redefined as the mass of a specific sample of a platinum alloy. The sample is called "Le Grand K" and is kept in a humidity-controlled vault just outside of Paris, France. This definition was inadequate because it was not universally reproducible, so they considered a few alternate definitions, and, in 2019, redefined it to the current definition. [https://www.youtube.com/watch?v=ZMByI4s-D-Y This video from 2013], also by Veritasium, explains the shortcomings of Le Grand K and describes a proposed alternate definition that was being considered at that time (although it did not end up being the chosen definition).

====Ampere (A)====

The ampere measures electrical current.

Today the ampere is defined by fixing the value of the elementary charge <math>e</math> at exactly 1.602176634e-19 C. This defines the Coulomb, and since an ampere is 1 Coulomb per second (and the second is already defined), it also defines the Ampere.

In 1881, the ampere was defined as a tenth of the current required to create a magnetic field of one Oersted at the center of a 1cm arc of wire with curve radius 1cm. In 1946, the ampere was defined as the current required to create a force of 2e-7 Newtons per meter of length between two straight parallel conductors of infinite length placed 1m apart in a vacuum. [https://www.youtube.com/watch?v=YmCrFPC1qi4 This video] by the physics YouTube channel Kaustubhan describes the 1946 definition of the Ampere.

====Kelvin (K)====

The Kelvin measures temperature.

Today, the Kelvin is defined by fixing the value of the Boltzmann constant <math>k</math> at exactly 1.380649 x 10–23 J/K. Since the joule is a derived unit derived from the second, the meter, and the kilogram, each of which has already been defined, this leaves only one possible value for the Kelvin.

====Mole (mol)====

The mole measures the amount of a substance present by number of particles.

Today, the mole is defined by fixing the value of the Avogadro constant <math>N_A</math> at exactly 6.022140857 × 10<sup>23</sup> mol<sup>-1</sup>. This defines the mol to be 6.022140857 × 10<sup>23</sup> mol particles.

The mol was originally defined in 1967 the mole as the number of carbon-12 atoms necessary to comprise a 12 gram sample.

====Candela (cd)====

The candela measures luminous intensity.

Today, the candela is defined by fixing the value of the luminous efficacy constant <math>K_{cd}</math> at exactly 683 lumens per watt. This defines the candela as the luminous instensity of a source that emits radiation of a frequency 5.4e14 Hz in 1/683 of a steradian. (A steradian is the 3D analog for the radian; it is a part of the sphere's surface area equalling the radius squared in area.) This is about the luminous intensity of a candle.

===Derived Units===

There are many useful quantities in physics that cannot be measured in a single base unit. These must be expressed in derived units, which are obtained by multiplying and dividing the base units. For example, the meter (m) is the base unit for the dimension distance and the second (s) is the base unit for the dimension time. The meter per second (m/s) measures the dimension speed and is a derived unit because it can be obtained by dividing the meter by the second. Speed does not get its own base unit because it does not need one; speed is fundamentally a relationship between distance and time (specifically, distance traveled and the time it takes to travel that distance). Other examples of derived SI units are the kg/m<sup>3</sup>, which measures density, and the kgm/s, which measures momentum.

Some derived units are used so frequently in physics that they are given their own names. Below are several such derived units:

<table border="1">
<tr>
<th>Unit</th><th>Symbol</th><th>In Terms of Base Units</th><th>Dimension</th>
</tr>
<tr>
<td>hertz</td><td>Hz</td><td><math>\mathrm{s}^{-1}</math></td><td>frequency</td>
</tr>
<tr>
<td>newton</td><td>N</td><td><math>\frac{\mathrm{kg}\mathrm{m}}{\mathrm{s}^2}</math></td><td>force</td>
</tr>
<tr>
<td>pascal</td><td>Pa</td><td><math>\frac{\mathrm{kg}}{\mathrm{m}\mathrm{s}^2}</math></td><td>pressure</td>
</tr>
<tr>
<td>joule</td><td>J</td><td><math>\frac{\mathrm{kg}\mathrm{m}^2}{\mathrm{s}^2}</math></td><td>energy</td>
</tr>
<tr>
<td>watt</td><td>W</td><td><math>\frac{\mathrm{kg}\mathrm{m}^2}{\mathrm{s}^3}</math></td><td>power</td>
</tr>
<tr>
<td>coulomb</td><td>C</td><td><math>\mathrm{As}</math></td><td>charge</td>
</tr>
<tr>
<td>volt</td><td>V</td><td><math>\frac{\mathrm{kg}\mathrm{m}^2}{\mathrm{A}\mathrm{s}^3}</math></td><td>electric potential</td>
</tr>
</table>

This list is not comprehensive, and there are many unnamed derived units that are commonly used as well.

===Prefixes===

Because measurements must often be made on many different scales, the International System of Units also defines a variety of prefixes for use with its units. The prefixes scale the values of each unit by powers of 10. For example, the kilometer is 1000 times as long as the meter because of the prefix "kilo." The use of prefixes allows even large and small measurements to be reported with reasonable and easy to read numbers. For example, it is easier to express the atomic radius of the hydrogen atom as 53 picometers than 5.3 x 10<sup>-11</sup> meters.

Each prefix has a symbol, which can be combined with the symbol of a unit. For example, the symbol for pico is "p", so "pm" is the symbol for picometer.

The following table shows the common prefixes used with SI units.

<table border="1">
<tr>
<th>Prefix</th><th>Symbol</th><th>Multiplier</th>
</tr>
<tr>
<td>peta</td><td>P</td><td>10<sup>15</sup></td>
</tr>
<tr>
<td>tera</td><td>T</td><td>10<sup>12</sup></td>
</tr>
<tr>
<td>giga</td><td>G</td><td>10<sup>9</sup></td>
</tr>
<tr>
<td>mega</td><td>M</td><td>10<sup>6</sup></td>
</tr>
<tr>
<td>kilo</td><td>k</td><td>1000 (10<sup>3</sup>)</td>
</tr>
<tr>
<td>hecto</td><td>h</td><td>100 (10<sup>2</sup>)</td>
</tr>
<tr>
<td>deka</td><td>da</td><td>10 (10<sup>1</sup>)</td>
</tr>
<tr>
<td>deci</td><td>d</td><td>.1 (10<sup>-1</sup>)</td>
</tr>
<tr>
<td>centi</td><td>c</td><td>.01 (10<sup>-2</sup>)</td>
</tr>
<tr>
<td>milli</td><td>m</td><td>.001 (10<sup>-3</sup>)</td>
</tr>
<tr>
<td>micro</td><td><math>\mu</math></td><td>10<sup>-6</sup></td>
</tr>
<tr>
<td>nano</td><td>n</td><td>10<sup>-9</sup></td>
</tr>
<tr>
<td>pico</td><td>p</td><td>10<sup>-12</sup></td>
</tr>
<tr>
<td>femto</td><td>f</td><td>10<sup>-15</sup></td>
</tr>
</table>

There are several popular mnemonic devices for remembering some of the prefixes. A common example is "King Henry Died by Drinking Chocolate Milk." The first letter of each word represents a prefix, starting at kilo and ending at mili. The "b" in "by" stands for "base unit," which in this context means a unit with no prefixes.

===Dimensional Analysis===

Dimensional analysis is the analysis of the units used in expressions and equations. Dimensional analysis revolves around the fact that when two quantities, each measured with a specific unit, are multiplied or divided, their units are multiplied or divided as well. For example, consider car traveling in a straight line at a constant speed of 20m/s. Suppose you want to know how far it travels in 30 seconds. To calculate the answer, simply multiply 20m/s by 30s, which is 600m. Notice that the meter per second times the second equals the meter.

A result of the above rule is the fact that both sides of an equation must have the same units. This makes sense because two quantities in different dimensions can't be compared. To illustrate this fact, consider the ideal gas law:

<math>PV=nRT</math>

where <math>P</math> is the pressure of the gas, <math>V</math> is its volume, <math>n</math> is the number moles of the gas, <math>R</math> is the ideal gas constant, and <math>T</math> is the temperature of the gas.

For the ideal gas law to be numerically true, the ideal gas constant <math>R</math> must be in the correct units. The SI units for pressure, volume, and temperature respectively are the kilopascal (kPa), the cubic meter (m<sup>3</sup>), and Kelvin (K). If these units are used, <math>R</math> must be given in m<sup>3</sup> kPa / mol K. That way, both sides of the equation are in m<sup>3</sup> kPa. R can also be given in other units, as long as they are a unit of volume times a unit of pressure divided by moles and a unit of temperature. Its value depends on the units used. In fact, all natural constants can have different values depending on the units in which they are given. In this class, they are typically given in their SI units.

All equations in physics, like <math>PV=nRT</math>, are true for all units, as long as their two sides agree in units, which requires the correct versions of any natural constants. Generally, SI units are preferred.

The fact that both sides of an equation must have the same units can be useful for determining the units of a quantity, or, if the units of all quantities are known, for verifying that a derived equation makes sense and is possible.

===Unit Conversion===

Measured quantities can be converted from one unit to another as long as the relationship between the two units is known. To do so, the quantity should be multiplied by the ratio of the two units' values, with the quantity's current unit in the denominator and the target unit in the numerator. This ratio is called a conversion factor. For example, suppose you want to convert 7cm to inches, given that an inch is equivalent to 2.54cm. 7cm should be multiplied by the ratio of the inch to the centimeter, which is <math>\frac{1\mathrm{in}}{2.54\mathrm{cm}}</math>:

<math>7\mathrm{cm} * \frac{1\mathrm{in}}{2.54\mathrm{cm}} = 2.76\mathrm{in}</math>.

It is also possible to use this method to convert between units that measure different dimensions, as long as they are proportional to each other and the relationship between them is known. For example, it is possible to convert from the volume of a sample of a substance to its mass if its density is known:

<math>.125\mathrm{m}^3\mathrm{ steel} * \frac{8050\mathrm{kg steel}}{1 \mathrm{m}^3\mathrm{ steel}} = 1006.25\mathrm{kg steel}</math>

Sometimes, it is necessary to perform many conversions in sequence. For example, consider a person who wants to convert from meters to feet, but the only relationship they know between SI and customary length units is that an inch is equivalent to 2.54cm. They would first have to convert meters to centimeters, then centimeters to inches, then inches to feet. In such a situation, it is often helpful to use a notation such as the one below to keep track of the conversions:

[[File:SIUnitsFenceConversion.png]]

Each column represents the conversion from one unit to another using a different conversion factor. Each unit in the top (except the target unit) should cancel with a different unit in the bottom.

It is also possible to convert between complex units with both a numerator and a denominator. The following chart converts from meters per second to miles per hour. Some columns are for converting from meters to miles and others are for converting from seconds to hours.

[[File:SIUnitsFenceConversion2.png]]

Finally, it is worth noting that when converting between higher powers of units, such as cubic inches to cubic cm, their conversion factors are raised to that power also. That is, one inch is equivalent to 2.54 cm, so one cubic inch is equivalent to 2.54<sup>3</sup> = 16.39 cubic cm.

==Examples==

===Simple===

The law of universal gravitation states that the magnitude of the gravitational force between two objects of masses <math>m_1</math> and <math>m_2</math> is given by

<math>|\vec{F}_{grav}| = \frac{Gm_1m_2}{r^2}</math>,

where <math>G</math> is the universal gravitational force constant and <math>r</math> is the distance between the objects' centers of mass.

Assuming that <math>|\vec{F}_{grav}|</math>, <math>m_1</math>, <math>m_2</math>, and <math>r</math> are given in their respective standard SI units, what units should G have?

Solution:

The left side of the equation is in N because that is the SI unit for force. The right side of the equation must therefore also be in N. Ignoring the G, the right side of the equation has units of kg<sup>2</sup>/m<sup>2</sup>. G must introduce the unit N in the numerator while canceling the kg<sup>2</sup>/m<sup>2</sup>, so G has units of N*m<sup>2</sup>/kg<sup>2</sup>. This is equivalent to m<sup>3</sup>/kgs<sup>2</sup> because 1N = 1kgm/s<sup>2</sup>. These two answers can also be written respectively as N*m<sup>2</sup>*kg<sup>-2</sup> and m<sup>3</sup>*kg<sup>-1</sup>*s<sup>-2</sup>. Any of the 4 answers is correct.

===Middling===

A rectangular prism made out of aluminum (density: 2,710kg/m<sup>3</sup>) has sides of length 1in, 2in, and 6in. 1 inch is equivalent to 2.54cm. What is the weight of the sample near the surface of the earth?

Solution:

The rectangular prism has a volume of 1in x 2in x 6in = 12in<sup>3</sup>. Let us use the unit conversion notation discussed earlier in the page to convert this volume to force:

[[File:SIUnitsAluminumBlockSoln.png]]

Note that although g is normally thought of as having as units m/s<sup>2</sup>, we used it above as having units N/kg. These units make more sense when it serves as a constant of proportionality between weight (measured in N) and mass (measured in kg). These units are equivalent to m/s<sup>2</sup>.

===Difficult===

A garage door has a width of 2.5m, a width of 2.2m, and a thickness of 4cm. The average density of the materials that comprise it is 273 kg/m<sup>3</sup>.

The garage works like most home garages; it can slide up and down along a rail to open and close respectively, and the rail has a 90<math>^\circ</math> turn (of small radius) in it so that when the garage door is closed, it is vertical, but when it is open, it hangs horizontally below the ceiling. When the garage is partially opened, the horizontal part of it is supported by the rail while the vertical part hangs freely. Because the garage door is intended to move at a constant speed, it is fitted with a spring designed to cancel the weight of the vertical part of the garage door regardless of the door's position. This makes it easier for the cable system to move the garage door. The spring obeys [[Hooke's Law]]. It is at its equilibrium length when the garage door is open and entirely supported by the rail, and as the garage door closes and more of its weight becomes unsupported, the spring extends to apply a force equal to the unsupported weight in the opposite direction.

[[File:SIUnitsGarageDoor.png]]

What should be the spring constant k of the spring?

Solution:

Let d be the length of the vertical part of the garage door, in m. The volume of the vertical part of the garage door is 2.5 x .04 x d m<sup>3</sup> = .1d m<sup>3</sup>. Its mass is .1d m<sup>3</sup> x 273 kg/m<sup>3</sup> = 27.3d kg. Its weight is 27.3d kg x 9.8 N/kg = 268d N. The displacement of the spring from its equilibrium length is also d due to the geometry of the garage door system. According to Hooke's Law, the spring must exert a force whose magnitude is given by f = kd, where k is the spring constant of the spring. The magnitude of the force of the spring must equal the weight of the vertical part of the garage door, which is 268d Newtons, so k must equal 268 N/m.

==Connectedness==

Physics can be thought of as the study of quantifying natural phenomena, which could not be done without some system of units allowing for measurements and calculations to be made. The SI unit system is a good candidate for a number of reasons: its units can apply to quantities on many different scales, new and existing units can be defined and redefined as necessary, and each unit is defined in such a way that they can be replicated with high precision by anyone at any place and time, provided they have the resources to perform certain experiments. The SI system is connected to all of physics (and, indeed, all of science) because practically the entire scientific community has agreed to use it for all measurements and calculations. A basic understanding of the SI system is necessary for any type of technical industry (engineering, architecture, etc.) in most countries outside the united states, where the SI system is used for such industries (the United States uses the customary system for all industries except pure science). An in-depth understanding of the SI system is necessary for certain industries with very small tolerances, such as the creation of precision machine tools.

==History==

The Metric System was created around the time of the French Revolution. On June 22, 1799, two platinum standards defining the meter and the kilogram were deposited in the Archives de la Republic in Paris. This can be seen as the first step in the development of the present International System of Units. The values of the meter and the kg were based on the physical world; the meter was based on the dimensions of the earth and the kilogram was based on the density of water. These two units formed the foundation of what was then called the metric system. The metric system was abandoned and then readopted by France, and slowly gained popularity in the scientific communities of other countries. [[James Maxwell]] proposed the addition of the second as a third base unit for time, which was quickly instated, and Giovanni Giorgi, an Italian physicist and electrical engineer, advocated for the addition of a fourth base unit to describe electromagnetic phenomena. The ampere was defined in 1935 for this purpose. By the 20th century, the metric system was commonplace across scientific communities across the world.

The SI unit system was established in 1960 by the 11th General Conference on Weights and Measures (CGPM), a part of the International Bureau of Weights and Measures (BIPM). The CGPM is an international authority on units recognized by most scientists which aims to standardized the units used around the world. The SI system adopted the four base units of the metric system and added three additional base units: the Kelvin, the candela, and the mole. Since then, the CGPM has continued to occasionally redefine the base SI units.

==See Also==

===Further reading===

SI Units for Clinical Measurement 1st Edition by Donald S. Young

===External Links===

https://www.youtube.com/watch?v=c_e1wITe_ig&t=521s (Video describing the redefinition of SI base units of May 20 2019)

http://wps.prenhall.com/wps/media/objects/165/169061/blb9ch0104.html (Pearson educational site)

http://physics.nist.gov/ (National institute of standards and Technology)

https://www.bipm.org/en/measurement-units/ (International Bureau of Weights and Measures site)

==References==

Matter & Interactions, Vol. I: Modern Mechanics, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

Tutorial & Drill Problems for General Chemistry (and Intro) By Walter S. Hamilton, Ph.D.

Base units of the SI, fundamental constants and modern quantum physics By Christian J Bordé, Published 15 September 2005

http://physics.nist.gov/.

https://www.bipm.org/en/measurement-units/

[[Category: The Scientific Method]]

SI Units

2021-11-29T17:08:19Z

Jnation8:

This page describes the International System of Units and lists its units.
'''Claimed By: Jared Nation'''
==The Main Idea==

The International System of Units is a set of units allowing for the quantification of each physical dimension. It was invented in France as an extension of the metric system under the name "Système Internationale d'Unités," which is why it is abbreviated to "SI". SI Units are practically universally used for scientific applications, and are used in most countries in everyday life as well, although the United States, which uses customary units, is a notable exception. Most of quantities in this course will be given in SI units.

The International System of Units has seven base units with specific definitions and countless derived units which are created by multiplying and dividing the base units. For example, the meter (m) is the base unit for the dimension distance and the second (s) is the base unit for the dimension time. The meter per second (m/s) measures the dimension speed and is a derived unit because it can be obtained by dividing the meter by the second.

Because measurements must often be made on many different scales, the International System of Units also defines a variety of [[#Prefixes|prefixes]] for use with its units. The prefixes scale the values of each unit by powers of 10.

The International System of Units is adaptable and can change over time to meet the evolving needs of scientists. Over the last several centuries, the number of base units has grown from 3 to 7 as a result of the discovery of new dimensions that could not be expressed in terms of existing base units. The number of derived units recognized as meaningful has also grown substantially and will likely continue to as the human understanding of the physical world improves. Additionally, the seven base units have undergone redefinition at several points in time, most recently on May 20 2019. The purpose of each redefinition is not to change the units' values (in fact, their values are changed as little as possible), but to make their definitions more universally constant and reproducible.

===The Seven Base Units===

The values of the seven base units are historically motivated, but most of their current definitions have been changed from the original. Today, each of the 7 base unit is defined by fixing a specific natural universal constant at a certain numerical value.

====Second (s)====

The second measures time.

Today, the second is defined by fixing the value of the cesium hyperfine transition frequency <math>\Delta v_{Cs}</math> at exactly 9192631770 hz (transitions per second). This defines the second as the 9192631770 times the period of the transition between the two levels of the ground state of cesium-133 atom. Atomic clocks make use of this phenomenon by counting the cesium transitions as a mechanism for keeping time! This definition was instated in 1967.

The second was initially defined as a fraction (1/86400) of a day of 24 hours of 60 minutes of 60 seconds. This definition is poor primarily because the duration of the day changes slightly over time. In 1956 the second was defined as 1/31556925.9747 of the tropical year. (The tropical year is the time it takes the sun to return to the same position in the season cycle, which is about 20 minutes shorter than the causal year.) However, this is still not as elegant as the current definition because it is based on our particular solar system, which may not always exist or be available for measuring.

====Meter (m)====

The meter measures distance.

Today, the meter is defined by fixing the value of the speed of light in a vacuum <math>c</math> at exactly 299,792,458 m/s. Since the second is already defined, this defines the meter as the distance traveled by light in a vacuum in 1/29979248<sup>th</sup> of a second.

The meter was originally defined in 1973 as 1/10,000,000 of the meridian through Paris between the North Pole and the Equator. In 1960 it was redefined as 1650763.73 wavelengths of the radiation in a vacuum corresponding to the transition between 2p10 and 5d5 transition levels of the krypton-86 atom.

====Kilogram (kg)====

The kilogram measures mass.

Today, the kilogram is defined by fixing the value of Planck's constant <math>h</math> at exactly 6.62607015×10−34 J s. Recall that the joule is not a base unit, but a derived unit equal to the kg m<sup>2</sup>/s<sup>2</sup>. Since the meter and the second are already defined, this leaves only one possible value for the kilogram. This definition was instated as part of the SI unit redefinition of May 20, 2019. [https://www.youtube.com/watch?v=c_e1wITe_ig&t=521s This video] by the science YouTube channel Veritasium describes the redefinition of base units of May 20, 2019, with an emphasis on the new definition of the kilogram.

In 1973, the SI unit for mass was called the grave. It was defined as being the mass of 1 liter of pure water at its freezing temperature. The grave was renamed to the kilogram, and later, in 1889, it was redefined as the mass of a specific sample of a platinum alloy. The sample is called "Le Grand K" and is kept in a humidity-controlled vault just outside of Paris, France. This definition was inadequate because it was not universally reproducible, so they considered a few alternate definitions, and, in 2019, redefined it to the current definition. [https://www.youtube.com/watch?v=ZMByI4s-D-Y This video from 2013], also by Veritasium, explains the shortcomings of Le Grand K and describes a proposed alternate definition that was being considered at that time (although it did not end up being the chosen definition).

====Ampere (A)====

The ampere measures electrical current.

Today the ampere is defined by fixing the value of the elementary charge <math>e</math> at exactly 1.602176634e-19 C. This defines the Coulomb, and since an ampere is 1 Coulomb per second (and the second is already defined), it also defines the Ampere.

In 1881, the ampere was defined as a tenth of the current required to create a magnetic field of one Oersted at the center of a 1cm arc of wire with curve radius 1cm. In 1946, the ampere was defined as the current required to create a force of 2e-7 Newtons per meter of length between two straight parallel conductors of infinite length placed 1m apart in a vacuum. [https://www.youtube.com/watch?v=YmCrFPC1qi4 This video] by the physics YouTube channel Kaustubhan describes the 1946 definition of the Ampere.

====Kelvin (K)====

The Kelvin measures temperature.

Today, the Kelvin is defined by fixing the value of the Boltzmann constant <math>k</math> at exactly 1.380649 x 10–23 J/K. Since the joule is a derived unit derived from the second, the meter, and the kilogram, each of which has already been defined, this leaves only one possible value for the Kelvin.

====Mole (mol)====

The mole measures the amount of a substance present by number of particles.

Today, the mole is defined by fixing the value of the Avogadro constant <math>N_A</math> at exactly 6.022140857 × 10<sup>23</sup> mol<sup>-1</sup>. This defines the mol to be 6.022140857 × 10<sup>23</sup> mol particles.

The mol was originally defined in 1967 the mole as the number of carbon-12 atoms necessary to comprise a 12 gram sample.

====Candela (cd)====

The candela measures luminous intensity.

Today, the candela is defined by fixing the value of the luminous efficacy constant <math>K_{cd}</math> at exactly 683 lumens per watt. This defines the candela as the luminous instensity of a source that emits radiation of a frequency 5.4e14 Hz in 1/683 of a steradian. (A steradian is the 3D analog for the radian; it is a part of the sphere's surface area equalling the radius squared in area.) This is about the luminous intensity of a candle.

===Derived Units===

There are many useful quantities in physics that cannot be measured in a single base unit. These must be expressed in derived units, which are obtained by multiplying and dividing the base units. For example, the meter (m) is the base unit for the dimension distance and the second (s) is the base unit for the dimension time. The meter per second (m/s) measures the dimension speed and is a derived unit because it can be obtained by dividing the meter by the second. Speed does not get its own base unit because it does not need one; speed is fundamentally a relationship between distance and time (specifically, distance traveled and the time it takes to travel that distance). Other examples of derived SI units are the kg/m<sup>3</sup>, which measures density, and the kgm/s, which measures momentum.

Some derived units are used so frequently in physics that they are given their own names. Below are several such derived units:

<table border="1">
<tr>
<th>Unit</th><th>Symbol</th><th>In Terms of Base Units</th><th>Dimension</th>
</tr>
<tr>
<td>hertz</td><td>Hz</td><td><math>\mathrm{s}^{-1}</math></td><td>frequency</td>
</tr>
<tr>
<td>newton</td><td>N</td><td><math>\frac{\mathrm{kg}\mathrm{m}}{\mathrm{s}^2}</math></td><td>force</td>
</tr>
<tr>
<td>pascal</td><td>Pa</td><td><math>\frac{\mathrm{kg}}{\mathrm{m}\mathrm{s}^2}</math></td><td>pressure</td>
</tr>
<tr>
<td>joule</td><td>J</td><td><math>\frac{\mathrm{kg}\mathrm{m}^2}{\mathrm{s}^2}</math></td><td>energy</td>
</tr>
<tr>
<td>watt</td><td>W</td><td><math>\frac{\mathrm{kg}\mathrm{m}^2}{\mathrm{s}^3}</math></td><td>power</td>
</tr>
<tr>
<td>coulomb</td><td>C</td><td><math>\mathrm{As}</math></td><td>charge</td>
</tr>
<tr>
<td>volt</td><td>V</td><td><math>\frac{\mathrm{kg}\mathrm{m}^2}{\mathrm{A}\mathrm{s}^3}</math></td><td>electric potential</td>
</tr>
</table>

This list is not comprehensive, and there are many unnamed derived units that are commonly used as well.

===Prefixes===

Because measurements must often be made on many different scales, the International System of Units also defines a variety of prefixes for use with its units. The prefixes scale the values of each unit by powers of 10. For example, the kilometer is 1000 times as long as the meter because of the prefix "kilo." The use of prefixes allows even large and small measurements to be reported with reasonable and easy to read numbers. For example, it is easier to express the atomic radius of the hydrogen atom as 53 picometers than 5.3 x 10<sup>-11</sup> meters.

Each prefix has a symbol, which can be combined with the symbol of a unit. For example, the symbol for pico is "p", so "pm" is the symbol for picometer.

The following table shows the common prefixes used with SI units.

<table border="1">
<tr>
<th>Prefix</th><th>Symbol</th><th>Multiplier</th>
</tr>
<tr>
<td>peta</td><td>P</td><td>10<sup>15</sup></td>
</tr>
<tr>
<td>tera</td><td>T</td><td>10<sup>12</sup></td>
</tr>
<tr>
<td>giga</td><td>G</td><td>10<sup>9</sup></td>
</tr>
<tr>
<td>mega</td><td>M</td><td>10<sup>6</sup></td>
</tr>
<tr>
<td>kilo</td><td>k</td><td>1000 (10<sup>3</sup>)</td>
</tr>
<tr>
<td>hecto</td><td>h</td><td>100 (10<sup>2</sup>)</td>
</tr>
<tr>
<td>deka</td><td>da</td><td>10 (10<sup>1</sup>)</td>
</tr>
<tr>
<td>deci</td><td>d</td><td>.1 (10<sup>-1</sup>)</td>
</tr>
<tr>
<td>centi</td><td>c</td><td>.01 (10<sup>-2</sup>)</td>
</tr>
<tr>
<td>milli</td><td>m</td><td>.001 (10<sup>-3</sup>)</td>
</tr>
<tr>
<td>micro</td><td><math>\mu</math></td><td>10<sup>-6</sup></td>
</tr>
<tr>
<td>nano</td><td>n</td><td>10<sup>-9</sup></td>
</tr>
<tr>
<td>pico</td><td>p</td><td>10<sup>-12</sup></td>
</tr>
<tr>
<td>femto</td><td>f</td><td>10<sup>-15</sup></td>
</tr>
</table>

There are several popular mnemonic devices for remembering some of the prefixes. A common example is "King Henry Died by Drinking Chocolate Milk." The first letter of each word represents a prefix, starting at kilo and ending at mili. The "b" in "by" stands for "base unit," which in this context means a unit with no prefixes.

===Dimensional Analysis===

Dimensional analysis is the analysis of the units used in expressions and equations. Dimensional analysis revolves around the fact that when two quantities, each measured with a specific unit, are multiplied or divided, their units are multiplied or divided as well. For example, consider car traveling in a straight line at a constant speed of 20m/s. Suppose you want to know how far it travels in 30 seconds. To calculate the answer, simply multiply 20m/s by 30s, which is 600m. Notice that the meter per second times the second equals the meter.

A result of the above rule is the fact that both sides of an equation must have the same units. This makes sense because two quantities in different dimensions can't be compared. To illustrate this fact, consider the ideal gas law:

<math>PV=nRT</math>

where <math>P</math> is the pressure of the gas, <math>V</math> is its volume, <math>n</math> is the number moles of the gas, <math>R</math> is the ideal gas constant, and <math>T</math> is the temperature of the gas.

For the ideal gas law to be numerically true, the ideal gas constant <math>R</math> must be in the correct units. The SI units for pressure, volume, and temperature respectively are the kilopascal (kPa), the cubic meter (m<sup>3</sup>), and Kelvin (K). If these units are used, <math>R</math> must be given in m<sup>3</sup> kPa / mol K. That way, both sides of the equation are in m<sup>3</sup> kPa. R can also be given in other units, as long as they are a unit of volume times a unit of pressure divided by moles and a unit of temperature. Its value depends on the units used. In fact, all natural constants can have different values depending on the units in which they are given. In this class, they are typically given in their SI units.

All equations in physics, like <math>PV=nRT</math>, are true for all units, as long as their two sides agree in units, which requires the correct versions of any natural constants. Generally, SI units are preferred.

The fact that both sides of an equation must have the same units can be useful for determining the units of a quantity, or, if the units of all quantities are known, for verifying that a derived equation makes sense and is possible.

===Unit Conversion===

Measured quantities can be converted from one unit to another as long as the relationship between the two units is known. To do so, the quantity should be multiplied by the ratio of the two units' values, with the quantity's current unit in the denominator and the target unit in the numerator. This ratio is called a conversion factor. For example, suppose you want to convert 7cm to inches, given that an inch is equivalent to 2.54cm. 7cm should be multiplied by the ratio of the inch to the centimeter, which is <math>\frac{1\mathrm{in}}{2.54\mathrm{cm}}</math>:

<math>7\mathrm{cm} * \frac{1\mathrm{in}}{2.54\mathrm{cm}} = 2.76\mathrm{in}</math>.

It is also possible to use this method to convert between units that measure different dimensions, as long as they are proportional to each other and the relationship between them is known. For example, it is possible to convert from the volume of a sample of a substance to its mass if its density is known:

<math>.125\mathrm{m}^3\mathrm{ steel} * \frac{8050\mathrm{kg steel}}{1 \mathrm{m}^3\mathrm{ steel}} = 1006.25\mathrm{kg steel}</math>

Sometimes, it is necessary to perform many conversions in sequence. For example, consider a person who wants to convert from meters to feet, but the only relationship they know between SI and customary length units is that an inch is equivalent to 2.54cm. They would first have to convert meters to centimeters, then centimeters to inches, then inches to feet. In such a situation, it is often helpful to use a notation such as the one below to keep track of the conversions:

[[File:SIUnitsFenceConversion.png]]

Each column represents the conversion from one unit to another using a different conversion factor. Each unit in the top (except the target unit) should cancel with a different unit in the bottom.

It is also possible to convert between complex units with both a numerator and a denominator. The following chart converts from meters per second to miles per hour. Some columns are for converting from meters to miles and others are for converting from seconds to hours.

[[File:SIUnitsFenceConversion2.png]]

Finally, it is worth noting that when converting between higher powers of units, such as cubic inches to cubic cm, their conversion factors are raised to that power also. That is, one inch is equivalent to 2.54 cm, so one cubic inch is equivalent to 2.54<sup>3</sup> = 16.39 cubic cm.

==Examples==

===Simple===

The law of universal gravitation states that the magnitude of the gravitational force between two objects of masses <math>m_1</math> and <math>m_2</math> is given by

<math>|\vec{F}_{grav}| = \frac{Gm_1m_2}{r^2}</math>,

where <math>G</math> is the universal gravitational force constant and <math>r</math> is the distance between the objects' centers of mass.

Assuming that <math>|\vec{F}_{grav}|</math>, <math>m_1</math>, <math>m_2</math>, and <math>r</math> are given in their respective standard SI units, what units should G have?

Solution:

The left side of the equation is in N because that is the SI unit for force. The right side of the equation must therefore also be in N. Ignoring the G, the right side of the equation has units of kg<sup>2</sup>/m<sup>2</sup>. G must introduce the unit N in the numerator while canceling the kg<sup>2</sup>/m<sup>2</sup>, so G has units of N*m<sup>2</sup>/kg<sup>2</sup>. This is equivalent to m<sup>3</sup>/kgs<sup>2</sup> because 1N = 1kgm/s<sup>2</sup>. These two answers can also be written respectively as N*m<sup>2</sup>*kg<sup>-2</sup> and m<sup>3</sup>*kg<sup>-1</sup>*s<sup>-2</sup>. Any of the 4 answers is correct.

===Middling===

A rectangular prism made out of aluminum (density: 2,710kg/m<sup>3</sup>) has sides of length 1in, 2in, and 6in. 1 inch is equivalent to 2.54cm. What is the weight of the sample near the surface of the earth?

Solution:

The rectangular prism has a volume of 1in x 2in x 6in = 12in<sup>3</sup>. Let us use the unit conversion notation discussed earlier in the page to convert this volume to force:

[[File:SIUnitsAluminumBlockSoln.png]]

Note that although g is normally thought of as having as units m/s<sup>2</sup>, we used it above as having units N/kg. These units make more sense when it serves as a constant of proportionality between weight (measured in N) and mass (measured in kg). These units are equivalent to m/s<sup>2</sup>.

===Difficult===

A garage door has a width of 2.5m, a width of 2.2m, and a thickness of 4cm. The average density of the materials that comprise it is 273 kg/m<sup>3</sup>.

The garage works like most home garages; it can slide up and down along a rail to open and close respectively, and the rail has a 90<math>^\circ</math> turn (of small radius) in it so that when the garage door is closed, it is vertical, but when it is open, it hangs horizontally below the ceiling. When the garage is partially opened, the horizontal part of it is supported by the rail while the vertical part hangs freely. Because the garage door is intended to move at a constant speed, it is fitted with a spring designed to cancel the weight of the vertical part of the garage door regardless of the door's position. This makes it easier for the cable system to move the garage door. The spring obeys [[Hooke's Law]]. It is at its equilibrium length when the garage door is open and entirely supported by the rail, and as the garage door closes and more of its weight becomes unsupported, the spring extends to apply a force equal to the unsupported weight in the opposite direction.

[[File:SIUnitsGarageDoor.png]]

What should be the spring constant k of the spring?

Solution:

Let d be the length of the vertical part of the garage door, in m. The volume of the vertical part of the garage door is 2.5 x .04 x d m<sup>3</sup> = .1d m<sup>3</sup>. Its mass is .1d m<sup>3</sup> x 273 kg/m<sup>3</sup> = 27.3d kg. Its weight is 27.3d kg x 9.8 N/kg = 268d N. The displacement of the spring from its equilibrium length is also d due to the geometry of the garage door system. According to Hooke's Law, the spring must exert a force whose magnitude is given by f = kd, where k is the spring constant of the spring. The magnitude of the force of the spring must equal the weight of the vertical part of the garage door, which is 268d Newtons, so k must equal 268 N/m.

==Connectedness==

Physics can be thought of as the study of quantifying natural phenomena, which could not be done without some system of units allowing for measurements and calculations to be made. The SI unit system is a good candidate for a number of reasons: its units can apply to quantities on many different scales, new and existing units can be defined and redefined as necessary, and each unit is defined in such a way that they can be replicated with high precision by anyone at any place and time, provided they have the resources to perform certain experiments. The SI system is connected to all of physics (and, indeed, all of science) because practically the entire scientific community has agreed to use it for all measurements and calculations. A basic understanding of the SI system is necessary for any type of technical industry (engineering, architecture, etc.) in most countries outside the united states, where the SI system is used for such industries (the United States uses the customary system for all industries except pure science). An in-depth understanding of the SI system is necessary for certain industries with very small tolerances, such as the creation of precision machine tools.

==History==

The Metric System was created around the time of the French Revolution. On June 22, 1799, two platinum standards defining the meter and the kilogram were deposited in the Archives de la Republic in Paris. This can be seen as the first step in the development of the present International System of Units. The values of the meter and the kg were based on the physical world; the meter was based on the dimensions of the earth and the kilogram was based on the density of water. These two units formed the foundation of what was then called the metric system. The metric system was abandoned and then readopted by France, and slowly gained popularity in the scientific communities of other countries. [[James Maxwell]] proposed the addition of the second as a third base unit for time, which was quickly instated, and Giovanni Giorgi, an Italian physicist and electrical engineer, advocated for the addition of a fourth base unit to describe electromagnetic phenomena. The ampere was defined in 1935 for this purpose. By the 20th century, the metric system was commonplace across scientific communities across the world.

The SI unit system was established in 1960 by the 11th General Conference on Weights and Measures (CGPM), a part of the International Bureau of Weights and Measures (BIPM). The CGPM is an international authority on units recognized by most scientists which aims to standardized the units used around the world. The SI system adopted the four base units of the metric system and added three additional base units: the Kelvin, the candela, and the mole. Since then, the CGPM has continued to occasionally redefine the base SI units.

==See Also==

===Further reading===

SI Units for Clinical Measurement 1st Edition by Donald S. Young

===External Links===

https://www.youtube.com/watch?v=c_e1wITe_ig&t=521s (Video describing the redefinition of SI base units of May 20 2019)

http://wps.prenhall.com/wps/media/objects/165/169061/blb9ch0104.html (Pearson educational site)

http://physics.nist.gov/ (National institute of standards and Technology)

https://www.bipm.org/en/measurement-units/ (International Bureau of Weights and Measures site)

==References==

Matter & Interactions, Vol. I: Modern Mechanics, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

Tutorial & Drill Problems for General Chemistry (and Intro) By Walter S. Hamilton, Ph.D.

Base units of the SI, fundamental constants and modern quantum physics By Christian J Bordé, Published 15 September 2005

http://physics.nist.gov/.

https://www.bipm.org/en/measurement-units/

[[Category: The Scientific Method]]

Free Body Diagram

2021-11-29T17:07:31Z

Jnation8:

==Main Idea==
A free body diagram, or force diagram, is a rough sketch that shows the relative magnitude and direction of all the forces acting on a system. There are various forces that can be acting on the object, such as applied force, frictional force, normal force, and gravitational force. A free body diagram allows for analysis in a steady state condition, where there can be no acceleration or a constant acceleration on a system.

[[File:FBDIntroductoryImage.jpg|center|400px|An Example Sketch and Free Body Diagram]]

All forces in a free body diagram are due to the system's interactions with its surroundings. Especially when problems become complicated and involve different forces acting on multiple objects, free body diagrams can be extremely effective in making problems simpler to handle. Using a free body diagram, one can solve for an unknown force acting on the body or the net acceleration after identifying all of the forces acting on it.

===Mathematical Model===
Since a free body diagram is a snapshot of a steady state, there is no net acceleration or a constant net acceleration when we are modeling a free body diagram. Therefore, we can use Newton's First Law (an object in an inertial frame at rest or moving with constant velocity, will continue to do so until acted upon by a net force), Newton's Second Law (the net force on an object is equivalent to the object's mass multiplied by its net acceleration) and Newton's Third Law (for every force there is an equal and opposite reactionary force) to create a system of equations to describe and gain new information about the steady state:

<math>\mathbf{F_{net}} = \sum \mathbf{F} = m \mathbf{a} = m \frac{d\mathbf{v}}{dt} </math> ('''Newton's Second Law''')

Additionally, if the net acceleration is equivalent to 0, which means the object is either at rest or moving with constant velocity, then:

<math>\mathbf{F_{net}} = \sum \mathbf{F} = m \mathbf{a} = m \frac{d\mathbf{v}}{dt} = \mathbf{0} </math> ('''Newton's First Law''')

Also, if it is known the net acceleration is '''0''' or one or more components of the net acceleration are 0, for example a<sub>x</sub> and a<sub>y</sub>, then the components of the forces along those directions must be balanced. ('''Newton's First and Third Laws''')

Using these strategies, a large set of force related problems can be solved.
===Computational Model===
The following is a VPython model of the acceleration of a box as it moves down an inclined plane with negligible friction. The box starts at the top of the inclined plane, which is given by <math>pos = (5,5,5)</math>, as shown by the accompanying diagram.

IMPORT STATEMENTS

<code>

from __future__ import division

from visual import *

import math

</code>

INITIAL CONDITIONS

<code>

scene.width = 1024

scene.height = 760

time = 0 #Time in seconds

dtime = .1 #The time will be iterated through every .1 seconds

init_pos = vector(0, 5, 5) #The initial position is at (x, y, z) = (0, 5, 5) meters

init_vel = vector(0, 0, 0) #The intial velocity is (0, 0, 0)

cube = box(pos = init_pos, length = 1, height = 1, width = 1) #Create a cube at the initial position

mass_cube = 10 #The mass of the cube is 10 kg

theta = 45 #The angle between the horizontal and the hypotneuse of the inclined plane is 45 degrees

force_gravity = mass_cube * 9.81 #Force of gravity on the cube in Newtons

thetaRads = math.radians(theta) #Convert degrees to radians

force_normal = force_gravity * math.cos(thetaRads)#Newton's First and Second Laws #The x-axis is chosen to be along the hypotenuse of the inclined plane, and the y-axis is chosen to be along the normal to the hypotenuse.

F_y = force_normal - force_gravity * math.cos(thetaRads) #This is equal to 0

F_x = force_gravity * math.sin(thetaRads) #The only force acting along the chosen x-axis is the component of the gravitational force parallel to the hypotenuse

a_x = F_x / mass_cube

</code>

ITERATE THROUGH TIME TO FOLLOW THE MOVEMENT OF THE CUBE DOWN THE INCLINED PLANE

<code>

while cube.pos.x < 5:
rate(200)
velocity = vector(init_vel.x + (dtime * a_x), init_vel.y, init_vel.z)
position = vector(cube.pos.x + (dtime * velocity.x), cube.pos.y + (dtime * velocity.y), cube.pos.z + (dtime * velocity.z))
cube.pos = position
trail.append(pos = cube.pos)
time = time + dtime
print(cube.pos)

</code>

=== How To Draw a Free Body Diagram ===

==== Establish a Convenient Coordinate System ====
Before beginning to analyze a system, it is important to choose an appropriate coordinate system (an x and y plane) that will be the most convenient to avoid dealing with complicated angles of forces and difficult algebra and trigonometry. This is especially true when dealing with angled forces where tilting the x and y plane could make solving for resulting forces a lot easier, such as on an inclined plane. Also, in order to maintain consistency with direction amongst various people solving the same system, an x-y coordinate system can ensure that anyone solving the same system will end up with the same directions on their forces.
[[File:Angled Axis.png|right|This is an example of how establishing a coordinate system can make solving/identifying forces easier]]

==== Identify the System ====
Often times, you will be faced with single and multi particle systems where bodies within that system will be interacting with/exerting forces on each other. For example, a block is resting on top of a larger block, and both are moving down a ramp. In that case, analyzing the top block versus the bottom block will result in different free body diagrams. Therefore, identifying which part of the system you are analyzing before beginning to draw a free body diagram will allow for a more efficient solving process.

==== List and identify all surroundings that interact with the system ====
Although we usually think of these interactions in terms of force names, it's best to get in the habit of identifying the force AND the object that specifically causes that force. For example, the force of gravity on a block could be caused by the earth or a normal force could be caused by a second block and another normal force could be caused by the table. Identifying forces by the specific objects that cause them helps us not forget where the forces belong.

=====Types of Forces to Consider for Free Body Diagrams=====

'''Disclaimer: Not all of these forces will be present in every situation. These are not all of the possible forces; These are very common ones.'''

*'''Applied Force''':
**Symbol: <math style="inline">\mathbf{F_A}</math>
**Force applied to the system by a person or other object.
*'''Force of Friction''':
**Symbol: <math>\mathbf{F_f}</math>
**Force that a surface applies on the system that is moving (or trying to move) on that surface.
**Formula: <math>\mathbf{F_{f}} = \mu \mathbf{N}</math>
***<math>\mathbf{N} =</math> Normal Force
***<math>\mu =</math> Coefficient of Friction
*'''Force of Gravity''':
**Symbol: <math>\mathbf{F_g}</math>
**Force that, on Earth, will act downward toward the center of the Earth. In general, gravity is the attraction of any mass to any other mass.
**Formula: <math>\mathbf{F_g} = \frac{G M_1 M_2}{r^2}{\hat{\mathbf{r}}}</math>
***This can be shown to be equal to <math>m\mathbf{g}</math> near the Earth's surface.
***<math>G = 6.7 \times 10^{-11}\ \frac{Nm^2}{{kg}^2}</math> (Gravitational Constant)
***<math>M_1 =</math> Mass 1
***<math>M_2 =</math> Mass 2
***<math>r =</math> Displacement between Masses 1 and 2
***<math>{\hat{\mathbf{r}}} =</math> Unit vector in direction of Displacement between the Masses
***<math>m =</math> Mass of object on Earth
***<math>\mathbf{g} = 9.8\ m/s^2 </math> (on earth)
*'''Normal Force''':
**Symbol: <math>\mathbf{F_N}</math>
**Force that is present when the system is on another object or surface, and the object or surface is exerting a force on the system as support.
*'''Spring Force'''
**Symbol: <math>\mathbf{F_s}</math>
**Force that is exerted by a spring on any system that is attached to it.
**Formula: <math>\mathbf{F_s} = -k\mathbf{x}</math> (Hookian Spring)
***<math>k =</math> Spring constant (<math>\ \frac{N}{m}</math>)
***<math>\mathbf{x} =</math> Displacement from the spring's equilibrium position
*'''Force of Tension''':
**Symbol: <math>\mathbf{F_T}</math> or <math>\mathbf{T}</math>
**Force that exists when a rope, string, wire, etc. is pulling on the system.

'''Side Note:'''
Now that we've touched on frictional forces, it is important to understand the types of frictional forces. Frictional forces include two types: static friction and dynamic (kinetic) friction. These two types of friction have different coefficients of friction, usually represented by <math>\mu_s</math> for static friction and <math>\mu_k</math> for dynamic (kinetic friction.
*'''Static Friction''' refers to the frictional force while the object is stationary. An external net force higher than this static frictional force is required to move the object.
*'''Dynamic (Kinetic) Friction''' refers to the frictional force while the object is in motion. For example, the object may experience a frictional force with the ground while it is moving along the ground.

==== Draw a diagram with the system at the center ====
#Use a dot to represent the system, OR
#You can draw the details of the system (Draw a block, car, etc.)
#Draw the conveniently oriented coordinate system on the free body diagram or next to the it
#Draw the system's relevant force vectors with correct relative magnitude and direction, pointing away from the dot or detailed drawing
[[File:ContactForce.JPG|300px|thumb|right|]]

==== Label all Forces With a Symbol ====
Symbols are useful to quickly represent the name of the force and identify them by the object causing the force. Refer to [[#Types of Forces to Consider for Free Body Diagrams| Common Symbols for Forces]].

==== Break forces into their components as needed. ====
If a force is acting diagonal to the system, create a dashed line parallel and perpendicular to the system and label it as the x and y components of that force. Use sine(θ) or cosine(θ) as needed.

'''General Tips:'''
*If an object has constant speed, it means the object has no acceleration. Since net force is the rate of change of acceleration, net force in that direction would then be zero. This means that there are either no forces currently acting on the object, or there are equal, opposite forces acting on the object in that direction. To represent this in a free body diagram, draw forces as arrows pointing in opposite directions with equal lengths.
*Don't be confused by contact forces. Most of the time, contact force is an umbrella term that includes other types of forces. If you had a block on a ramp, you could draw the contact force as being diagonal to the ramp. Or, if you wanted to break it into its components which are easier to consider, you would draw the Normal force caused by the ramp pushing up as a perpendicular arrow, and the force of Friction caused by the ramp as a parallel arrow (see image to the upper right).

==== Sum the forces ====
Once you have drawn the free body diagram and broken the forces into components along the convenient x-y axis (if needed), then it is time to use [[#Mathematical Model | Mathematical Model]] to create a system of equations summing the forces in the x and y directions and then set them equal to the correct value, such as 0 if there is no acceleration in that direction or <math>m\mathbf{a}</math>.

==Examples==
In this section we will go step by step through 3 illustrative examples, increasing in difficulty as we go.

=== Simple ===
A person with a mass of <math>M</math> rides up to the 20th floor of their apartment building in an elevator moving at constant velocity. Create a free body diagram to represent this situation. Using your free body diagram, use Newton's Second Law to create a force equation for the system. Make sure to keep track of the relative magnitudes of the forces and indicate these magnitudes by the length of the arrow.

'''''Answer'''''

We start by drawing a free body diagram with correct relative magnitudes and directions of the Normal Force and the Gravitational Force. The normal force is always normal (perpendicular or orthogonal) to the surface, whether the surface is angled or not. Gravity always acts downwards on Earth. Therefore, there is no need to create a different, convenient coordinate plane. A horizontal x-axis and vertical y-axis will suffice. These are the only two forces acting on the person because nothing else is pushing on the person, and friction is not being taken into account. We know the normal force must be present due to the person not falling through the floor (due to the always present gravitational force). We can also reason that these forces must be equal in magnitude and opposite in direction.

[[File:FBDSimpleSketch.jpeg|right|400px|A Sketch]]

[[File:FBDSimple.jpeg|right|200px|A Free Body Diagram]]

Next, we use Newton's Second Law to begin our analysis:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a}</math>

Next, we notice the problem states that the person moves with a constant velocity upwards, therefore there is no acceleration in any direction, and the net force must be equal to <math>\mathbf{0}</math>:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a} = \mathbf{0}</math>

The normal force and gravitational force can be described as:

:<math display="block">\mathbf{F_g} = \begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(180^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(180^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ -|\mathbf{F_g}| \end{bmatrix}</math>

:<math display="block">\mathbf{F_N} = \begin{bmatrix} |\mathbf{F_N}| \ \text{cos}(90^\text{o}) \\ |\mathbf{F_N}| \ \text{sin}(90^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ |\mathbf{F_N}| \end{bmatrix}</math>

Next, putting the free body diagram and Newton's Second Law together gives:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = \begin{bmatrix} 0 \\ \mathbf{|F_N|} \end{bmatrix} + \begin{bmatrix} 0 \\ -\mathbf{|F_g|} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}</math>

Therefore, it is possible to simplify the above to show that the gravitational force is exactly opposite in direction and equal in magnitude:

:<math display="block">|\mathbf{F_g}| = |\mathbf{F_N}|</math>
:<math display="block">|\mathbf{F_N}| = Mg </math>

We have fully done what the problem asked.

=== Middling ===
A ball with mass <math>m</math> is hanging tautly from two strings of negligible mass, each of which are connected to the ceiling. To the left of the ball, one string has a tension <math>\mathbf{T_1}</math> and makes an angle of <math>20^\text{o}</math> with the ceiling, and on the right of the ball, the other string has a tension <math>\mathbf{T_2}</math> and creates an angle of <math>45^\text{o}</math> with the ceiling. Both angles are measured from the ceiling to the inner side of the string. Refer to the diagram for any confusion. Create a free body diagram to model this situation. Also, create a system of force equations. '''Hint''': x and y components will be needed for this problem.

'''''Answer'''''

To draw the free body diagram it is better to first draw a sketch of the system. In the sketch we see the string to the left (<math>\mathbf{T_1}</math>) of the ball makes an angle of <math>20^\text{o}</math> with the ceiling, and the string to the right (<math>\mathbf{T_2}</math>) of the ball makes an angle <math>45^\text{o}</math> with the ceiling. For the free body diagram, we first draw the force of gravity pointing straight downwards. This force should have the longest arrow due to the fact that the strings are supporting this force...therefore, it can be reasoned that this force should be greater than either of the tensions in the strings for the given angles. Next, we draw the tension forces, which point away from the ball along each respective string. <math>\mathbf{T_2}</math> should be longer than <math>\mathbf{T_1}</math> due to the fact that the tensions must balance each other along the x-axis and the y-component of <math>\mathbf{T_2}</math> must be greater than the y-component of <math>\mathbf{T_1}</math>.

[[File:FBDMiddlingSketch.jpeg|right| 400px|A Sketch]]

[[File:FBDMiddling.jpeg|right|300px|A Free Body Diagram]]

Now, we are ready to use Newton's First and Second Laws to create a system of force equations:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a}</math>

Since the ball is hanging tautly, it must NOT be moving. Therefore, the velocity and acceleration must both be <math>\mathbf{0}</math>:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a} = \mathbf{0}</math>

Next, we choose the x-axis to be horizontal, and the y-axis to be vertical, as in a typical coordinate system, since it is convenient.

Now, we begin specifically identifying the forces and their components using right triangles:

:<math display="block">\mathbf{F_g} = \begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(180^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(180^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ -|\mathbf{F_g}| \end{bmatrix}</math>

:<math display="block">\mathbf{T_1} = \begin{bmatrix} -|\mathbf{T_1}| \ \text{cos}(20^\text{o}) \\ |\mathbf{T_1}| \ \text{sin}(20^\text{o})\end{bmatrix} </math>

:<math display="block">\mathbf{T_2} = \begin{bmatrix} -|\mathbf{T_2}| \ \text{cos}(45^\text{o}) \\ |\mathbf{T_2}| \ \text{sin}(45^\text{o})\end{bmatrix}</math>

Next, we use Newton's Second Law to sum these forces. Since the ball is not moving, we can set this summation equal to <math>\mathbf{0}</math>:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a} = \mathbf{0} = \mathbf{F_g} + \mathbf{T_1} + \mathbf{T_2}</math>

This is equivalent to:

:<math display="block"> \begin{bmatrix} 0 \\ -|\mathbf{F_g}| \end{bmatrix} + \begin{bmatrix} -|\mathbf{T_1}| \ \text{cos}(20^\text{o}) \\ |\mathbf{T_1}| \ \text{sin}(20^\text{o})\end{bmatrix} + \begin{bmatrix} -|\mathbf{T_2}| \ \text{cos}(45^\text{o}) \\ |\mathbf{T_2}| \ \text{sin}(45^\text{o})\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}</math>

Hence, we have done what the question asked and are done with this problem.

=== Difficult ===
A father pushes his son's sled horizontally with a force <math>\mathbf{F_A}</math> all the way down a hill on a sled of mass <math>M</math>. The son has a mass <math>m</math>. The son/sled system is accelerating down the hill. Assume the hill is shaped like an inclined plane with an angle to the horizontal ground of <math>35^{\text{o}}</math> and that friction works against the sled with a coefficient of dynamic (kinetic) friction <math>\mu_k</math>. Draw a free-body diagram. Using your free body diagram, write a system of equations that could be used to solve for the acceleration of the son/sled system.

'''''Answer'''''

We start with a sketch. In this sketch, we create an inclined plane with the the specified <math>35^{\text{o}}</math> angle, with the dad behind the kid pushing the sled down the hill. Some forces to be thinking about in this situation are the normal force, the applied force of the dad, the frictional force opposing the motion, and the gravitational force pulling the system straight down. To begin the free body diagram, we draw a circle to symbolize the son/sled system. Next, we draw the gravitational force as straight down, as it will always be in these Earth situations. Its length should be longer than the normal force. Otherwise, it shouldn't be compared to the other forces. Next are the frictional force and the normal force. The normal force should be normal to the inclined plane, and the frictional force should be parallel to the inclined plane, but opposite the direction of motion. Finally, the applied force of the dad should be parallel to the flat ground. The free body diagram has been created. Now, we must create a system of equations that could be solved for the acceleration of the son/sled system.

[[File:FBDSketchDifficult.jpeg|right|400px|A Sketch]]

[[File:FBDDifficult.jpeg|right|400px|A Free Body Diagram]]

To begin, we draw an x-y plane, where the y-axis runs along the the normal force and the x-axis runs along the frictional force. This will simplify the representation of these two forces, as well as represent the gravitational force with just a couple trigonometric functions. The applied force will have the same done to it.

Newton's Second Law states the net force on a system is equal to the mass of the system times the acceleration:

:<math display="block">\mathbf{F_{net}} = \sum \mathbf{F} = m_{system}\mathbf{a}</math>

We are told the son/sled system is accelerating, therefore the above equation can NOT be set equal to <math>\mathbf{0}</math>.

We also know the son/sled system should have a combined mass of <math>M + m</math>, where <math>M</math> is the mass of the sled and <math>m</math> is the mass of the son, as stated in the problem:

:<math display="block">\mathbf{F_{net}} = \sum \mathbf{F} = m_{system}\mathbf{a} = (M + m) \mathbf{a}</math>

The angle <math>\mathbf{F_A}</math> makes with the chosen x-axis is <math>35^{\text{o}}</math>. This can be seen from the sketch. The angle the gravitational force makes with the chosen y-axis is also <math>35^{\text{o}}</math>. This can also be seen in the sketch. Therefore, the four forces considered in this system can be described as:

:<math display="block">\mathbf{F_g} = \begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(35^\text{o})\end{bmatrix}</math>

:<math display="block">\mathbf{F_A} = \begin{bmatrix} |\mathbf{F_A}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_A}| \ \text{sin}(35^\text{o}) \end{bmatrix}</math>

:<math display="block">\mathbf{F_f} = \begin{bmatrix} |\mathbf{F_f}| \ \text{cos}(180^\text{o}) \\ |\mathbf{F_f}| \ \text{sin}(180^\text{o}) \end{bmatrix} = \begin{bmatrix} -|\mathbf{F_f}| \\ 0 \end{bmatrix}</math>

:<math display="block">\mathbf{F_N} = \begin{bmatrix} |\mathbf{F_N}| \ \text{cos}(90^\text{o}) \\ |\mathbf{F_N}| \ \text{sin}(90^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ |\mathbf{F_N}| \end{bmatrix}</math>

Using Newton's Second Law, we know that:

:<math display="block">\mathbf{F_{net}} = \sum \mathbf{F} = m_{system}\mathbf{a} = (M + m) \mathbf{a} = \mathbf{F_g} + \mathbf{F_A} + \mathbf{F_f} + \mathbf{F_N}</math>

This is equivalent to:

:<math display="block">\mathbf{a} = \frac{\begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(35^\text{o})\end{bmatrix} + \begin{bmatrix} |\mathbf{F_A}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_A}| \ \text{sin}(35^\text{o}) \end{bmatrix} + \begin{bmatrix} -|\mathbf{F_A}| \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ |\mathbf{F_N}| \end{bmatrix}}{M+m}</math>

Thus we have done what the problem asked.

==Connectedness==
Free body diagrams are the building blocks for many scientists, physicists, engineers and have many applications in the real world. They form the basis for determining a device's or building's structural stability, safety, and overall practicality.

'''Medical Applications'''
[[File:Leg.gif|right|]]
There have recently been many advances in medicine in terms of wearables and prosthetic limbs that function just like normal body parts for either congenital defects or accident recovery. Let's take a look at a leg prosthetic. While walking, a prosthetic exerts many forces on your leg such as: the force of gravity, normal force from the ground, frictional force, etc.
Prosthetics move like springs, while your body moves forward it also moves up and down which causes changes in kinetic energy due to movement and changes in gravitational potential energy. While it contracts and relaxes, different energies are converted back and forth into kinetic and potential energy. In order to analyze all the forces on a prosthetic to ensure it's safety and usability, the most basic step when designing one is a free body diagram.

'''Buildings and Trusses'''

Many civil engineers and architects are responsible for building trusses, bridges, and buildings which function using the basic concepts of physics. It is important to ensure that components of a truss and bridge are in equilibrium so that one day they don't collapse.
When building a truss bridge, engineers put straight members in place that form the bridge's top and bottom, and they are linked by a structure of diagonals and vertical posts. For trusses especially, all the components must remain in equilibrium, the magnitudes of forces exerted must be equal, and the components in tension and compression must be identified. In order to create a proper structure, a free body diagram must be used by engineers in the preliminary stages for design and safety purposes.

[[File:Truss.gif]]

==History==

Free Body Diagrams were developed alongside Newtonian physics, as a graphical way to analyze what forces were acting on a system, and how the system would react.
Because of their widespread usefulness, free body diagrams have not fallen out of favor with scientists and engineers.

Free body diagrams have been used as a teaching and real world tool for well over a century.

==See Also==

===Further Reading===
[[Inclined Plane]]

[[Compression or Normal Force]]

[[Tension]]

[[Gravitational Force]]

[[Spring Force]]

[[Hooke's Law]]

[[Net Force]]

===External Links===
[http://hyperphysics.phy-astr.gsu.edu/hbase/force.html Forces Mental Map]

[http://www.physicsclassroom.com/class/newtlaws/Lesson-3/Newton-s-Second-Law Newton's Second Law]

[https://www.youtube.com/watch?v=nDis6HbXxjg Using Free-Body Diagrams to Solve Kinematics Problems]

== References ==
Matter and Interactions: Modern Mechanics. Volume One. 4th Edition.

http://demos.smu.ca/index.php/demos/mechanics/141-free-body-diagram

http://hyperphysics.phy-astr.gsu.edu/hbase/freeb.html

https://www.wisc-online.com/learn/natural-science/physics/tp1502/construction-of-free-body-diagrams

http://www.physicsclassroom.com/Class/newtlaws

https://sites.google.com/a/cpsdigital.org/peraplegic/human-prosthetics

http://slideplayer.com/slide/6086550/ - Tilted Axis Image

http://www.revistas.unal.edu.co/index.php/dyna/rt/printerFriendly/30749/39025 - Prosthetic Limb Image

http://www.mathalino.com/reviewer/engineering-mechanics/241-finding-resulatnt-vertical-forces-acting-fink-truss - Truss Image

Free Body Diagram

2021-11-29T17:03:39Z

Jnation8:

'''Claimed By: Jared Nation Fall 2021
'''
==Main Idea==
A free body diagram, or force diagram, is a rough sketch that shows the relative magnitude and direction of all the forces acting on a system. There are various forces that can be acting on the object, such as applied force, frictional force, normal force, and gravitational force. A free body diagram allows for analysis in a steady state condition, where there can be no acceleration or a constant acceleration on a system.

[[File:FBDIntroductoryImage.jpg|center|400px|An Example Sketch and Free Body Diagram]]

All forces in a free body diagram are due to the system's interactions with its surroundings. Especially when problems become complicated and involve different forces acting on multiple objects, free body diagrams can be extremely effective in making problems simpler to handle. Using a free body diagram, one can solve for an unknown force acting on the body or the net acceleration after identifying all of the forces acting on it.

===Mathematical Model===
Since a free body diagram is a snapshot of a steady state, there is no net acceleration or a constant net acceleration when we are modeling a free body diagram. Therefore, we can use Newton's First Law (an object in an inertial frame at rest or moving with constant velocity, will continue to do so until acted upon by a net force), Newton's Second Law (the net force on an object is equivalent to the object's mass multiplied by its net acceleration) and Newton's Third Law (for every force there is an equal and opposite reactionary force) to create a system of equations to describe and gain new information about the steady state:

<math>\mathbf{F_{net}} = \sum \mathbf{F} = m \mathbf{a} = m \frac{d\mathbf{v}}{dt} </math> ('''Newton's Second Law''')

Additionally, if the net acceleration is equivalent to 0, which means the object is either at rest or moving with constant velocity, then:

<math>\mathbf{F_{net}} = \sum \mathbf{F} = m \mathbf{a} = m \frac{d\mathbf{v}}{dt} = \mathbf{0} </math> ('''Newton's First Law''')

Also, if it is known the net acceleration is '''0''' or one or more components of the net acceleration are 0, for example a<sub>x</sub> and a<sub>y</sub>, then the components of the forces along those directions must be balanced. ('''Newton's First and Third Laws''')

Using these strategies, a large set of force related problems can be solved.
===Computational Model===
The following is a VPython model of the acceleration of a box as it moves down an inclined plane with negligible friction. The box starts at the top of the inclined plane, which is given by <math>pos = (5,5,5)</math>, as shown by the accompanying diagram.

IMPORT STATEMENTS

<code>

from __future__ import division

from visual import *

import math

</code>

INITIAL CONDITIONS

<code>

scene.width = 1024

scene.height = 760

time = 0 #Time in seconds

dtime = .1 #The time will be iterated through every .1 seconds

init_pos = vector(0, 5, 5) #The initial position is at (x, y, z) = (0, 5, 5) meters

init_vel = vector(0, 0, 0) #The intial velocity is (0, 0, 0)

cube = box(pos = init_pos, length = 1, height = 1, width = 1) #Create a cube at the initial position

mass_cube = 10 #The mass of the cube is 10 kg

theta = 45 #The angle between the horizontal and the hypotneuse of the inclined plane is 45 degrees

force_gravity = mass_cube * 9.81 #Force of gravity on the cube in Newtons

thetaRads = math.radians(theta) #Convert degrees to radians

force_normal = force_gravity * math.cos(thetaRads)#Newton's First and Second Laws #The x-axis is chosen to be along the hypotenuse of the inclined plane, and the y-axis is chosen to be along the normal to the hypotenuse.

F_y = force_normal - force_gravity * math.cos(thetaRads) #This is equal to 0

F_x = force_gravity * math.sin(thetaRads) #The only force acting along the chosen x-axis is the component of the gravitational force parallel to the hypotenuse

a_x = F_x / mass_cube

</code>

ITERATE THROUGH TIME TO FOLLOW THE MOVEMENT OF THE CUBE DOWN THE INCLINED PLANE

<code>

while cube.pos.x < 5:
rate(200)
velocity = vector(init_vel.x + (dtime * a_x), init_vel.y, init_vel.z)
position = vector(cube.pos.x + (dtime * velocity.x), cube.pos.y + (dtime * velocity.y), cube.pos.z + (dtime * velocity.z))
cube.pos = position
trail.append(pos = cube.pos)
time = time + dtime
print(cube.pos)

</code>

=== How To Draw a Free Body Diagram ===

==== Establish a Convenient Coordinate System ====
Before beginning to analyze a system, it is important to choose an appropriate coordinate system (an x and y plane) that will be the most convenient to avoid dealing with complicated angles of forces and difficult algebra and trigonometry. This is especially true when dealing with angled forces where tilting the x and y plane could make solving for resulting forces a lot easier, such as on an inclined plane. Also, in order to maintain consistency with direction amongst various people solving the same system, an x-y coordinate system can ensure that anyone solving the same system will end up with the same directions on their forces.
[[File:Angled Axis.png|right|This is an example of how establishing a coordinate system can make solving/identifying forces easier]]

==== Identify the System ====
Often times, you will be faced with single and multi particle systems where bodies within that system will be interacting with/exerting forces on each other. For example, a block is resting on top of a larger block, and both are moving down a ramp. In that case, analyzing the top block versus the bottom block will result in different free body diagrams. Therefore, identifying which part of the system you are analyzing before beginning to draw a free body diagram will allow for a more efficient solving process.

==== List and identify all surroundings that interact with the system ====
Although we usually think of these interactions in terms of force names, it's best to get in the habit of identifying the force AND the object that specifically causes that force. For example, the force of gravity on a block could be caused by the earth or a normal force could be caused by a second block and another normal force could be caused by the table. Identifying forces by the specific objects that cause them helps us not forget where the forces belong.

=====Types of Forces to Consider for Free Body Diagrams=====

'''Disclaimer: Not all of these forces will be present in every situation. These are not all of the possible forces; These are very common ones.'''

*'''Applied Force''':
**Symbol: <math style="inline">\mathbf{F_A}</math>
**Force applied to the system by a person or other object.
*'''Force of Friction''':
**Symbol: <math>\mathbf{F_f}</math>
**Force that a surface applies on the system that is moving (or trying to move) on that surface.
**Formula: <math>\mathbf{F_{f}} = \mu \mathbf{N}</math>
***<math>\mathbf{N} =</math> Normal Force
***<math>\mu =</math> Coefficient of Friction
*'''Force of Gravity''':
**Symbol: <math>\mathbf{F_g}</math>
**Force that, on Earth, will act downward toward the center of the Earth. In general, gravity is the attraction of any mass to any other mass.
**Formula: <math>\mathbf{F_g} = \frac{G M_1 M_2}{r^2}{\hat{\mathbf{r}}}</math>
***This can be shown to be equal to <math>m\mathbf{g}</math> near the Earth's surface.
***<math>G = 6.7 \times 10^{-11}\ \frac{Nm^2}{{kg}^2}</math> (Gravitational Constant)
***<math>M_1 =</math> Mass 1
***<math>M_2 =</math> Mass 2
***<math>r =</math> Displacement between Masses 1 and 2
***<math>{\hat{\mathbf{r}}} =</math> Unit vector in direction of Displacement between the Masses
***<math>m =</math> Mass of object on Earth
***<math>\mathbf{g} = 9.8\ m/s^2 </math> (on earth)
*'''Normal Force''':
**Symbol: <math>\mathbf{F_N}</math>
**Force that is present when the system is on another object or surface, and the object or surface is exerting a force on the system as support.
*'''Spring Force'''
**Symbol: <math>\mathbf{F_s}</math>
**Force that is exerted by a spring on any system that is attached to it.
**Formula: <math>\mathbf{F_s} = -k\mathbf{x}</math> (Hookian Spring)
***<math>k =</math> Spring constant (<math>\ \frac{N}{m}</math>)
***<math>\mathbf{x} =</math> Displacement from the spring's equilibrium position
*'''Force of Tension''':
**Symbol: <math>\mathbf{F_T}</math> or <math>\mathbf{T}</math>
**Force that exists when a rope, string, wire, etc. is pulling on the system.

'''Side Note:'''
Now that we've touched on frictional forces, it is important to understand the types of frictional forces. Frictional forces include two types: static friction and dynamic (kinetic) friction. These two types of friction have different coefficients of friction, usually represented by <math>\mu_s</math> for static friction and <math>\mu_k</math> for dynamic (kinetic friction.
*'''Static Friction''' refers to the frictional force while the object is stationary. An external net force higher than this static frictional force is required to move the object.
*'''Dynamic (Kinetic) Friction''' refers to the frictional force while the object is in motion. For example, the object may experience a frictional force with the ground while it is moving along the ground.

==== Draw a diagram with the system at the center ====
#Use a dot to represent the system, OR
#You can draw the details of the system (Draw a block, car, etc.)
#Draw the conveniently oriented coordinate system on the free body diagram or next to the it
#Draw the system's relevant force vectors with correct relative magnitude and direction, pointing away from the dot or detailed drawing
[[File:ContactForce.JPG|300px|thumb|right|]]

==== Label all Forces With a Symbol ====
Symbols are useful to quickly represent the name of the force and identify them by the object causing the force. Refer to [[#Types of Forces to Consider for Free Body Diagrams| Common Symbols for Forces]].

==== Break forces into their components as needed. ====
If a force is acting diagonal to the system, create a dashed line parallel and perpendicular to the system and label it as the x and y components of that force. Use sine(θ) or cosine(θ) as needed.

'''General Tips:'''
*If an object has constant speed, it means the object has no acceleration. Since net force is the rate of change of acceleration, net force in that direction would then be zero. This means that there are either no forces currently acting on the object, or there are equal, opposite forces acting on the object in that direction. To represent this in a free body diagram, draw forces as arrows pointing in opposite directions with equal lengths.
*Don't be confused by contact forces. Most of the time, contact force is an umbrella term that includes other types of forces. If you had a block on a ramp, you could draw the contact force as being diagonal to the ramp. Or, if you wanted to break it into its components which are easier to consider, you would draw the Normal force caused by the ramp pushing up as a perpendicular arrow, and the force of Friction caused by the ramp as a parallel arrow (see image to the upper right).

==== Sum the forces ====
Once you have drawn the free body diagram and broken the forces into components along the convenient x-y axis (if needed), then it is time to use [[#Mathematical Model | Mathematical Model]] to create a system of equations summing the forces in the x and y directions and then set them equal to the correct value, such as 0 if there is no acceleration in that direction or <math>m\mathbf{a}</math>.

==Examples==
In this section we will go step by step through 3 illustrative examples, increasing in difficulty as we go.

=== Simple ===
A person with a mass of <math>M</math> rides up to the 20th floor of their apartment building in an elevator moving at constant velocity. Create a free body diagram to represent this situation. Using your free body diagram, use Newton's Second Law to create a force equation for the system. Make sure to keep track of the relative magnitudes of the forces and indicate these magnitudes by the length of the arrow.

'''''Answer'''''

We start by drawing a free body diagram with correct relative magnitudes and directions of the Normal Force and the Gravitational Force. The normal force is always normal (perpendicular or orthogonal) to the surface, whether the surface is angled or not. Gravity always acts downwards on Earth. Therefore, there is no need to create a different, convenient coordinate plane. A horizontal x-axis and vertical y-axis will suffice. These are the only two forces acting on the person because nothing else is pushing on the person, and friction is not being taken into account. We know the normal force must be present due to the person not falling through the floor (due to the always present gravitational force). We can also reason that these forces must be equal in magnitude and opposite in direction.

[[File:FBDSimpleSketch.jpeg|right|400px|A Sketch]]

[[File:FBDSimple.jpeg|right|200px|A Free Body Diagram]]

Next, we use Newton's Second Law to begin our analysis:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a}</math>

Next, we notice the problem states that the person moves with a constant velocity upwards, therefore there is no acceleration in any direction, and the net force must be equal to <math>\mathbf{0}</math>:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a} = \mathbf{0}</math>

The normal force and gravitational force can be described as:

:<math display="block">\mathbf{F_g} = \begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(180^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(180^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ -|\mathbf{F_g}| \end{bmatrix}</math>

:<math display="block">\mathbf{F_N} = \begin{bmatrix} |\mathbf{F_N}| \ \text{cos}(90^\text{o}) \\ |\mathbf{F_N}| \ \text{sin}(90^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ |\mathbf{F_N}| \end{bmatrix}</math>

Next, putting the free body diagram and Newton's Second Law together gives:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = \begin{bmatrix} 0 \\ \mathbf{|F_N|} \end{bmatrix} + \begin{bmatrix} 0 \\ -\mathbf{|F_g|} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}</math>

Therefore, it is possible to simplify the above to show that the gravitational force is exactly opposite in direction and equal in magnitude:

:<math display="block">|\mathbf{F_g}| = |\mathbf{F_N}|</math>
:<math display="block">|\mathbf{F_N}| = Mg </math>

We have fully done what the problem asked.

=== Middling ===
A ball with mass <math>m</math> is hanging tautly from two strings of negligible mass, each of which are connected to the ceiling. To the left of the ball, one string has a tension <math>\mathbf{T_1}</math> and makes an angle of <math>20^\text{o}</math> with the ceiling, and on the right of the ball, the other string has a tension <math>\mathbf{T_2}</math> and creates an angle of <math>45^\text{o}</math> with the ceiling. Both angles are measured from the ceiling to the inner side of the string. Refer to the diagram for any confusion. Create a free body diagram to model this situation. Also, create a system of force equations. '''Hint''': x and y components will be needed for this problem.

'''''Answer'''''

To draw the free body diagram it is better to first draw a sketch of the system. In the sketch we see the string to the left (<math>\mathbf{T_1}</math>) of the ball makes an angle of <math>20^\text{o}</math> with the ceiling, and the string to the right (<math>\mathbf{T_2}</math>) of the ball makes an angle <math>45^\text{o}</math> with the ceiling. For the free body diagram, we first draw the force of gravity pointing straight downwards. This force should have the longest arrow due to the fact that the strings are supporting this force...therefore, it can be reasoned that this force should be greater than either of the tensions in the strings for the given angles. Next, we draw the tension forces, which point away from the ball along each respective string. <math>\mathbf{T_2}</math> should be longer than <math>\mathbf{T_1}</math> due to the fact that the tensions must balance each other along the x-axis and the y-component of <math>\mathbf{T_2}</math> must be greater than the y-component of <math>\mathbf{T_1}</math>.

[[File:FBDMiddlingSketch.jpeg|right| 400px|A Sketch]]

[[File:FBDMiddling.jpeg|right|300px|A Free Body Diagram]]

Now, we are ready to use Newton's First and Second Laws to create a system of force equations:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a}</math>

Since the ball is hanging tautly, it must NOT be moving. Therefore, the velocity and acceleration must both be <math>\mathbf{0}</math>:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a} = \mathbf{0}</math>

Next, we choose the x-axis to be horizontal, and the y-axis to be vertical, as in a typical coordinate system, since it is convenient.

Now, we begin specifically identifying the forces and their components using right triangles:

:<math display="block">\mathbf{F_g} = \begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(180^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(180^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ -|\mathbf{F_g}| \end{bmatrix}</math>

:<math display="block">\mathbf{T_1} = \begin{bmatrix} -|\mathbf{T_1}| \ \text{cos}(20^\text{o}) \\ |\mathbf{T_1}| \ \text{sin}(20^\text{o})\end{bmatrix} </math>

:<math display="block">\mathbf{T_2} = \begin{bmatrix} -|\mathbf{T_2}| \ \text{cos}(45^\text{o}) \\ |\mathbf{T_2}| \ \text{sin}(45^\text{o})\end{bmatrix}</math>

Next, we use Newton's Second Law to sum these forces. Since the ball is not moving, we can set this summation equal to <math>\mathbf{0}</math>:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a} = \mathbf{0} = \mathbf{F_g} + \mathbf{T_1} + \mathbf{T_2}</math>

This is equivalent to:

:<math display="block"> \begin{bmatrix} 0 \\ -|\mathbf{F_g}| \end{bmatrix} + \begin{bmatrix} -|\mathbf{T_1}| \ \text{cos}(20^\text{o}) \\ |\mathbf{T_1}| \ \text{sin}(20^\text{o})\end{bmatrix} + \begin{bmatrix} -|\mathbf{T_2}| \ \text{cos}(45^\text{o}) \\ |\mathbf{T_2}| \ \text{sin}(45^\text{o})\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}</math>

Hence, we have done what the question asked and are done with this problem.

=== Difficult ===
A father pushes his son's sled horizontally with a force <math>\mathbf{F_A}</math> all the way down a hill on a sled of mass <math>M</math>. The son has a mass <math>m</math>. The son/sled system is accelerating down the hill. Assume the hill is shaped like an inclined plane with an angle to the horizontal ground of <math>35^{\text{o}}</math> and that friction works against the sled with a coefficient of dynamic (kinetic) friction <math>\mu_k</math>. Draw a free-body diagram. Using your free body diagram, write a system of equations that could be used to solve for the acceleration of the son/sled system.

'''''Answer'''''

We start with a sketch. In this sketch, we create an inclined plane with the the specified <math>35^{\text{o}}</math> angle, with the dad behind the kid pushing the sled down the hill. Some forces to be thinking about in this situation are the normal force, the applied force of the dad, the frictional force opposing the motion, and the gravitational force pulling the system straight down. To begin the free body diagram, we draw a circle to symbolize the son/sled system. Next, we draw the gravitational force as straight down, as it will always be in these Earth situations. Its length should be longer than the normal force. Otherwise, it shouldn't be compared to the other forces. Next are the frictional force and the normal force. The normal force should be normal to the inclined plane, and the frictional force should be parallel to the inclined plane, but opposite the direction of motion. Finally, the applied force of the dad should be parallel to the flat ground. The free body diagram has been created. Now, we must create a system of equations that could be solved for the acceleration of the son/sled system.

[[File:FBDSketchDifficult.jpeg|right|400px|A Sketch]]

[[File:FBDDifficult.jpeg|right|400px|A Free Body Diagram]]

To begin, we draw an x-y plane, where the y-axis runs along the the normal force and the x-axis runs along the frictional force. This will simplify the representation of these two forces, as well as represent the gravitational force with just a couple trigonometric functions. The applied force will have the same done to it.

Newton's Second Law states the net force on a system is equal to the mass of the system times the acceleration:

:<math display="block">\mathbf{F_{net}} = \sum \mathbf{F} = m_{system}\mathbf{a}</math>

We are told the son/sled system is accelerating, therefore the above equation can NOT be set equal to <math>\mathbf{0}</math>.

We also know the son/sled system should have a combined mass of <math>M + m</math>, where <math>M</math> is the mass of the sled and <math>m</math> is the mass of the son, as stated in the problem:

:<math display="block">\mathbf{F_{net}} = \sum \mathbf{F} = m_{system}\mathbf{a} = (M + m) \mathbf{a}</math>

The angle <math>\mathbf{F_A}</math> makes with the chosen x-axis is <math>35^{\text{o}}</math>. This can be seen from the sketch. The angle the gravitational force makes with the chosen y-axis is also <math>35^{\text{o}}</math>. This can also be seen in the sketch. Therefore, the four forces considered in this system can be described as:

:<math display="block">\mathbf{F_g} = \begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(35^\text{o})\end{bmatrix}</math>

:<math display="block">\mathbf{F_A} = \begin{bmatrix} |\mathbf{F_A}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_A}| \ \text{sin}(35^\text{o}) \end{bmatrix}</math>

:<math display="block">\mathbf{F_f} = \begin{bmatrix} |\mathbf{F_f}| \ \text{cos}(180^\text{o}) \\ |\mathbf{F_f}| \ \text{sin}(180^\text{o}) \end{bmatrix} = \begin{bmatrix} -|\mathbf{F_f}| \\ 0 \end{bmatrix}</math>

:<math display="block">\mathbf{F_N} = \begin{bmatrix} |\mathbf{F_N}| \ \text{cos}(90^\text{o}) \\ |\mathbf{F_N}| \ \text{sin}(90^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ |\mathbf{F_N}| \end{bmatrix}</math>

Using Newton's Second Law, we know that:

:<math display="block">\mathbf{F_{net}} = \sum \mathbf{F} = m_{system}\mathbf{a} = (M + m) \mathbf{a} = \mathbf{F_g} + \mathbf{F_A} + \mathbf{F_f} + \mathbf{F_N}</math>

This is equivalent to:

:<math display="block">\mathbf{a} = \frac{\begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(35^\text{o})\end{bmatrix} + \begin{bmatrix} |\mathbf{F_A}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_A}| \ \text{sin}(35^\text{o}) \end{bmatrix} + \begin{bmatrix} -|\mathbf{F_A}| \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ |\mathbf{F_N}| \end{bmatrix}}{M+m}</math>

Thus we have done what the problem asked.

==Connectedness==
Free body diagrams are the building blocks for many scientists, physicists, engineers and have many applications in the real world. They form the basis for determining a device's or building's structural stability, safety, and overall practicality.

'''Medical Applications'''
[[File:Leg.gif|right|]]
There have recently been many advances in medicine in terms of wearables and prosthetic limbs that function just like normal body parts for either congenital defects or accident recovery. Let's take a look at a leg prosthetic. While walking, a prosthetic exerts many forces on your leg such as: the force of gravity, normal force from the ground, frictional force, etc.
Prosthetics move like springs, while your body moves forward it also moves up and down which causes changes in kinetic energy due to movement and changes in gravitational potential energy. While it contracts and relaxes, different energies are converted back and forth into kinetic and potential energy. In order to analyze all the forces on a prosthetic to ensure it's safety and usability, the most basic step when designing one is a free body diagram.

'''Buildings and Trusses'''

Many civil engineers and architects are responsible for building trusses, bridges, and buildings which function using the basic concepts of physics. It is important to ensure that components of a truss and bridge are in equilibrium so that one day they don't collapse.
When building a truss bridge, engineers put straight members in place that form the bridge's top and bottom, and they are linked by a structure of diagonals and vertical posts. For trusses especially, all the components must remain in equilibrium, the magnitudes of forces exerted must be equal, and the components in tension and compression must be identified. In order to create a proper structure, a free body diagram must be used by engineers in the preliminary stages for design and safety purposes.

[[File:Truss.gif]]

==History==

Free Body Diagrams were developed alongside Newtonian physics, as a graphical way to analyze what forces were acting on a system, and how the system would react.
Because of their widespread usefulness, free body diagrams have not fallen out of favor with scientists and engineers.

Free body diagrams have been used as a teaching and real world tool for well over a century.

==See Also==

===Further Reading===
[[Inclined Plane]]

[[Compression or Normal Force]]

[[Tension]]

[[Gravitational Force]]

[[Spring Force]]

[[Hooke's Law]]

[[Net Force]]

===External Links===
[http://hyperphysics.phy-astr.gsu.edu/hbase/force.html Forces Mental Map]

[http://www.physicsclassroom.com/class/newtlaws/Lesson-3/Newton-s-Second-Law Newton's Second Law]

[https://www.youtube.com/watch?v=nDis6HbXxjg Using Free-Body Diagrams to Solve Kinematics Problems]

== References ==
Matter and Interactions: Modern Mechanics. Volume One. 4th Edition.

http://demos.smu.ca/index.php/demos/mechanics/141-free-body-diagram

http://hyperphysics.phy-astr.gsu.edu/hbase/freeb.html

https://www.wisc-online.com/learn/natural-science/physics/tp1502/construction-of-free-body-diagrams

http://www.physicsclassroom.com/Class/newtlaws

https://sites.google.com/a/cpsdigital.org/peraplegic/human-prosthetics

http://slideplayer.com/slide/6086550/ - Tilted Axis Image

http://www.revistas.unal.edu.co/index.php/dyna/rt/printerFriendly/30749/39025 - Prosthetic Limb Image

http://www.mathalino.com/reviewer/engineering-mechanics/241-finding-resulatnt-vertical-forces-acting-fink-truss - Truss Image

Free Body Diagram

2021-11-29T17:03:23Z

Jnation8:

'''Claimed By: Jared Nation Fall 2021
'''==Main Idea==
A free body diagram, or force diagram, is a rough sketch that shows the relative magnitude and direction of all the forces acting on a system. There are various forces that can be acting on the object, such as applied force, frictional force, normal force, and gravitational force. A free body diagram allows for analysis in a steady state condition, where there can be no acceleration or a constant acceleration on a system.

[[File:FBDIntroductoryImage.jpg|center|400px|An Example Sketch and Free Body Diagram]]

All forces in a free body diagram are due to the system's interactions with its surroundings. Especially when problems become complicated and involve different forces acting on multiple objects, free body diagrams can be extremely effective in making problems simpler to handle. Using a free body diagram, one can solve for an unknown force acting on the body or the net acceleration after identifying all of the forces acting on it.

===Mathematical Model===
Since a free body diagram is a snapshot of a steady state, there is no net acceleration or a constant net acceleration when we are modeling a free body diagram. Therefore, we can use Newton's First Law (an object in an inertial frame at rest or moving with constant velocity, will continue to do so until acted upon by a net force), Newton's Second Law (the net force on an object is equivalent to the object's mass multiplied by its net acceleration) and Newton's Third Law (for every force there is an equal and opposite reactionary force) to create a system of equations to describe and gain new information about the steady state:

<math>\mathbf{F_{net}} = \sum \mathbf{F} = m \mathbf{a} = m \frac{d\mathbf{v}}{dt} </math> ('''Newton's Second Law''')

Additionally, if the net acceleration is equivalent to 0, which means the object is either at rest or moving with constant velocity, then:

<math>\mathbf{F_{net}} = \sum \mathbf{F} = m \mathbf{a} = m \frac{d\mathbf{v}}{dt} = \mathbf{0} </math> ('''Newton's First Law''')

Also, if it is known the net acceleration is '''0''' or one or more components of the net acceleration are 0, for example a<sub>x</sub> and a<sub>y</sub>, then the components of the forces along those directions must be balanced. ('''Newton's First and Third Laws''')

Using these strategies, a large set of force related problems can be solved.
===Computational Model===
The following is a VPython model of the acceleration of a box as it moves down an inclined plane with negligible friction. The box starts at the top of the inclined plane, which is given by <math>pos = (5,5,5)</math>, as shown by the accompanying diagram.

IMPORT STATEMENTS

<code>

from __future__ import division

from visual import *

import math

</code>

INITIAL CONDITIONS

<code>

scene.width = 1024

scene.height = 760

time = 0 #Time in seconds

dtime = .1 #The time will be iterated through every .1 seconds

init_pos = vector(0, 5, 5) #The initial position is at (x, y, z) = (0, 5, 5) meters

init_vel = vector(0, 0, 0) #The intial velocity is (0, 0, 0)

cube = box(pos = init_pos, length = 1, height = 1, width = 1) #Create a cube at the initial position

mass_cube = 10 #The mass of the cube is 10 kg

theta = 45 #The angle between the horizontal and the hypotneuse of the inclined plane is 45 degrees

force_gravity = mass_cube * 9.81 #Force of gravity on the cube in Newtons

thetaRads = math.radians(theta) #Convert degrees to radians

force_normal = force_gravity * math.cos(thetaRads)#Newton's First and Second Laws #The x-axis is chosen to be along the hypotenuse of the inclined plane, and the y-axis is chosen to be along the normal to the hypotenuse.

F_y = force_normal - force_gravity * math.cos(thetaRads) #This is equal to 0

F_x = force_gravity * math.sin(thetaRads) #The only force acting along the chosen x-axis is the component of the gravitational force parallel to the hypotenuse

a_x = F_x / mass_cube

</code>

ITERATE THROUGH TIME TO FOLLOW THE MOVEMENT OF THE CUBE DOWN THE INCLINED PLANE

<code>

while cube.pos.x < 5:
rate(200)
velocity = vector(init_vel.x + (dtime * a_x), init_vel.y, init_vel.z)
position = vector(cube.pos.x + (dtime * velocity.x), cube.pos.y + (dtime * velocity.y), cube.pos.z + (dtime * velocity.z))
cube.pos = position
trail.append(pos = cube.pos)
time = time + dtime
print(cube.pos)

</code>

=== How To Draw a Free Body Diagram ===

==== Establish a Convenient Coordinate System ====
Before beginning to analyze a system, it is important to choose an appropriate coordinate system (an x and y plane) that will be the most convenient to avoid dealing with complicated angles of forces and difficult algebra and trigonometry. This is especially true when dealing with angled forces where tilting the x and y plane could make solving for resulting forces a lot easier, such as on an inclined plane. Also, in order to maintain consistency with direction amongst various people solving the same system, an x-y coordinate system can ensure that anyone solving the same system will end up with the same directions on their forces.
[[File:Angled Axis.png|right|This is an example of how establishing a coordinate system can make solving/identifying forces easier]]

==== Identify the System ====
Often times, you will be faced with single and multi particle systems where bodies within that system will be interacting with/exerting forces on each other. For example, a block is resting on top of a larger block, and both are moving down a ramp. In that case, analyzing the top block versus the bottom block will result in different free body diagrams. Therefore, identifying which part of the system you are analyzing before beginning to draw a free body diagram will allow for a more efficient solving process.

==== List and identify all surroundings that interact with the system ====
Although we usually think of these interactions in terms of force names, it's best to get in the habit of identifying the force AND the object that specifically causes that force. For example, the force of gravity on a block could be caused by the earth or a normal force could be caused by a second block and another normal force could be caused by the table. Identifying forces by the specific objects that cause them helps us not forget where the forces belong.

=====Types of Forces to Consider for Free Body Diagrams=====

'''Disclaimer: Not all of these forces will be present in every situation. These are not all of the possible forces; These are very common ones.'''

*'''Applied Force''':
**Symbol: <math style="inline">\mathbf{F_A}</math>
**Force applied to the system by a person or other object.
*'''Force of Friction''':
**Symbol: <math>\mathbf{F_f}</math>
**Force that a surface applies on the system that is moving (or trying to move) on that surface.
**Formula: <math>\mathbf{F_{f}} = \mu \mathbf{N}</math>
***<math>\mathbf{N} =</math> Normal Force
***<math>\mu =</math> Coefficient of Friction
*'''Force of Gravity''':
**Symbol: <math>\mathbf{F_g}</math>
**Force that, on Earth, will act downward toward the center of the Earth. In general, gravity is the attraction of any mass to any other mass.
**Formula: <math>\mathbf{F_g} = \frac{G M_1 M_2}{r^2}{\hat{\mathbf{r}}}</math>
***This can be shown to be equal to <math>m\mathbf{g}</math> near the Earth's surface.
***<math>G = 6.7 \times 10^{-11}\ \frac{Nm^2}{{kg}^2}</math> (Gravitational Constant)
***<math>M_1 =</math> Mass 1
***<math>M_2 =</math> Mass 2
***<math>r =</math> Displacement between Masses 1 and 2
***<math>{\hat{\mathbf{r}}} =</math> Unit vector in direction of Displacement between the Masses
***<math>m =</math> Mass of object on Earth
***<math>\mathbf{g} = 9.8\ m/s^2 </math> (on earth)
*'''Normal Force''':
**Symbol: <math>\mathbf{F_N}</math>
**Force that is present when the system is on another object or surface, and the object or surface is exerting a force on the system as support.
*'''Spring Force'''
**Symbol: <math>\mathbf{F_s}</math>
**Force that is exerted by a spring on any system that is attached to it.
**Formula: <math>\mathbf{F_s} = -k\mathbf{x}</math> (Hookian Spring)
***<math>k =</math> Spring constant (<math>\ \frac{N}{m}</math>)
***<math>\mathbf{x} =</math> Displacement from the spring's equilibrium position
*'''Force of Tension''':
**Symbol: <math>\mathbf{F_T}</math> or <math>\mathbf{T}</math>
**Force that exists when a rope, string, wire, etc. is pulling on the system.

'''Side Note:'''
Now that we've touched on frictional forces, it is important to understand the types of frictional forces. Frictional forces include two types: static friction and dynamic (kinetic) friction. These two types of friction have different coefficients of friction, usually represented by <math>\mu_s</math> for static friction and <math>\mu_k</math> for dynamic (kinetic friction.
*'''Static Friction''' refers to the frictional force while the object is stationary. An external net force higher than this static frictional force is required to move the object.
*'''Dynamic (Kinetic) Friction''' refers to the frictional force while the object is in motion. For example, the object may experience a frictional force with the ground while it is moving along the ground.

==== Draw a diagram with the system at the center ====
#Use a dot to represent the system, OR
#You can draw the details of the system (Draw a block, car, etc.)
#Draw the conveniently oriented coordinate system on the free body diagram or next to the it
#Draw the system's relevant force vectors with correct relative magnitude and direction, pointing away from the dot or detailed drawing
[[File:ContactForce.JPG|300px|thumb|right|]]

==== Label all Forces With a Symbol ====
Symbols are useful to quickly represent the name of the force and identify them by the object causing the force. Refer to [[#Types of Forces to Consider for Free Body Diagrams| Common Symbols for Forces]].

==== Break forces into their components as needed. ====
If a force is acting diagonal to the system, create a dashed line parallel and perpendicular to the system and label it as the x and y components of that force. Use sine(θ) or cosine(θ) as needed.

'''General Tips:'''
*If an object has constant speed, it means the object has no acceleration. Since net force is the rate of change of acceleration, net force in that direction would then be zero. This means that there are either no forces currently acting on the object, or there are equal, opposite forces acting on the object in that direction. To represent this in a free body diagram, draw forces as arrows pointing in opposite directions with equal lengths.
*Don't be confused by contact forces. Most of the time, contact force is an umbrella term that includes other types of forces. If you had a block on a ramp, you could draw the contact force as being diagonal to the ramp. Or, if you wanted to break it into its components which are easier to consider, you would draw the Normal force caused by the ramp pushing up as a perpendicular arrow, and the force of Friction caused by the ramp as a parallel arrow (see image to the upper right).

==== Sum the forces ====
Once you have drawn the free body diagram and broken the forces into components along the convenient x-y axis (if needed), then it is time to use [[#Mathematical Model | Mathematical Model]] to create a system of equations summing the forces in the x and y directions and then set them equal to the correct value, such as 0 if there is no acceleration in that direction or <math>m\mathbf{a}</math>.

==Examples==
In this section we will go step by step through 3 illustrative examples, increasing in difficulty as we go.

=== Simple ===
A person with a mass of <math>M</math> rides up to the 20th floor of their apartment building in an elevator moving at constant velocity. Create a free body diagram to represent this situation. Using your free body diagram, use Newton's Second Law to create a force equation for the system. Make sure to keep track of the relative magnitudes of the forces and indicate these magnitudes by the length of the arrow.

'''''Answer'''''

We start by drawing a free body diagram with correct relative magnitudes and directions of the Normal Force and the Gravitational Force. The normal force is always normal (perpendicular or orthogonal) to the surface, whether the surface is angled or not. Gravity always acts downwards on Earth. Therefore, there is no need to create a different, convenient coordinate plane. A horizontal x-axis and vertical y-axis will suffice. These are the only two forces acting on the person because nothing else is pushing on the person, and friction is not being taken into account. We know the normal force must be present due to the person not falling through the floor (due to the always present gravitational force). We can also reason that these forces must be equal in magnitude and opposite in direction.

[[File:FBDSimpleSketch.jpeg|right|400px|A Sketch]]

[[File:FBDSimple.jpeg|right|200px|A Free Body Diagram]]

Next, we use Newton's Second Law to begin our analysis:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a}</math>

Next, we notice the problem states that the person moves with a constant velocity upwards, therefore there is no acceleration in any direction, and the net force must be equal to <math>\mathbf{0}</math>:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a} = \mathbf{0}</math>

The normal force and gravitational force can be described as:

:<math display="block">\mathbf{F_g} = \begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(180^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(180^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ -|\mathbf{F_g}| \end{bmatrix}</math>

:<math display="block">\mathbf{F_N} = \begin{bmatrix} |\mathbf{F_N}| \ \text{cos}(90^\text{o}) \\ |\mathbf{F_N}| \ \text{sin}(90^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ |\mathbf{F_N}| \end{bmatrix}</math>

Next, putting the free body diagram and Newton's Second Law together gives:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = \begin{bmatrix} 0 \\ \mathbf{|F_N|} \end{bmatrix} + \begin{bmatrix} 0 \\ -\mathbf{|F_g|} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}</math>

Therefore, it is possible to simplify the above to show that the gravitational force is exactly opposite in direction and equal in magnitude:

:<math display="block">|\mathbf{F_g}| = |\mathbf{F_N}|</math>
:<math display="block">|\mathbf{F_N}| = Mg </math>

We have fully done what the problem asked.

=== Middling ===
A ball with mass <math>m</math> is hanging tautly from two strings of negligible mass, each of which are connected to the ceiling. To the left of the ball, one string has a tension <math>\mathbf{T_1}</math> and makes an angle of <math>20^\text{o}</math> with the ceiling, and on the right of the ball, the other string has a tension <math>\mathbf{T_2}</math> and creates an angle of <math>45^\text{o}</math> with the ceiling. Both angles are measured from the ceiling to the inner side of the string. Refer to the diagram for any confusion. Create a free body diagram to model this situation. Also, create a system of force equations. '''Hint''': x and y components will be needed for this problem.

'''''Answer'''''

To draw the free body diagram it is better to first draw a sketch of the system. In the sketch we see the string to the left (<math>\mathbf{T_1}</math>) of the ball makes an angle of <math>20^\text{o}</math> with the ceiling, and the string to the right (<math>\mathbf{T_2}</math>) of the ball makes an angle <math>45^\text{o}</math> with the ceiling. For the free body diagram, we first draw the force of gravity pointing straight downwards. This force should have the longest arrow due to the fact that the strings are supporting this force...therefore, it can be reasoned that this force should be greater than either of the tensions in the strings for the given angles. Next, we draw the tension forces, which point away from the ball along each respective string. <math>\mathbf{T_2}</math> should be longer than <math>\mathbf{T_1}</math> due to the fact that the tensions must balance each other along the x-axis and the y-component of <math>\mathbf{T_2}</math> must be greater than the y-component of <math>\mathbf{T_1}</math>.

[[File:FBDMiddlingSketch.jpeg|right| 400px|A Sketch]]

[[File:FBDMiddling.jpeg|right|300px|A Free Body Diagram]]

Now, we are ready to use Newton's First and Second Laws to create a system of force equations:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a}</math>

Since the ball is hanging tautly, it must NOT be moving. Therefore, the velocity and acceleration must both be <math>\mathbf{0}</math>:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a} = \mathbf{0}</math>

Next, we choose the x-axis to be horizontal, and the y-axis to be vertical, as in a typical coordinate system, since it is convenient.

Now, we begin specifically identifying the forces and their components using right triangles:

:<math display="block">\mathbf{F_g} = \begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(180^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(180^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ -|\mathbf{F_g}| \end{bmatrix}</math>

:<math display="block">\mathbf{T_1} = \begin{bmatrix} -|\mathbf{T_1}| \ \text{cos}(20^\text{o}) \\ |\mathbf{T_1}| \ \text{sin}(20^\text{o})\end{bmatrix} </math>

:<math display="block">\mathbf{T_2} = \begin{bmatrix} -|\mathbf{T_2}| \ \text{cos}(45^\text{o}) \\ |\mathbf{T_2}| \ \text{sin}(45^\text{o})\end{bmatrix}</math>

Next, we use Newton's Second Law to sum these forces. Since the ball is not moving, we can set this summation equal to <math>\mathbf{0}</math>:

:<math display="block">\mathbf{F_{net}} = \sum\mathbf{F} = m\mathbf{a} = \mathbf{0} = \mathbf{F_g} + \mathbf{T_1} + \mathbf{T_2}</math>

This is equivalent to:

:<math display="block"> \begin{bmatrix} 0 \\ -|\mathbf{F_g}| \end{bmatrix} + \begin{bmatrix} -|\mathbf{T_1}| \ \text{cos}(20^\text{o}) \\ |\mathbf{T_1}| \ \text{sin}(20^\text{o})\end{bmatrix} + \begin{bmatrix} -|\mathbf{T_2}| \ \text{cos}(45^\text{o}) \\ |\mathbf{T_2}| \ \text{sin}(45^\text{o})\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}</math>

Hence, we have done what the question asked and are done with this problem.

=== Difficult ===
A father pushes his son's sled horizontally with a force <math>\mathbf{F_A}</math> all the way down a hill on a sled of mass <math>M</math>. The son has a mass <math>m</math>. The son/sled system is accelerating down the hill. Assume the hill is shaped like an inclined plane with an angle to the horizontal ground of <math>35^{\text{o}}</math> and that friction works against the sled with a coefficient of dynamic (kinetic) friction <math>\mu_k</math>. Draw a free-body diagram. Using your free body diagram, write a system of equations that could be used to solve for the acceleration of the son/sled system.

'''''Answer'''''

We start with a sketch. In this sketch, we create an inclined plane with the the specified <math>35^{\text{o}}</math> angle, with the dad behind the kid pushing the sled down the hill. Some forces to be thinking about in this situation are the normal force, the applied force of the dad, the frictional force opposing the motion, and the gravitational force pulling the system straight down. To begin the free body diagram, we draw a circle to symbolize the son/sled system. Next, we draw the gravitational force as straight down, as it will always be in these Earth situations. Its length should be longer than the normal force. Otherwise, it shouldn't be compared to the other forces. Next are the frictional force and the normal force. The normal force should be normal to the inclined plane, and the frictional force should be parallel to the inclined plane, but opposite the direction of motion. Finally, the applied force of the dad should be parallel to the flat ground. The free body diagram has been created. Now, we must create a system of equations that could be solved for the acceleration of the son/sled system.

[[File:FBDSketchDifficult.jpeg|right|400px|A Sketch]]

[[File:FBDDifficult.jpeg|right|400px|A Free Body Diagram]]

To begin, we draw an x-y plane, where the y-axis runs along the the normal force and the x-axis runs along the frictional force. This will simplify the representation of these two forces, as well as represent the gravitational force with just a couple trigonometric functions. The applied force will have the same done to it.

Newton's Second Law states the net force on a system is equal to the mass of the system times the acceleration:

:<math display="block">\mathbf{F_{net}} = \sum \mathbf{F} = m_{system}\mathbf{a}</math>

We are told the son/sled system is accelerating, therefore the above equation can NOT be set equal to <math>\mathbf{0}</math>.

We also know the son/sled system should have a combined mass of <math>M + m</math>, where <math>M</math> is the mass of the sled and <math>m</math> is the mass of the son, as stated in the problem:

:<math display="block">\mathbf{F_{net}} = \sum \mathbf{F} = m_{system}\mathbf{a} = (M + m) \mathbf{a}</math>

The angle <math>\mathbf{F_A}</math> makes with the chosen x-axis is <math>35^{\text{o}}</math>. This can be seen from the sketch. The angle the gravitational force makes with the chosen y-axis is also <math>35^{\text{o}}</math>. This can also be seen in the sketch. Therefore, the four forces considered in this system can be described as:

:<math display="block">\mathbf{F_g} = \begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(35^\text{o})\end{bmatrix}</math>

:<math display="block">\mathbf{F_A} = \begin{bmatrix} |\mathbf{F_A}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_A}| \ \text{sin}(35^\text{o}) \end{bmatrix}</math>

:<math display="block">\mathbf{F_f} = \begin{bmatrix} |\mathbf{F_f}| \ \text{cos}(180^\text{o}) \\ |\mathbf{F_f}| \ \text{sin}(180^\text{o}) \end{bmatrix} = \begin{bmatrix} -|\mathbf{F_f}| \\ 0 \end{bmatrix}</math>

:<math display="block">\mathbf{F_N} = \begin{bmatrix} |\mathbf{F_N}| \ \text{cos}(90^\text{o}) \\ |\mathbf{F_N}| \ \text{sin}(90^\text{o}) \end{bmatrix} = \begin{bmatrix} 0 \\ |\mathbf{F_N}| \end{bmatrix}</math>

Using Newton's Second Law, we know that:

:<math display="block">\mathbf{F_{net}} = \sum \mathbf{F} = m_{system}\mathbf{a} = (M + m) \mathbf{a} = \mathbf{F_g} + \mathbf{F_A} + \mathbf{F_f} + \mathbf{F_N}</math>

This is equivalent to:

:<math display="block">\mathbf{a} = \frac{\begin{bmatrix} |\mathbf{F_g}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_g}| \ \text{sin}(35^\text{o})\end{bmatrix} + \begin{bmatrix} |\mathbf{F_A}| \ \text{cos}(35^\text{o}) \\ |\mathbf{F_A}| \ \text{sin}(35^\text{o}) \end{bmatrix} + \begin{bmatrix} -|\mathbf{F_A}| \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ |\mathbf{F_N}| \end{bmatrix}}{M+m}</math>

Thus we have done what the problem asked.

==Connectedness==
Free body diagrams are the building blocks for many scientists, physicists, engineers and have many applications in the real world. They form the basis for determining a device's or building's structural stability, safety, and overall practicality.

'''Medical Applications'''
[[File:Leg.gif|right|]]
There have recently been many advances in medicine in terms of wearables and prosthetic limbs that function just like normal body parts for either congenital defects or accident recovery. Let's take a look at a leg prosthetic. While walking, a prosthetic exerts many forces on your leg such as: the force of gravity, normal force from the ground, frictional force, etc.
Prosthetics move like springs, while your body moves forward it also moves up and down which causes changes in kinetic energy due to movement and changes in gravitational potential energy. While it contracts and relaxes, different energies are converted back and forth into kinetic and potential energy. In order to analyze all the forces on a prosthetic to ensure it's safety and usability, the most basic step when designing one is a free body diagram.

'''Buildings and Trusses'''

Many civil engineers and architects are responsible for building trusses, bridges, and buildings which function using the basic concepts of physics. It is important to ensure that components of a truss and bridge are in equilibrium so that one day they don't collapse.
When building a truss bridge, engineers put straight members in place that form the bridge's top and bottom, and they are linked by a structure of diagonals and vertical posts. For trusses especially, all the components must remain in equilibrium, the magnitudes of forces exerted must be equal, and the components in tension and compression must be identified. In order to create a proper structure, a free body diagram must be used by engineers in the preliminary stages for design and safety purposes.

[[File:Truss.gif]]

==History==

Free Body Diagrams were developed alongside Newtonian physics, as a graphical way to analyze what forces were acting on a system, and how the system would react.
Because of their widespread usefulness, free body diagrams have not fallen out of favor with scientists and engineers.

Free body diagrams have been used as a teaching and real world tool for well over a century.

==See Also==

===Further Reading===
[[Inclined Plane]]

[[Compression or Normal Force]]

[[Tension]]

[[Gravitational Force]]

[[Spring Force]]

[[Hooke's Law]]

[[Net Force]]

===External Links===
[http://hyperphysics.phy-astr.gsu.edu/hbase/force.html Forces Mental Map]

[http://www.physicsclassroom.com/class/newtlaws/Lesson-3/Newton-s-Second-Law Newton's Second Law]

[https://www.youtube.com/watch?v=nDis6HbXxjg Using Free-Body Diagrams to Solve Kinematics Problems]

== References ==
Matter and Interactions: Modern Mechanics. Volume One. 4th Edition.

http://demos.smu.ca/index.php/demos/mechanics/141-free-body-diagram

http://hyperphysics.phy-astr.gsu.edu/hbase/freeb.html

https://www.wisc-online.com/learn/natural-science/physics/tp1502/construction-of-free-body-diagrams

http://www.physicsclassroom.com/Class/newtlaws

https://sites.google.com/a/cpsdigital.org/peraplegic/human-prosthetics

http://slideplayer.com/slide/6086550/ - Tilted Axis Image

http://www.revistas.unal.edu.co/index.php/dyna/rt/printerFriendly/30749/39025 - Prosthetic Limb Image

http://www.mathalino.com/reviewer/engineering-mechanics/241-finding-resulatnt-vertical-forces-acting-fink-truss - Truss Image