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<math>Where \frac{1}{4 \pi \epsilon_0 } \text{is a constant representing the permittivity of free space, and is approximately }  9*10^{9} \frac{N m^2}{C^2} </math>
<math>Where \frac{1}{4 \pi \epsilon_0 } \text{is a constant representing the permittivity of free space, and is approximately }  9*10^{9} \frac{N m^2}{C^2} </math>
, r is the magnitude of the distance  
, r is the magnitude of the distance between the observation location and the source location
, q is the charge of the particle between the point charge and the observation point
, q is the charge of the particle  
and <math>\hat r \text { is the direction of the distance from the point charge to the observation point.} </math>
and <math>\hat r \text { is the direction of the distance from the point charge to the observation point.} </math>


It is very easy to derive Coulombs Law using the equation for the electric field of a point charge, as all you have to do is multiply the value obtained by the charge of the second particle.  
It is very easy to derive Coulombs Law using the equation for the electric field of a point charge, as all you have to do is multiply the value obtained by the charge of the second particle.


===A Computational Model===
===A Computational Model===

Revision as of 17:22, 27 November 2016

This page is all about the Electric Field due to a Point Charge.CLAIMED BY DEJAN TOJCIC 10-31-2016 == Electric Field==

A Work In Progress by Dejan Tojcic. dtojcic3 (talk)

The Main Idea

Looking at one of the four fundamental interactions of physics, electromagnetism, you can derive the expression for Coulombs Force Law. Using this expression, there is just a simple equation that is used to derive the the exact value of the electric field. The electric field of a point charge can be directly coordinated with a couple of independent variables. The variables that are directly coordinated with the value of an electric field is the direction of the particle and the charge of the particle. Electric field of a point charge is also inversely proportional to distance.

A Mathematical Model of Electric Field due to Point Charge

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

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

[math]\displaystyle{ Where \frac{1}{4 \pi \epsilon_0 } \text{is a constant representing the permittivity of free space, and is approximately } 9*10^{9} \frac{N m^2}{C^2} }[/math] , r is the magnitude of the distance between the observation location and the source location , q is the charge of the particle and [math]\displaystyle{ \hat r \text { is the direction of the distance from the point charge to the observation point.} }[/math]

It is very easy to derive Coulombs Law using the equation for the electric field of a point charge, as all you have to do is multiply the value obtained by the charge of the second particle.

A Computational Model

Here is a link to some code which can help visualize and compute the Electric Field due to an electron at different points. You can adjust the values of the observation locations in order to observe how the effects of distance, both magnitude and direction, can affect the electric field.

https://trinket.io/glowscript/8d66c87ced

Examples

Simple

Problem 1: There is an electron at <0,0,0>. Calculate the electric field at <4,-3,1>.

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

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

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

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

Step 2: Find the magnitude of the Electric Field

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

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

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

Middling

Problem 1: A particle of unknown charge is located at <-.21, .02, .11>. Its electric field at point <-.02, .31, .28> is <.124, .188, .109> N/C . Find the value of the particles charge.

While this problem looks challenging, lets think about it for a second. We are given both an observation location and a source location, so we can find both r and [math]\displaystyle{ \hat{r} }[/math] Since we are given the value of the electric field, we can simply find the magnitude of the electric field. Then, we can use the equation for the magnitude of electric field of a point charge,[math]\displaystyle{ E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} }[/math] to find the charge.

Step 1: Find [math]\displaystyle{ \vec r_{obs} - \vec r_{particle} \lt -.02,.31,.28\gt - \lt -.21,.02,.11\gt = \lt .19,.29,.17\gt }[/math]

To find r_{mag}, we simply find the magnitude of <.19,.29,.17>.

([math]\displaystyle{ \sqrt{.19^2+.29^2+.17^2}=\sqrt{.1491} }[/math]= .39

Step 2: Find the magnitude of the Electric Field

E= <.124, .188, .109> N/C

E_{mag} = ([math]\displaystyle{ \sqrt{.124^2+.188^2+.109^2}=\sqrt{.0626} }[/math]=.25

Step 3: Find q by rearranging the equation for E_{mag}

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

By rearranging this equation we get

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

[math]\displaystyle{ q= {1/(9*10^9)} *{.39^2}*.25 }[/math]

q= 4.3*10^-12 C



Difficult

Problem 2:

Once the basic equation and execution of Coulomb’s Law is understood, it can seem that problems with point charges cannot possibly get any harder. The three variables are Electric Field, charge, and distance between the charges. Any problem must provide two of the three, and the student “plugs and chugs” to get the third. It may seem that these problems could not possibly be made harder without looking at a bunch of point charges together in the shape of a disk or ring or rod. However, this is not true.

The variable that one often does not consider is that the third variable, distance between charges, is actually the negative summation of two variables, observation location and location of charge (or source location). Meaning, to know the distance between the charges (or the charge and wherever you are calculating the field), you must know both the observation location and where the charge is. The example problem below elucidates how this can become quite tricky.

Problem: There is a charged particle at an unknown location. Its charge is [math]\displaystyle{ Q }[/math] Coulombs. At location ([math]\displaystyle{ obs_{x}, obs_{y}, obs_{z}) }[/math] meters, an Electric Field of [math]\displaystyle{ (E_{x},E_{y},E_{z}) N/C }[/math] is observed. At what location is the charged particle? Assume the Electric Field is caused only by the single charged particle, no other charges are close. You could use your own made up values for the three givens, but for the solution we will us [math]\displaystyle{ Q = 2*10^{-8} }[/math] Coulombs, location is [math]\displaystyle{ (.26, .75, .09) }[/math] meters, and Electric Field is [math]\displaystyle{ (400,800,500) N/C }[/math].

Solution: First, I will summarize the steps. Then I will give a detailed explanation of why each step is the correct one, plugging in the numbers for the student to follow along with. The last part of the solution guide is a Matlab code that solves any inputs to this problem so the student can visualize the solving of the problem in a slightly different way, as well as generate nir own problems.

Summary of Steps:

Step 1: Find Magnitude of Electric Field by using Pythagorean Theorem on the components

Step 2: Find Magnitude of [math]\displaystyle{ \vec r }[/math] (which is distance between charge and observation location) by rearranging Coulomb’s Law

Step 3: Find unit vector of Electric Field

Step 4: Using sign of the charge and unit vector of Electric Field, find unit vector of [math]\displaystyle{ \vec r }[/math]

Step 5: Find the vector [math]\displaystyle{ \vec r }[/math] by multiplying the unit vector and magnitude of [math]\displaystyle{ \vec r }[/math]

Step 6: Solve for position of charge

Detail Explanation with numbers:

First, let’s clarify what [math]\displaystyle{ \vec r }[/math] is. We observe an Electric Field, [math]\displaystyle{ \vec E }[/math], at a given point; thus, this is the observation location. The distance from the charged particle to this observation location is [math]\displaystyle{ \vec r }[/math]. So for this problem, we know the observation location, but not where the charge is. Observation location is called [math]\displaystyle{ (obs_{x}, obs_{y}, obs_{z} }[/math]).

So: [math]\displaystyle{ r = (obs_{x}, obs_{y}, obs_{z}) – (x, y, z) }[/math]

Or in this case:

[math]\displaystyle{ r = (.26,.75, .09) – (x,y,z) }[/math]

The problem is asking us what (x,y,z) is, so to solve it, we must find [math]\displaystyle{ \vec r }[/math].

However, the issue is Coulomb’s Law doesn’t use r in vector form. It uses the magnitude of r and the unit vector of r, which from now on will be referred to as [math]\displaystyle{ r_{mag} }[/math] and [math]\displaystyle{ \hat{r} }[/math] respectively.

So to find r vector, we need both [math]\displaystyle{ r_{mag} }[/math] and [math]\displaystyle{ \hat{r} }[/math]

Magnitude is easy because we can just use the magnitude version of Coulomb’s Law:

[math]\displaystyle{ E_{mag} = \frac{1}{4 \pi \epsilon_0} * \frac{q}{(rmag)^{2}} }[/math]

We are given [math]\displaystyle{ q }[/math], and [math]\displaystyle{ E_{mag} }[/math] can easily be calculated from the given [math]\displaystyle{ \vec E }[/math] using Pythagorean Theorem. And then we can rearrange Coulomb’s Law from above so that rmag is on one side of the equation.

So:

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

And with our numbers:

[math]\displaystyle{ r_{mag} = \sqrt{9*10^{9} * \frac{2*10^{-8}}{1024.7}} = .42 }[/math]

But how do we find [math]\displaystyle{ \hat{r} }[/math] so we can finish calculating the [math]\displaystyle{ \vec r }[/math]?

Well, think about this. A unit vector is simply a way of mathematically describing direction. All of the direction of the Electric Field comes from the direction of the vector between the two locations, or [math]\displaystyle{ \vec r }[/math]. [math]\displaystyle{ Q }[/math] is a scalar and alters direction in no way, with one exception. All of the direction of [math]\displaystyle{ \vec E }[/math] comes from [math]\displaystyle{ \vec r }[/math],then. Said mathematically, [math]\displaystyle{ \hat{E} }[/math] equals the absolute value of [math]\displaystyle{ \hat{r} }[/math]. Why absolute value? It is necessary because of the exception mentioned above. The charge can flip the direction of the field 180 degrees depending on its sign. Replacing a charged particle with a particle of the opposite sign results in an [math]\displaystyle{ \vec E }[/math] in the exact opposite direction, despite [math]\displaystyle{ \hat{r} }[/math] remaining the same.

So: [math]\displaystyle{ \hat{E} = \frac{\vec E}{E_{mag}} = (400,800,500)/1024.7 = (.39,.78,.488) }[/math] And q is positive: [math]\displaystyle{ \hat{E} = (.39,.78,.488) }[/math] [math]\displaystyle{ \vec r = \hat{r} * r_{mag} = (.39,.78,.488) * .42 = (.1638, .3276, .2058) }[/math]

Going back to our original equation: [math]\displaystyle{ (.1638, .3276, .2058) = (.26,.75,.09) – (x,y,z) }[/math] So the location of the charge is [math]\displaystyle{ (.096, .422, -.115) }[/math]

This problem can be solved instantaneously with this Matlab code, which take the inputs of E and observation location as six different scalars:

   function [positionOfCharge] = Elec(Ex, Ey, Ez, obs_x, obs_y, obs_z, q)
   Emag = (Ex^2 + Ey^2 + Ez^2)^(1/2);
   rmag = abs((9e9*q/Emag)^(1/2));
   Ehatx = Ex./Emag;
   Ehaty = Ey./Emag;
   Ehatz = Ez./Emag;
       if q < 0
           rhatx = -(Ehatx);
           rhaty = -(Ehaty);
           rhatz = -(Ehatz);
       elseif q > 0
           rhatx = Ehatx;
           rhaty = Ehaty;
           rhatz = Ehatz;
       end
   rx = rhatx * rmag;
   ry = rhaty * rmag;
   rz = rhatz * rmag;
   xpos = obs_x - rx;
   ypos = obs_y - ry;
   zpos = obs_z - rz;
   positionOfCharge = [xpos ypos zpos];
   end

Connectedness

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

I am very interested in the idea of forces and how objects interact with each other. After you calculate the Electric Field you can easily find the Electric Force one particle exerts on another.

2.How is it connected to your major?

I am a CompE major and so Electric Fields have to do with my major because when you integrate them with respect to dL, and swap the sign, you get potential difference(voltage), which is very important in circuits. As ECE majors take circuits classes, this topic is relevant to me.

3.Is there an interesting industrial application?

An interesting application is that electric fields of point charges can be used to find forces. Then you can predict the motion of various particles by the forces acting on them.

History

In the 1780s a French scientist named Charles Coulomb published many scientific papers on electricity and magnetism. While doing experiments, Coulomb had discovered an inverse square relationship between the amount of electric field and the distance between two particles and the electric field pointed in a line between the particles.

Also, he discovered that the charge of an particle (ie. positive or negative) determined the direction of the electric field (either a repulsion or attraction).

From these observations, as well as the use of fundamental constants, the equation of the electric field due to a point charge was created.

See also

Electric Field More general ideas about electric fields
Electric Force One application of electric fields due to point charges deals with finding electric force

Further reading

Principles of Electrodynamics by Melvin Schwartz ISBN: 9780486134673

External links

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

References

Matter and Interactions Vol. II


Charles-Augustin de Coulomb. (n.d.). Retrieved December 3, 2015, from https://nationalmaglab.org/education/magnet-academy/history-of-electricity-magnetism/pioneers/charles-augustin-de-coulomb


Shech, E., & Hatleback, E. (n.d.). The Material Intricacies of Coulomb’s 1785 Electric Torsion Balance Experiment. Retrieved December 3, 2015, from http://philsci-archive.pitt.edu/11048/1/The_Material_Intricacies_of_Coulomb's_1785_Electric_Torsion_Balance_Experiment_(EV).pdf