Point Charge: Difference between revisions

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This page is all about the Electric Field due to a Point Charge.

Electric Field

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A Mathematical Model of Electric Field due to Point Charge

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

[math]\displaystyle{ \vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r,\text{where } \frac{1}{4 \pi \epsilon_0 } \text{is approximately } 9*10^{9} \frac{N m^2}{C^2} \text{, q is the charge of the particle,} \text{r is the magnitude of the distance} \text{ between the point charge and the observation point, and } }[/math] [math]\displaystyle{ \hat r \text { is the direction of the distance from } \text{the point charge to the observation point.} \text{ This equation becomes Coulomb's Law when multiplied by a second particles's charge. } }[/math]

A Computational Model

Here is a link to some code which shows the Electric Field due to an Proton at different points.

Examples

Problem 1: There is a proton at <1,2,3>. Calculate the electric field at <2,-1,3>.

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

Find [math]\displaystyle{ \vec r_{obs} - \vec r_{proton} (\lt 2,-1,3\gt - \lt 1,2,3\gt = \lt 1,-3,0\gt ) }[/math]

Calculate the magnitude of r. ([math]\displaystyle{ \sqrt{1^2+(-3)^2+0^2}=\sqrt{10} }[/math]

From r, find the unit vector [math]\displaystyle{ \hat{r}. }[/math] [math]\displaystyle{ \lt \frac{1}{\sqrt{10}},\frac{-3}{\sqrt{10}},\frac{0}{\sqrt{10}}\gt }[/math]

Step 2: Find the magnitude of the Electric Field

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

Step 3: Multiply the magnitude 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}}{10}*\lt \frac{1}{\sqrt{10}},\frac{-3}{\sqrt{10}},\frac{0}{\sqrt{10}}\gt =\lt 4.554*10^{-11},-1.366*10^{-10},0\gt N/C }[/math]

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.

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

@@ Line 9: / Line 9: @@
 The Electric Field of a Point Charge can be found by the formula:
-<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r,\text{where } \frac{1}{4 \pi \epsilon_0 } \text{is approximately }  9*10^{9} \text{, q is the charge of the particle,}
+<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r,\text{where } \frac{1}{4 \pi \epsilon_0 } \text{is approximately }  9*10^{9} \frac{N m^2}{C^2} \text{, q is the charge of the particle,}
 \text{r is the magnitude of the distance}

Point Charge: Difference between revisions

Revision as of 15:42, 7 February 2016

Contents

Electric Field

A Mathematical Model of Electric Field due to Point Charge

A Computational Model

Examples

Connectedness

History

See also

Further reading

External links

References

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