Introduction to Electrodynamics/Chapter 2

2.1 The Electrostatic Field

2.1.1 Introduction

The question of Electrodynamics is, given point charges distributed and moving around, what effect will they have on a test charge placed somewhere in space?

This is not an easy question to answer. To do so, we have to tackle the problem in stages. First, we'll consider what happens when the charges are not moving around.

2.1.2 Coulomb's Law

<math> \mathbf{F} = {1 \over 4 \pi \epsilon_0} {q Q \over \mathbf{r}^2 } \hat \mathbf{r} </math>

Where:

• <math>\mathbf{F}</math> is the force on charge Q
• <math>\epsilon_0 = 8.85 \times 10^{-12} {C^2 \over N \cdot m^2}</math>, the permittivity of free space.
• <math>q</math> is the charge of the other charge.
• <math>Q</math> is the charge of the charge in question.
• <math>\mathbf{r}</math> is the vector from <math>q</math> to <math>Q</math>

2.1.3 The Electric Field

When you have multiple charges pushing or pulling on one charge, it makes sense to pull the charge out and look at each contributing force individually. In math,

<math> \begin{align} \sum \mathbf{F} &= \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 + ... + \mathbf{F}_n \\

 &= {1 \over 4 \pi \epsilon_0} {q_1 Q \over \mathbf{r}_1^2 } \hat \mathbf{r}_1 +

{1 \over 4 \pi \epsilon_0} {q_2 Q \over \mathbf{r}_2^2 } \hat \mathbf{r}_2 + {1 \over 4 \pi \epsilon_0} {q_3 Q \over \mathbf{r}_3^2 } \hat \mathbf{r}_3 + ... + {1 \over 4 \pi \epsilon_0} {q_n Q \over \mathbf{r}_n^2 } \hat \mathbf{r}_n \\

&= \sum {1 \over 4 \pi \epsilon_0} {q_i Q \over \mathbf{r}_i^2 } \hat \mathbf{r}_i \\
&= Q \left ( \sum {1 \over 4 \pi \epsilon_0} {q_i \over \mathbf{r}_i^2 } \hat \mathbf{r}_i

\right ) \\

&= Q \mathbf{E} \\

\end{align} </math>

where <math>\mathbf{E}(\mathbf{x}) = \sum {1 \over 4 \pi \epsilon_0} {q_i \over \mathbf{r}_i^2 } \hat \mathbf{r}_i</math>.

Note that the electric field <math>\mathbf{E}</math> needs a position, <math>\mathbf{x}</math>.

Note that the position or charge of Q is irrelevant to the electric field. If you can calculate the electric field at any point, then the force on the particle at that point depends only on its charge and the electric field.

2.1.4 Continuous Charge Distributions

What about the case where you have an infinite number of point charges that come together to charge distributions? We simply use calculus to find the electric field.

Problem 2.7

This is rather difficult. The integrals I had to look up.

First, let's start with a formal definition of <math>\mathbf{r}</math>. This is the vector that points from a point on the surface of the sphere to the point z above the x-y plane.

So, it is simply:

<math> \begin{align} \mathbf{r} &= z \hat k - (R \sin \theta \cos \phi \hat i + R \sin \theta \sin \phi \hat j + R \cos \theta \hat k) \\

 &= -R \sin \theta \cos \phi \hat i -R \sin \theta \sin \phi \hat j + (z - R \cos \theta) \hat k \\

\end{align} </math>

The quantity <math>r^2</math> is simply <math>\mathbf{r} \dot \mathbf{r}</math>:

<math> \begin{align} r^2 &= R^2 \sin^2 \theta \cos^2 \phi + R^2 \sin^2 \theta \sin^2 \phi + (z - R \cos \theta)^2 \\

   &= R^2 \sin^2 \theta + z^2 + R^2 \cos ^2 \theta - 2 z R \cos \theta \\
   &= R^2 + z^2 - 2 z R \cos \theta \\

\end{align} </math>

Since this is what the Law of Cosines would've given us, it's not surprising.

Now, to calculate the elusive quantity <math>\hat r / r^2 = \mathbf{r} / (r^2)^{3/2}</math>:

<math> \begin{align} {\hat r \over r^2} &= {-R \sin \theta \cos \phi \hat i -R \sin \theta \sin \phi \hat j + (z - R \cos \theta) \hat k

 \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\

\end{align} </math>

With this, we can calculate the infinitesimal <math>d\mathbf{E}</math>.

<math> \begin{align} da &= R^2 \sin \theta d\theta d\phi \\ d\mathbf{E} &= {1 \over 4 \pi \epsilon_0} \sigma da {\hat r \over r^2} \\ dE_x &= -{1 \over 4 \pi \epsilon_0} \sigma R^2 \sin \theta d\theta d\phi {R \sin \theta \cos \phi \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\

    &= -{1 \over 4 \pi \epsilon_0} { \sigma R^3 \sin^2 \theta d\theta \cos \phi d\phi \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\

dE_y &= -{1 \over 4 \pi \epsilon_0} \sigma R^2 \sin \theta d\theta d\phi {R \sin \theta \sin \phi \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\

    &= -{1 \over 4 \pi \epsilon_0} { \sigma R^3 \sin^2 \theta d\theta \sin \phi d\phi \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\

dE_z &= {1 \over 4 \pi \epsilon_0} \sigma R^2 \sin \theta d\theta d\phi {z - R \cos \theta \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\

    &= {1 \over 4 \pi \epsilon_0} {\sigma R^2 \sin \theta d\theta d\phi (z - R \cos \theta) \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\

\end{align} </math>

Of note, the x and y components of the electric field will be zero when integrating over the sphere. This is because they will cancel out with the <math>\phi</math> integral.

<math> \begin{align} \int dE_x &= \int -{1 \over 4 \pi \epsilon_0} { \sigma R^3 \sin^2 \theta d\theta \cos \phi d\phi \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\

         &= -{1 \over 4 \pi \epsilon_0} \sigma R^3  \int_{\phi=0}^{2\pi} \cos \phi d\phi \int_{\theta=0}^{\pi} {\sin^2 \theta d\theta \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\
         &= -{1 \over 4 \pi \epsilon_0} \sigma R^3  (\sin \phi) \big|_{0}^{2\pi} \int_{\theta=0}^{\pi} {\sin^2 \theta d\theta \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\
         &= -{1 \over 4 \pi \epsilon_0} \sigma R^3  (\sin 2\pi - \sin 0) \int_{\theta=0}^{\pi} {\sin^2 \theta d\theta \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\
         &= -{1 \over 4 \pi \epsilon_0} \sigma R^3  (0) \int_{\theta=0}^{\pi} {\sin^2 \theta d\theta \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\
         &= 0 \\

\int dE_y &= 0 \text{ by similar reasoning}\\ \end{align} </math>

The z component, of course, is where the electric field will end up.

<math> \begin{align} \mathbf{E} &= \int dE_z \hat k \\ \int dE_z &= \int {1 \over 4 \pi \epsilon_0} {\sigma R^2 \sin \theta d\theta d\phi (z - R \cos \theta) \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\

         &= {1 \over 4 \pi \epsilon_0} \sigma R^2 \int_{0}^{2\pi} d\phi \int_{0}^{\pi} {\sin \theta (z - R \cos \theta) d\theta \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\
         &= {1 \over 4 \pi \epsilon_0} 2 \pi \sigma R^2 \int_{0}^{\pi} {\sin \theta (z - R \cos \theta) d\theta \over (R^2 + z^2 - 2 z R \cos \theta)^{3/2}} \\

\end{align} </math>

The integral is pretty scary. If you look it up, you may not find it. So we will have to substitute <math>x = - \cos \theta; \ dx = \sin \theta d\theta</math>. Don't forget to change the boundaries.

<math> \begin{align} \mathbf{E} &= \int dE_z \hat k \\ \int dE_z &= {1 \over 4 \pi \epsilon_0} 2 \pi \sigma R^2 \int_{-1}^{1} {(z + R x) dx \over (R^2 + z^2 + 2 z R x)^{3/2}} \\ \end{align} </math>

The integral is really the sum of two different integrals now. The first one:

<math> \begin{align} \int_{-1}^{1} {z dx \over (R^2 + z^2 + 2 z R x)^{3/2}}

&= z \int_{-1}^{1} (R^2 + z^2 + 2 z R x)^{-3/2} dx \\
&= z \int_{-1}^{1} (a + b x)^{n} dx
   && \text{ where } a = R^2 + z^2,\  b = 2zR,\  n = -3/2 \\
&= z \left ( {(a + bx)^{n+1} \over b(n+1)} \right ) \bigg |_{-1}^1 \\
&= {z \over b(n+1)} \left ( {1 \over \sqrt{a+b}} - {1 \over \sqrt{a-b}} \right) \\
&= -{1 \over R} {\sqrt{a-b} - \sqrt{a+b} \over \sqrt{(a+b)} \sqrt{(a-b)}} \\
&= {\sqrt{a+b} - \sqrt{a-b} \over R \sqrt{(a+b)} \sqrt{(a-b)}} \\

\end{align} </math>

It is at this point that we must pause. Since we have square roots, the question is do we take the positive or the negative roots?

We'll try the other half of the integral now.

<math> \begin{align} \int_{-1}^{1} {R x dx \over (R^2 + z^2 + 2 z R x)^{3/2}}

 &= R \int_{-1}^{1}  x (R^2 + z^2 + 2 z R x)^{-3/2} dx \\
 &= R \int_{-1}^{1}  x (a + b x)^{n} dx && \text{ same values of } a, b, n \\
 &= R \left( {(a + bx)^{n+2} \over b^2(n+2)} \right) \bigg |_{-1}^{1} \\
 &= R {\sqrt{a + b} - \sqrt{a - b} \over 2 z^2 R^2} \\
 &= {\sqrt{a + b} - \sqrt{a - b} \over 2 z^2 R} \\

\end{align} </math>

Once again, we must pause.

At this point, let's combine the two integrals.

<math> \begin{align} \int_{-1}^{1} {(z + R x) dx \over (R^2 + z^2 + 2 z R x)^{3/2}}

 &= {\sqrt{a+b} - \sqrt{a-b} \over R \sqrt{(a+b)} \sqrt{(a-b)}} +  {\sqrt{a+b} - \sqrt{a-b} \over 2 z^2 R} \\

\end{align} </math>

Note that:

<math> \begin{align} (a+b) &= R^2 + z^2 + 2zR \\

     &= (R+z)^2 \\

(a-b) &= R^2 + z^2 - 2zR \\

     &= (R-z)^2

\end{align} </math>

Let's talk about the square roots. We have only two square roots that appear above.

<math> \begin{align} \sqrt{a+b} &= \sqrt{(R+z)^2} \\

          &= \pm (R+z) \\

\sqrt{a-b} &= \sqrt{(R-z)^2} \\

          &= \pm (R-z) \\

\end{align} </math>

Substituting that in:

<math> \begin{align} & {\sqrt{a+b} - \sqrt{a-b} \over R \sqrt{(a+b)} \sqrt{(a-b)}} + {\sqrt{a+b} - \sqrt{a-b} \over 2 z^2 R} \\ &= {\pm (R+z) \pm(R-z) \over R (R+z)(R-z)} + {\pm(R+z) \pm(R-z) \over 2z^2R} \\ &= {\pm 2z^2 (R+z) \pm 2z^2 (R-z) \pm (R+z)^2(R-z) \pm (R+z)(R-z)^2 \over 2z^2R(R+z)(R-z)} \\ \end{align} </math>

Now my bias is going to show. I am going to expect that the electric field is going to be:

<math> \begin{align} \mathbf{E} &= {1 \over 4 \pi \epsilon_0} {4 \pi R^2 \sigma \over z^2} \hat k && \text{ when } z > R \text{ (point charge) } \\ \mathbf{E} &= 0 && \text{ when } z < R \\ \end{align} </math>

Why is this the case? Because of Gauss's Law.

The question is: Can I change the mess I have above into the two equations I described?

When <math>z > R</math>, then the integral I tried to solve should equal <math>2 / z^2</math>.

When <math>z < R</math>, all I have to show is that the integral we tried so hard to solve is 0.