Electrodynamics/Tutorials/3/3/2/Example 9

Video Intro

Hi, this is Jonathan Gardner.

We're covering [section reference] of Griffiths Introduction to Electrodynamics.

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As always, questions in a video response or comments.

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Intro

A surface charge <math>\sigma_0(\theta)\;</math> is glued onto a sphere of radius <math>R\;</math>. Find the potential inside and outside the sphere.

We get to find the potential inside AND outside, so the boundary conditions are important. (See 2.3.5 for what a surface charge does to the potential.)

Middle

We are going to calculate <math>V_{\text{out}}(r,\theta)\;</math> and <math>V_{\text{in}}(r,\theta)\;</math>. Both of them have the form:

<math>V(r,\theta) = \sum_{l=0}^\infty (A_lr^l + {B_l \over r^{l+1}})P_l(\cos \theta)</math>

Obviously, the A term blows up as r goes to infinity, so it's no good for the outside. The B term blows up at the origin, so it's no good for the inside. So we're left with:

<math>V_{\text{out}}(r,\theta) = \sum_{l=0}^\infty {B_l \over r^{l+1}}P_l(\cos \theta)</math>

and

<math>V_{\text{in}}(r,\theta) = \sum_{l=0}^\infty A_l r^l P_l(\cos \theta)</math>

Continuous

The potential must be continuous at <math>r = R\;</math>. That means, for both these potentials, we have to make them equal when they reach R.

<math>

\begin{align} V_{\text{out}}(R,\theta) &= V_{\text{in}}(R,\theta) \\ \sum_{l=0}^\infty {B_l \over R^{l+1}}P_l(\cos \theta)

&= \sum_{l=0}^\infty A_l R^l P_l(\cos \theta) \\

\end{align} </math>

Now, I'm going to wink at you and you are going to believe me when I say that this implies that the insides must be equal for each term. The book mentions how you can do this---simply multiply by <math>P_l(\cos \theta) \sin \theta\;</math> and integrate from 0 to pi, just like you would for grabbing the constants at the end.

Let's continue.

<math>

\begin{align} {B_l \over R^{l+1}} &= A_l R^l \\ B_l &= A_l R^l R^{l+1} \\ B_l &= A_l R^{2l+1} \\ \end{align} </math>

Note that this is big R, a constant, not little r, a variable.

Now we've effectively got 1 more constant to go.

First Derivative

Remember back when we discovered that when crossing a surface charge, the electric field and potential jump? That was section 2.3.5. Let me write down that equation for you:

<math>{\partial V_{\text{out}} \over \partial n} - {\partial V_{\text{in}} \over \partial n} = - {1 \over \epsilon_0} \sigma</math>

Here, normal to the sphere is r, so we'll be taking the partial derivative with respect to r.

Well, depending on the theta, at R, we will find this:

<math>

\begin{align} {\partial V_{\text{out}} \over \partial r}

  &- {\partial V_{\text{in}} \over \partial r}
&&= - {1 \over \epsilon_0} \sigma_0(\theta) \\

{\partial \over \partial r} \sum_{l=0}^\infty {A_l R^{2l+1} \over r^{l+1}}P_l(\cos \theta)

  &- {\partial  \over \partial r} \sum_{l=0}^\infty A_l r^l P_l(\cos \theta)
&&= - {1 \over \epsilon_0} \sigma_0(\theta) \\

\sum_{l=0}^\infty A_l R^{2l+1} {\partial \over \partial r} r^{-l-1} P_l(\cos \theta)

  &- \sum_{l=0}^\infty A_l {\partial \over \partial r} r^l P_l(\cos \theta) 
&&= - {1 \over \epsilon_0} \sigma_0(\theta) \\

\sum_{l=0}^\infty A_l R^{2l+1} ((-l-1) r^{-l-2}) P_l(\cos \theta)

  &- \sum_{l=0}^\infty A_l (l r^{l-1}) P_l(\cos \theta) 
&&= - {1 \over \epsilon_0} \sigma_0(\theta) \\

\sum_{l=0}^\infty A_l R^{2l+1} (l+1) r^{-l-2} P_l(\cos \theta)

  &+ \sum_{l=0}^\infty A_l l r^{l-1} P_l(\cos \theta) 
&&=  {1 \over \epsilon_0} \sigma_0(\theta) \\

\sum_{l=0}^\infty ( A_l R^{2l+1} (l+1) r^{-l-2}

  &+ A_l l r^{l-1} )P_l(\cos \theta) 
&&=  {1 \over \epsilon_0} \sigma_0(\theta) \\

\sum_{l=0}^\infty A_l( (l+1) R^{2l+1} R^{-l-2}

  &+ l R^{l-1} )P_l(\cos \theta) 
&&=  {1 \over \epsilon_0} \sigma_0(\theta) \\

\sum_{l=0}^\infty A_l( (l+1) R^{l-1}

  &+ l R^{l-1} )P_l(\cos \theta) 
&&=  {1 \over \epsilon_0} \sigma_0(\theta) \\

\sum_{l=0}^\infty A_l R^{l-1} ( (l+1)

  &+ l )P_l(\cos \theta) 
&&=  {1 \over \epsilon_0} \sigma_0(\theta) \\

\sum_{l=0}^\infty (2l + 1) A_l & R^{l-1} P_l(\cos \theta)

&&=  {1 \over \epsilon_0} \sigma_0(\theta) \\

\end{align} </math>

It is at this point that we can do Fourier's Trick with the Legendre Polynomials. Multiply both sides by <math>P_m(\cos \theta) \sin \theta</math>, then integrate from 0 to pi.

<math>

\begin{align} \int_0^\pi \sum_{l=0}^\infty (2l + 1) A_l R^{l-1} P_l(\cos \theta) P_m(\cos \theta) \sin \theta \ d\theta

&= \int_0^\pi {1 \over \epsilon_0} \sigma_0(\theta) P_m(\cos \theta) \sin \theta \ d\theta \\

\sum_{l=0}^\infty (2l + 1) A_l R^{l-1} \int_0^\pi P_l(\cos \theta) P_m(\cos \theta) \sin \theta \ d\theta

&= {1 \over \epsilon_0} \int_0^\pi \sigma_0(\theta) P_m(\cos \theta) \sin \theta \ d\theta \\

(2m + 1) A_m R^{m-1} {2 \over 2m + 1}

  &= {1 \over \epsilon_0} \int_0^\pi \sigma_0(\theta) P_m(\cos \theta) \sin \theta \ d\theta \\

A_m

 &= {1 \over 2 \epsilon_0 R^{m-1}} \int_0^\pi \sigma_0(\theta) P_m(\cos \theta) \sin \theta \ d\theta \\

\end{align} </math>

And there you go. Now you have the potential inside and outside.

Given a sigma

Let's suppose we're given <math>\sigma_0{\theta} = k \cos \theta\;</math>. Well, that's just the first Legendre polynomial.

So,

<math>

\begin{align} A_m &= {1 \over 2 \epsilon_0 R^{m-1}} \int_0^\pi k P_1(\cos \theta) P_m(\cos \theta) \sin \theta \ d\theta \\ A_1 &= {1 \over 2 \epsilon_0 R^{1-1}} {2k \over 2(1) + 1} \\ A_1 &= {k \over 3 \epsilon_0} \\ \end{align} </math>

Plug that in to the above:

<math>

\begin{align} V_{\text{out}}(r, \theta)

 &= {k \over 3 \epsilon_0} {R^{2+1} \over r^2}P_1(\cos \theta) \\
 &= {kR^3 \over 3 \epsilon_0}{ 1 \over r^2} \cos \theta \\

V_{\text{in}}(r, \theta)

 &= {k \over 3 \epsilon_0}r^1 P_1(\cos \theta) \\
 &= {k \over 3 \epsilon_0}r \cos \theta \\

\end{align} </math>

Note that if you chose <math>k=-3 \epsilon_0 E_0\;</math>, we'll get a field inside of:

<math>-E_0 r \cos \theta = -E_0 z\;</math>

This is the exact opposite of the field in Example 8. Put this sphere in that field, and we'll get an electric field of 0 inside.

What do you know? The surface charge is the same as we calculated.