View source for Electrodynamics/Tutorials/3/3/2/Example 9

== 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.