# Electric potential, getting coefficients, spherical harmonics

1. Jan 10, 2013

### fluidistic

1. The problem statement, all variables and given/known data
Consider 2 conductor spherical shells of radii a and b (where a>b). The inner shell is at zero potential and the outer shell is at a potential given by $V(\theta, \phi )=V_0 \sin \theta \cos \phi$ where $V_0$ is constant and theta and phi are the usual spherical coordinates. Find the potential in all the space.

2. Relevant equations
Laplace equation: $\triangle \Phi (r, \theta, \phi ) =0$ (for the region between the 2 shells.

3. The attempt at a solution
Inside the inner shell the potential is 0 because it must be constant since the E field is 0 there and it must equal the value of the potential of the inner sphere which is 0.
For the region between the 2 shells, I solve the Laplace equation in spherical coordinates.
This yields $\Phi (r, \theta, \phi ) = \sum _{l=0}^\infty \sum _{m=-l}^l [A_{lm} r^l+B_{lm} r^{-(l+1)}] Y_{lm}(\theta , \phi )$.
I apply the boundary condition $\Phi (b)=0$ which gives me that $B_{lm}=-A_{lm}b^{2l+1}$. So that $\Phi (r, \theta, \phi ) = \sum _{l=0}^\infty \sum _{m=-l}^l A_{lm} [r^l-b^{2l+1}r^{-(l+1)}]Y_{lm}(\theta , \phi )$.
The other boundary condition, $\Phi (a, \theta , \phi )=V_0 \sin \theta \cos \phi$ gives me that $\sum _{l=0}^\infty \sum _{m=-l}^l A_{lm} [a^l -b^{2l+1}a^{-(l+1)}]Y_{lm} (\theta , \phi ) =V_0 \sin \theta \cos \phi$. So I am left to calculate the $A_{lm}$ coefficients and I'd be done for the exercise if I didn't make any mistake.
I don't really know how to calculate those coefficients. I'm guessing some multiplication by $Y^*(\theta, \phi )$ and then some integration but the fact that the spherical harmonics are inside the sums make me unable to do it.

2. Jan 10, 2013

### TSny

Hello, fluidistic.

See if you can write $sin\theta cos\phi$ as a sum of spherical harmonics by inspection. See table of spherical harmonics

3. Jan 11, 2013

### fluidistic

I could not write it as a sum although I notice that $\cos \phi \sin \theta = \sqrt { -\frac{8\pi }{15} Y_{2,1}Y_{-2,1}}$.

4. Jan 11, 2013

### TSny

Note that $Y_{2,1}$ involves $sin\theta cos\theta$ rather than $sin\theta cos\phi$

It will help to rewrite $sin\theta cos\phi$ using the complex exponential representation for $cos\phi$.

5. Feb 17, 2013

### fluidistic

Sorry for my long silence, I was on a trip and now I'm back at home. First of all, thank you very much.
So I used the expression you suggested me to use and ended up with $A_{lm}=0$ for all $l \neq 1$.
The only 2 terms that were different from 0 ended up to be $A_{11}=\frac{\sqrt{\frac{2\pi}{3}}V_0}{\frac{b^3}{a^2}-a}$ and $A_{1-1}=\frac{\sqrt{\frac{2\pi}{3}}V_0}{a-\frac{b^3}{a^2}}$.
I might have made some arithmetic/algebra mistake but at least I've been totally unstuck by you.
For the region outside both shells, again I must solve Laplace equation and use the fact that the potential remains finite at infinity and apply the boundary condition over the surface of the sphere of radius a. It should not be that hard I guess, I'll do it if I have the time.