Electric potential, getting coefficients, spherical harmonics

[fluidistic]

Homework Statement

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.

Homework Equations

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

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.

 

[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

 

[fluidistic]

Hello, fluidistic.

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

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

 

[TSny]

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

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 $cos\phi$.

 

[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.