- #1

fluidistic

Gold Member

- 3,949

- 264

## 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.