# Homework Help: Field strength giving rise to a polarization charge of -s.cos(theta) on a cavity

1. Dec 9, 2008

### xn_

Hello,
I'm learning EM from Bleaney & Bleaney and got stuck on Ex2.1 (can do Ex2.2-2.7, though...) - If the polarization charge on the surface of a spherical cavity is -s.cos(theta), prove that the field strength at the centre is s/3e0. If I expand V(r) within the cavity as A.r.cos(theta) + B.r^-2.cos(theta), then I agree B=0 or else V(0) would be singular, but that leaves the A term and surface charge = e0Er = -e0(dV/dr) = -e0A.cos(theta) and hence A = s/e0 not s/3e0 - what am I doing wrong?
Thanks,
Julia

2. Dec 10, 2008

### gabbagabbahey

Check your boundary condition. Is that statement really true when the potential outside the boundary is not constant in 'r'?

3. Dec 11, 2008

### xn_

I've been thinking that the potential outside the cavity (radius R) could be written Vout(r) = B_1.r.cos(theta)+B_2.r^-2.cos(theta), then, and the boundary conditions are that Vout(R)=Vin(R) which implies:
A = B_1 + B_2 R^-2
and that e0E_in,radial(R) = eE_out,radial(R) (where e is the permittivity of the medium surrounding the cavity) which implies:
-e0.A = -e(B_1 - 2B_2.R^-3)
But your reply (thanks!) suggests that I'm doing something fundamentally wrong here - am I wrong to take Vin = A.r.cos(theta) because the Vout depends on r? I see that something of the form B.r would help, but I thought each power of r had do go with a certain power of cos(theta) to make a valid spherical harmonic?
Thanks, Julia