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

[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

 

[gabbagabbahey]

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

 

[xn_]

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

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