Conducting Spherical Shell Capacitor

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A conducting spherical shell is divided into upper and lower halves, with the top half at 10V and the bottom at -10V, requiring the application of Laplace's equation. The general solution involves spherical coordinates, but the B coefficients must be zero to ensure finite potential at the origin. The challenge lies in satisfying the boundary conditions at radius a, which requires determining the nonzero A coefficients. The discussion highlights the difficulty of applying the boundary conditions based on the angular dependence of the potential. This problem exemplifies the complexity of advanced electromagnetic theory typically encountered in graduate-level courses.
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Homework Statement


A conducting spherical shell is divided into upper and lower halves with a narrow insulating ring between them. The top half is at 10V and the bottom half is at -10V. Write down the appropriate expansion for Φ and use symmetry and the expected behavior at the origin to identify which coefficients are zero. Then solve for the nonzero coefficients which make Φ satisfy the values given at r = a. You will undoubtedly have to express the coefficients in integral form.


Homework Equations


No charge inside, so Laplace's equation applies:
\nabla^{2}\phi=0
Given the general solution for solving Laplaces equation in spherical coordinates:
\phi (r,\theta,\varphi)= \sum^{\infty}_{n=0}(A_{n}r^{n}+\frac{B^{n}}{r^{n+1}})P_{n}(cos\theta)

The Attempt at a Solution


I've only concluded so far that the B coefficients must all be 0 due to requiring finite potential at r=0. Past that I'm at a loss on how to tackle the function. I know:
\phi (r,\theta,\varphi)= \sum^{\infty}_{n=0}A_{n}r^{n}P_{n}(cos\theta)
But I don't know how I can tackle the boundary condition of plus and minus 10 at radius a, depending on angle theta.

Thanks in advance.
 
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Also I'm not quite sure which forum to put this in. It's a fourth year undergrad course, but all I've been told about the professor is that he gives us grad school type problems like this one, as previous graduates have come back and told us that their graduate EM course was actually easier.
 

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