The potential $\Phi(r, \vartheta)$, satisfies Laplace's equation everywhere except on a sphere of unit radius and $\Phi \rightarrow 0$ as $r \rightarrow \infty$. The potential is continuous at $r=1$, but the derivative of the potential satisfies

$$\lim _{r \rightarrow 1^{+}} \frac{\partial \Phi}{\partial r}-\lim _{r \rightarrow 1^{-}} \frac{\partial \Phi}{\partial r}=V \cos ^{2} \vartheta$$

where $V$ is a constant. Use the method of separation of variables to find $\Phi$ for both $r>1$ and $r<1$.

[The Laplacian in spherical polar coordinates for axisymmetric systems is

$$\nabla^{2} \equiv \frac{1}{r^{2}}\left(\frac{\partial}{\partial r} r^{2} \frac{\partial}{\partial r}\right)+\frac{1}{r^{2} \sin \vartheta}\left(\frac{\partial}{\partial \vartheta} \sin \vartheta \frac{\partial}{\partial \vartheta}\right)$$

You may assume that the equation

$$\left(\left(1-x^{2}\right) y^{\prime}\right)^{\prime}+\lambda y=0$$

has polynomial solutions of degree $n$, which are regular at $x=\pm 1$, if and only if $\lambda=n(n+1) .]$