I am trying to solve Problem 3.21 in Introduction to Electrodynamics, Griffiths where I am asked:
Find the potential outside a charged metal sphere of charge Q and radius R, placed in an otherwise uniform electric field $\mathbf E_0$.
Let us orient our coordinate system such that the electric field acts along the z-axis.
- BC 1: The sphere is conductive, thus set $V(R, \theta)=0$.
- BC 2: As $r \rightarrow \infty$, we notice that $V \rightarrow -E_0r \cos \theta- \frac{Q}{4\pi\epsilon_0r}$
Note the Laplace Equation solution in azimuthal-symmetric cases in spherical coordinates is given by:
$$V(r,\theta)=\sum_{l=0}^{\infty}{(A_l r^l+\frac{B_l}{r^{l+1}})P_l\cos(\theta)}$$
I am currently stuck at trying to make the two boundary conditions work together, all I get is a limit form of what the coefficients should be, and even an incompatibility.
Applying BC 1: $$V(r,\theta)=\sum_{l=0}^{\infty}{A_l( r^l-\frac{R^{2l+1}}{r^{l+1}})P_l\cos(\theta)}$$
But clearly for significantly large $r$, the $\frac{R^{2l+1}}{r^{l+1}}$ terms vanish, and now we can't use the part of the second boundary condition that scales as $\frac{Q}{4\pi\epsilon_0r}$, which is not a surprise, but the problem is that the second boundary condition is incompatible with the the first, due to the $\frac{Q}{4\pi\epsilon_0r}$ and $-E_0r \cos \theta$ terms not fitting the form required when we first applied BC 1.
Please could someone clarify on the issue of this incompatibility (Though not actually solve the problem using a different method, I am trying to understand where I went wrong with this method.)