Charged conducting sphere in a uniform electric field?

[TMFKAN64]

Hello, I've been working my way through Griffiths' "Introduction to Electrodynamics" book, and I'm slightly confused by Problem 3.20. For those of you without a copy of this book, given a conducting sphere of radius R and charge Q in a uniform electric field of strength E0, what is the potential outside the sphere?

Intuitively, this is virtually identical to example 3.8, which solved the problem for an uncharged sphere. The sphere is an equipotential which can be defined to be zero when r = R. Far away from the sphere, the field is just E0 in, say, the z direction, so the potential is -E0 z there. After playing games with separation of variables, you end up with V(r, theta) = -E0 (r - R^3/r^2) cos(theta).

In this problem, I can still say that the sphere is an equipotential zero and field is the same as above when we get far from the sphere. If we are to add a charge Q to an equipotential sphere so that it remains an equipotential, the charge must be uniform over the surface of the sphere. Therefore, the charged solution is V(r, theta) = -E0 (r - R^3/r^2) cos(theta) - Q/(4 pi e0 r).

I'm not happy with this solution though. Am I missing some way of setting up the problem and solving it directly without frantically waving my hands over the important bits?

 

[kuruman]

You shouldn't be happy. You can no longer consider the potential on the sphere zero because it has charge Q which raises the potential of the conductor to $V_0=\frac{Q}{4 \pi \epsilon_0 R}$. It also raises the potential every where outside the sphere by adding an additional $1/r$ term. Thus, the new potential outside the sphere is the old potential plus the potential of point charge $Q$ placed at the center of the sphere. This new potential is a solution to Laplace's equation and satisfies the boundary conditions. Therefore, by the uniqueness theorem is the solution.