# Homework Help: Potential Outside charged metal sphere in uniform E field

1. Oct 5, 2015

### Jonathan K

1. The problem statement, all variables and given/known data
Find the potential outside a charged metal sphere (charge Q, radius R) placed in an otherwise uniform electric field E0. Explain clearly where you are setting the zero of potential.

2. The attempt at a solution
So for this problem I figured I could exploit superposition and take the potential outside a neutral sphere in a uniform field in the z direction. This potential is V1(r, θ) = -E0(r - R3/r2)cosθ. Then I figured I could simply use the potential outside a uniformly charged sphere of charge Q which would be V2(r, θ) = 1/4πε0Q/r. Then the solution would become V(r, θ) = V1(r, θ) + V2(r, θ) where the zero would have to be along the xy plane where r →∞ since, that is where the potential vanishes by inspection. The only hangup I am having here is whether I am using superposition correctly. Is my argument valid here?

2. Oct 5, 2015

### Simon Bridge

In the field the sphere cannot be uniformly charged - it's a conductor.

3. Oct 6, 2015

### BvU

Hello Simon, I've been pondering over this one, too. Jon treats it as a superposition, so wouldn't that include the non-uniform charge redistribution to get the V1 ?

4. Oct 6, 2015

### rude man

I see nothing wrong with this approach.

One way to model the uniform E field is via two point charges of opposite polarity placed at + and - long distances from the sphere, and choosing q of each appropriately. In this way a uniform E field of any desired direction and magnitude can be modeled with just two point charges .

Then we take an uncharged sphere in this E field and place image charges within it arranged to satisfy the equipotential of the sphere's surface. Then, making the sphere charged just adds the extra point charge Q to the sphere's center. Then superposition is just adding the E fields due to the image charges (there are two of them) and extra charge Q, plus the two charges setting up the E field, all five of which being point charges and amenable to superposition.
.

Last edited: Oct 7, 2015
5. Oct 7, 2015

### rude man

Everybody lose interest? Too bad, I thought it was a good problem ...

Extra observation: how do we know placing the charge Q on the sphere will be at the center? Answer: because any other location would spoil the equipotential set up by the external E field and the image charges.

6. Oct 10, 2015

### Simon Bridge

A charge Q on the sphere won't be at the center because it is "on the sphere".
Its not so much a matter of losing interest as a matter of no further comment from OP.

7. Oct 11, 2015

### rude man

We are talking image charges here. One at the center = Q, the other two at distances +/- R^2/d from the center, d being the distance from the sphere to each of the charges setting up the external field. So, 5 image charges. Obviously, there are no charges within the sphere.