1. Feb 11, 2004

### AngelofMusic

In one of our exercises, there is a conducting sphere (S2) with a cavity inside (S1). A point charge is placed inside the cavity. We were told to try and find the electric field outside by following a set of steps.

1. Imagine that S2 is grounded. In that case, the charge on the cavity wall (S1) is -q and the charge on the outer surface is 0.

2. Imagine that only S2 exists, with no cavity and no point charge.

3. Use super-position from step 1 & 2 to determine the total electric field outside the original shell.

a) The charge on the inner cavity wall is -q.

b) The charge on the outer wall is +q (I assume this is because the conductor should remain neutral?).

c) This is where I get lost. The solution says that:

From superposition and uniqueness, it can be inferred that the charge over S2 is uniformly distributed. Hence, it can be treated as though a charge of q was placed at the centre of the sphere.

What does this mean, exactly? Superposition and uniqueness? How has the previous parts proved this, exactly?

2. Feb 11, 2004

### turin

The previous parts have not proved it. One of the things that is supposed to be learned in E&M is that superposition and uniqueness apply to solutions of Poisson's equation for electrostatics.

Superposition in this case means that the field generated by a charge distribution is the same as the vector sum of the fields that would be produced by all the individual pieces of the charge distribution. There are three pieces: the charge put inside, the charge on the inner surface, and the charge on the outer surface. By putting the charge inside and grounding the sphere, you account for two pieces. By having a charged sphere without a cavity, you account for the third.

Uniqueness is a principle, kind of like a rule of thumb that always works, that says, if you find one solution to the problem, then it is the solution. So, basically, uniqueness is the byword that transforms the result you get from the superposition approach into the answer to the problem.

It certainly isn't at all straightforward to me (i.e. I think it sounds pretty hand-wavey).

Last edited: Feb 11, 2004
3. Feb 11, 2004

### AngelofMusic

Ah, okay. Thanks a lot!

So, basically, to solve the problem of the electric field, I could sum up the previous two situations and see that the total charge enclosed would be:

+q (pt charge) -q (inner cavity) +q (outer surface) = +q. And then draw a Gaussian surface over the sphere, and use Gauss' Law to figure out the Electric field, right?

4. Feb 11, 2004

### turin

Yeah, kind of. But to use Gauss' Law, you need symmetry, and the location of the point charge that you put inside the cavity is unspecified (and the shape of the cavity too, if I remember correctly). So, you have to have some arguement (in order to do this somewhat rigorously) in order to show that the only thing that matters is the charge distribution on the outer surface and that this charge distribution is uniform.