$E$ outside a Metal Spherric Shell with a off-center Point Charge inside Shell

Question

If a charge +Q is placed inside a metal shell (NOT at the center), as shown below.

I can understand that E will be 0 between inner surface and outer surface due to this is a conductor. Then according to Gauss Law, -Q will be inducted on the inner surface , and +Q will be inducted on the outer surface.

And charge distribution will be non-uniform on the inner surface . But why the charge distribution will be uniform on the outer surface and E outside the shell will be as if the point charge Q is at the center? What is the logical steps to reach this conclusion? I'm looking for an answer that can be understood by students with AP Physics C knowledge. Thanks!

Note that the inner surface has non-uniform distribution. Why outer surface is different? And: E = 0 between inner and outer surface, this is the joint effect of the point charge, the charge on inner surface and the charge on outer surface.

Answer 1

The internal charge induces a charge in the inner surface to cancel the net flux immediately within the conductor. The offset charge produces a non-spherically symmetric field. This field in turn induces a non-spherically symmetric charge distribution on the internal surface. But the charge to create this field has to come from somewhere. There is no effect of the interior charge and the inner induced charge in all space beyond the inner surface. It effectively doesn't exist. But the charge accumulated on the interior surface had to come from some where. There were random changes in the previously neutral charge distribution within the volume. These charges affect each other even though they aren't effected by charges closer to the center. If you place charge in a conductor, it will distribute itself evenly on the outer surface seeking to minimize the potential energy.

This can be demonstrated more explicitly.

The proof of the following relies on calculus, but the relevant math is just algebra. It is called The Method of Images.

Since the electric field is zero within the conductor, the electric potential is constant within the conductor. We only care about differences in potential, so we can call this potential zero.

Given our internal charge, we can calculate the electric potential on the inner surface due to the internal charge.

Now consider an infinite line from the center of the sphere through the internal charge. Where on this line do you need to put a new charge so that its potential plus the potential of the internal charge is zero at the points where the line intersects the internal surface?

You have an equation at each point of intersection. Your equations will have two unknowns. Solve the equations for two unknowns and you will know where on the line to place the charge and how much charge to put.

Now calculate the electric field due to the placement of this Image Charge and the original charge and you have the field due to the original charge and the real induced charge. In some ways a point charge is equivalent to the induced charge.

Answer 2

Imagine a non-hollow metal sphere having a positive charge Q. Here, the distribution will be uniform. Now Imagine I created a cavity inside it, placed another charge Q inside the cavity (not necessarily at center) and gave an additional charge -Q to the sphere. The situation will be same as the one you asked for. But Electric Field inside the sphere will always be zero, so how can the outer charges know if any such cavity exist? They can't, and so the configuration of the outer charges is completely independent of the inner cavity.

Answer 3

I can understand that E will be 0 between inner surface and outer surface due to this is a conductor.

So you agree that the conducting shell is an equipotential volume and in particular the outer surface is all at the same potential?

That means that in bringing a test charge from a point outside the shell to any part of the outer surface of the shell the same amount of external work has to be done.

Now just suppose that the surface positive change density on the outer surface of the shell was greater on the right-hand side than the left hand side.
Because of the symmetrical nature of the outside surface of the shell this would mean that more external work would have to be done bringing the test charge to the right hand outer surface of the shell than the left hand outer surface.
In turn this would indicate that the potential of the right hand outer surface of the shell was greater than the left hand outer surface which you know is not the case, so the charges must be equally distributed over the outer surface of the shell.