How does a conductor shield the outside from its inside?

Question

If I have a charge of $+q$ placed arbitrarily within the spherical conducting shell, by Gauss' law, the $E$ field produced will be created outside the shell, as if the charge were placed at the center. If this charge is not placed in the center, I don't see how we can use symmetry to argue why the outside should not be able to 'see' how the inside of the shell looks like.

I'm aware that the charges in the conductor rearranges to make the Electric field within zero, but it is not apparent to me that this rearrangement should 'balance' out any perturbations created by the moving the enclosed charge off the center.

Answer 1

Here is my attempt to explain the final E-field pattern.

A point charge $+q$ produces radial E-field lines.

Diagram 1
Suppose that a conducting spherical shell is placed around the point charge with the point charge not at the centre of the conducting shell.
Since the E-field inside the conductor must be zero there must be another E-field inside the conductor (red) which is equal and opposite to that produced by the charge $+q$.
This induced field is produced by induced charges formed on the inside and the outside of the conducting shell.
For a number of reasons this cannot be the final state of the system one of them being that E-field lines must be perpendicular to the surface of a conductor.

Diagram 2
The E-field lines are now all perpendicular to the conducting shell.
As soon as this condition is satisfied (some) information about the position of the charge $+q$ when viewed from outside the conducting shell is lost because all the field near the outside surface of the conductor appear to come from the centre of the conducting shell. This diagram is again flawed one reason being that in this configuration the conductor is not an equipotential.
The work done in bringing a positive charge from infinity to the left hand surface surface of the conductor is more than the work done in bringing the same charge to a point on the right hand side of the conductor.
This cannot be so because the conducting shell is an equipotential.

Diagram 3
The induced charges on the outside (and inside?) rearrange themselves so that the work done in bringing a change to any point on the outside surface of the conducting shell is the same whilst at the same time making sure there is no E-field inside the conductor.
So to the outside world outside the conducting sphere the actual position of the charge $+q$ is unknown and the E-field is the same as that which would be produced by having charge $+q$ at the centre of the spherical conducting shell.

Answer 2

That's the beauty of inverse square laws: everything balances out for Newton's Shell Theorem, which Newton proved by geometric means, but which you can show quickly with Gauss' s theorem.

You may find it illuminating to work through Newton's proof, in a modern form.

Answer 3

If I have a charge of +q placed arbitrarily within the spherical conducting shell, by Gauss' law, the E field produced will be created outside the shell, as if the charge were placed at the center.

That isn't a law of physics. In fact, it only happens when the spherical conducting shell is electrically neutral.

Here's what happens:

Based on the charge inside some opposite charge arranges itself on the inside of the shell to shield the shell (and all the space outside) from the electric field due to the charge.

If the shell had a charge equal and opposite to the charge inside and there aren't any charges outside that would be it, there would be no electric field anywhere in the universe except on that inside surface and the charge inside. Laws of physics satisfied.

Now if there is some charge outside and there is +q charge on the outside that is free to move around, then the charge arranges itself on the outside of the shell to shield the entire inside (shell and the cavity inside) from the charge outside (just as if the shell were a solid conducting ball).

Now. What if the shell had some charge (possibly zero, but something fixed) and wasn't connected to a source of charge. Then every bit of charge that appeared on the inside shell or the outside shell needs to be added up and compare to the total charge. So if you have -q on the inside surface and +q on the outside surface you get a total charge of 0. But all you needed is that the inner surface needs to be -q and the two need to add up to the total charge.

Now. What happens if the charge inside is put somewhere else. The field on the inside surface will be the same magnitude but will be rearranged differently. But the two together produce a zero field outside. The charge on the outside surface doesn't change and that's what makes a field outside.