My textbook says two different things and I'm not sure how to reconcile these two:

  1. electric field inside a conductor is always 0.

  2. for a conductor with a cavity with a charge q inside it, the field due to the charge q for all exterior points is cancelled by the induced charge on the inner surface.

The charge q will induce charge of -q on the inner surface, and because the conductor is neutral, there will be charge q on the outer surface of the conductor (leftover charge). Due to (2), for all exterior points the field due to q and charges present on the inner surface is cancelled. However, the field due to the induced/leftover charge on the outer surface isn't, so there will be some field due to this charge inside the conductor, which contradicts (1).


The point is the charges on the outside reorganize themselves so the net field is $0$ inside the conductor. The charge distributions on the inside and outside surfaces need not be constant and in general will be quite messy unless the geometry is simple.

In the example below of a source charge off centre inside a hollow sphere, notice how the positive charges on the inside surface are not uniformly distributed, but how they are uniformly distributed on the outside surface.

enter image description here

The net result is that the field outside the cavity only depends on the distribution of the outside charges, but the field inside the conductor is still $0$ because of the skew charge distribution on the inside surface.

[Image source: Haliday Resnick Walker 10th edition]

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  • $\begingroup$ I don't think this is answering the question. Let me rephrase the question: when there is charge q in the cavity there should be -q charge on the inner surface and due to (2) this -q and q in the cavity shouldn't cause any field in the conductor, but because the conductor is charge-neutral there will be charge q on the outer surface. But this q on the outer surface will cause field in the conductor which violates (1) $\endgroup$ – RelativisticDolphin Sep 10 '17 at 14:01
  • $\begingroup$ In the figure it's a charge -q in the cavity. This doesn't matter but I'll stick to this to refer to the figure. The charge $-q$ inside and the charge $+q$ on the inside surface do cancel. The electric field inside a spherical shell with uniform charge distribution - this is exactly the distribution of the charge on the outside of surface of the figure - produces a net $0$ field anywhere inside the shell - i.e. inside the conductor - by Gauss's laws. In other words, the example precisely illustrates the situation. $\endgroup$ – ZeroTheHero Sep 10 '17 at 14:09
  • $\begingroup$ Another way to think about it: since the $-q$ inside the cavity and the $+q$ induced on the inside surface produce no net field inside the conductor, remove them. You are left with a spherically symmetric charge density on the outside of a surface, and by Gauss's law the field inside this surface is $0$. $\endgroup$ – ZeroTheHero Sep 10 '17 at 14:11

As you say, the hollow conductor with a charge suspended inside the hollow will have a charge on its outer surface. What seems to be worrying you is why this outer surface charge doesn't give rise to a field inside the conductor. So we can forget about the conductor being hollow with a charge inside it, and just consider a conductor with a charge on its (outer) surface?

"so there will be some field due to this charge inside the conductor, which contradicts (1)" No.

If there were a (macroscopic) electric field inside the conductor, it would make free electrons move, i.e. it would give rise to currents. But if we have an electrostatic situation, by definition there is no current! What happens is that surface charges arrange themselves (in a very short time) in such a way that there is no macroscopic resultant electric field anywhere inside the conductor.

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  • $\begingroup$ but due to (2) there shouldn't be any field in the conductor due to the charge in the cavity and the charge on the inner surface should there? $\endgroup$ – RelativisticDolphin Sep 10 '17 at 13:51
  • $\begingroup$ Indeed there shouldn't, but I'm not sure that's because of 2. Aren't the 'exterior points' referred to in 2 outside the conductor as a whole? If not, 2 would be just a special case of 1. I think 1 is all you need here, and I've tried to explain why it holds. $\endgroup$ – Philip Wood Sep 10 '17 at 15:48

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