When electrons go to the outside surface of a conductor why doesn't that create an electric field pointing radially outwards? $E$ is supposed to be 0

Question

If an isolated conductor has a net zero charge and is not in an external field, would the free charges still move to the surface of the conductor eg. a conducting sphere? Wouldn't this create an electric field inside the conductor pointing radially outwards as there would be positive ions within the conductor and negative charges on the surface? And then wouldn't the electrons move back to join the positive ions?

[I understand how free charges will move in a conductor to cancel any external field, cancelling the external field inside the conductor, but then wouldn't there still be a field pointing radially outwards?]

I also don't understand this in terms of energy - why is it a lower electric potential energy position for the charges to go to the surface - wouldn't it be lower energy for the electrons to go back to the positive ions?

Gauss's law talks about there being no charges inside a conductor, but wouldn't there still be all the positive ions left there if the conduction electrons have gone to the surface?

My guess is that people are usually talking about any NET charge going to the surface of a conductor. But wouldn't an isolated conductor at equilibrium (as in the general textbook questions on this) be net neutral and so then not have charge on the surface?

Answer 1

If an isolated conductor has a net zero charge and is not in an external field, would the free charges still move to the surface of the conductor eg. a conducting sphere?

No, they would not.

But it's not clear why you use the word "still". There's no situation where all the free charges go to the outer surface.

If you've been told that when there's an external field then free charge goes to the outer surface, that's true. But it only means that enough charge goes to the surface to terminate the field, not that all the free charge in the material goes to the surface. In everyday situations, this means a very small fraction of the available free charge. If it were otherwise (due to a very high field strength) you'd see arcing (the electrons freed from the material altogether) before you'd see all the charge bunched together at the surface.

Gauss's law talks about there being no charges inside a conductor, but wouldn't there still be all the positive ions left there if the conduction electrons have gone to the surface?

I'm not quite sure what you mean by this. Normally we say that there is no electric field inside a conductor. The reasoning is that if there were it would produce an extremely strong current due to the large amount of free charge in the conductor, and this would very quickly re-arrange the free charge so as to produce an opposite field, resulting in no net field.

This is exactly the behavior that causes (a tiny fraction of) the free charge to move to the surface of the conductor in an external field.

But I'm not clear how you relate this to Gauss's law. If anything I'd call it a result of Ohm's law (microscopic form).

[I understand how free charges will move in a conductor to cancel any external field, cancelling the external field inside the conductor, but then wouldn't there still be a field pointing radially outwards?]

There'd be exactly enough outward field produced by the displacement of the electrons as there was external field applied, resulting in no net field.

You'd also have a depletion of electrons on one side of the object producing a positive surface charge, balancing the negative surface charge caused by the accumulation of electrons on the other side of the object, so that the entire object remains electrically neutral.