So according to my notes, the field inside a conductor is zero. But what, exactly, is meant by inside?

I think we are in electrostatics for the purpose of this question.

The reason it is zero is because all the electrons are pushed to the edge of the conductor. So are these electrons assumed to be no longer inside the conductor (i.e. not strictly inside the conductor). If this is the case, does this mean that Gauss' Law applies to charges strictly inside a surface, and considers charges on the surface to be "outside" the surface, or more precisely, not strictly inside the surface?

Please let me know if I should write this more clearly.



Adding a bit of detail to Pygmalion's answer (at the risk of going down a rabbit hole):

1) The math definition of an "inside" point is that it is surrounded by other "inside" points, so the surface is not inside. (I like your "not strictly inside".) However, the surface charge electrons are still part of the conductor, and they can and do adjust so that the conductor surface is an equipotential (tangential electric field = 0). (In short, the conductor is composed of its inside plus its surface, and these two sets are disjoint.)

2) Gauss' law $\nabla \cdot E= \rho$ applies everywhere: in its differential form, $\nabla \cdot E=0$ both inside the conductor (where $E=0$) and outside. On the surface the divergence is non-zero (in fact a nasty discontinuous delta-function-type thingey). More tractably, you can use the integral version and consider Gaussian pill-boxes with one side inside and the other outside. I am not responsible if you put a pill-box side precisely on the surface.


The basic idea, at least as it is taught to the freshmen is as following:

In conductor you have plenty of electrons that are practically "free". You are interested in electrostatic situation, that is the situation in which all electrons are by definition practically still. This is possible only if electric field within conductor is zero. If that would not be so, there would exist electric force

$$\vec{F} = q \vec{E} = m \vec{a}$$

end electrons would accelerate.

By using Gauss law, you can prove, that the core of the conductor is actually charge neutral. Imagine that you make a Gauss surface that encloses all but surface of the conductor (i.e. Gauss surface is only tiny fraction smaller than surface of the conductor). Since electric field on the whole Gauss surface equals zero, that means that

$$\frac{Q}{\epsilon_0} = \int \vec{E} \cdot \text{d}\vec{A} = 0,$$

that is total enclosed charge $Q$ equals zero.

So we came to the final conclusion, which obviously claims that any excessive positive or negative charge (if such exists) must be on the surface of the conductor. Electric field can be (and actually usually is) larger than zero on the very surface of the conductor. This is because electrons, even if they are pushed out by electric field, cannot leave the material, they are not "free" to do that.

Generally all this idea of electrostatics seem to be a bit "over edge", isn't it? Well as it turns out electrons need only a fraction of a second to realign themselves in such a way that electrostatic situation is satisfied. So electrostatic situations are real.

  • $\begingroup$ Very good explanation, and I get most of it now. One thing though: "This is because electrons, even if they are pushed out by electric field, cannot leave the material, they are not "free" to do that." Why not? What stops them? And also, you say that the electric field can be larger than zero on the surface of the conductor. Does this mean the electrons are in some configuration where there is a balance between the electric field repulsion against one another (that each one produces) and some other force? If so, what force? $\endgroup$ – Adam Rubinson Apr 19 '12 at 21:47
  • $\begingroup$ This question requires the knowledge of solid state physics. From the perspective of the electrons, it would look like if they are free, but unfortunately within a huge and deep well. Atoms of the material are making this potential that has this well resemblance and, which does not allow electrons to leave. Well, electrons can leave, if you give them enough energy to jump out, and this is exactly what happens in a cathode tube. You heat the material and by heating provide electrons with enough energy to jump out of the well. $\endgroup$ – Pygmalion Apr 19 '12 at 21:53
  • $\begingroup$ How does this answer the op's question where the inside and outside of a conductor is? $\endgroup$ – John McAndrew Apr 19 '12 at 22:14
  • $\begingroup$ @JohnMcVirgo I think I understand op's question, but I don't quite understand yours. If I overlooked something, which is quite possible so late in the night, it would be great if you supplement me. $\endgroup$ – Pygmalion Apr 19 '12 at 22:28
  • $\begingroup$ @Pygmalion the question clearly asks: "But what, exactly, is meant by inside?" and I don't see your answer to this. $\endgroup$ – John McAndrew Apr 19 '12 at 22:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.