I would like to clear things up: How exactly the electric field inside a conductor is zero? Let a really "powerful" electric field be outside of it, how can the "few" charges in a conductor balance the field? From my point of view, it's the conductor surface which stops their movement. I'm obviously missing something fundamental here. Thanks in advance.

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    $\begingroup$ InRe '"few" charges'. What was Avogadro's Number again? Compare the number of atoms in a mole to the number of electrons in a Coulomb. $\endgroup$ – dmckee --- ex-moderator kitten Nov 9 '13 at 19:27
  • $\begingroup$ Yes, I understand, but what about the external field? $\endgroup$ – Danny Nov 9 '13 at 19:28
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    $\begingroup$ @Marshall If the field is too powerful, your conductor will be ripped apart. $\endgroup$ – David H Nov 9 '13 at 19:33
  • $\begingroup$ Ahha....it occurs to me that the questions may be "How can the conduction electrons be 'free' to move around the metal, but still bound to the bulk?" That's a better questions, and I think we already have it somewhere... Band-structure and other condensed matter things come into it. I know we have some condensed matter guys who could do it better justice that I could. $\endgroup$ – dmckee --- ex-moderator kitten Nov 9 '13 at 19:39
  • $\begingroup$ From the "Related" sidebar I find Electrons of conductors Free? which points to Are free electrons in a metal really free. Perhaps they can help. $\endgroup$ – dmckee --- ex-moderator kitten Nov 9 '13 at 19:41

The common phrase is that metals have "electrons to burn", in other words, a mind-bogglingly large amount of free electrons, as @dmckee pointed out. Now, the electric field near a sheet of charge density $\sigma$ is $E = \frac{\sigma}{2\epsilon_0}$. Now $\epsilon_0 \approx 8.85 \times 10^{-12} F/m$ and Faraday's constant gives us an order of magnitude for the number of Coulombs in a mole of electrons to be $\approx 9.6\times 10^4$, so a Coulomb of charge per square meter of metal will yield a surface field of $\approx 10^{12+5-1} \sim 10^{16} V/m$ if I did my calculations right.

How "powerful" an electric field did you have in mind? Because as @Marshall pointed out, we are already beyond the breakdown fieldstrength here, so the number of electrons is not the limiting factor. To give you an idea, at $10^{18}V/m$, the electric field in a vacuum is strong enough to create electron-positron pairs (Schwinger mechanism) causing it to conduct. That's only about two orders of magnitude away.

  • $\begingroup$ It's a shame they don't talk about what electron actually are in class, as they just name them as 'charges', let alone Avogadro's Number. Thank you. $\endgroup$ – Danny Nov 10 '13 at 14:37
  • $\begingroup$ Heh. Well, the theory doesn't really care if the charges are electrons or ions or what, so to keep it general, they speak in general terms. And nobody is really sure what electrons are... $\endgroup$ – lionelbrits Nov 10 '13 at 20:23

I'd like to add to Lionel Brit's answer: there are more than enough electrons to balance almost any field but there is one further ingredient to your answer: the special form of the Coulomb potential. As you say:

it's the conductor surface which stops their movement

so this simply means that if there is an electric field component along the surface, the electrons will keep shifting until they cancel it out. This tells you that there can be a field component normal to the surface. So how does this mean no field inside a conductor?

The field normal to the surface means the inside, closed surface is an equipotential surface. Unless there is charge inside the surface, this in turn means that the field must be nought inside the inside surface. This follows from the uniqueness theorem for solutions of Laplace's equation, the latter following from the very special form of the Coulomb potential. If the Coulomb potential were any other function, there would be a field inside a closed conductor, and the experimental fact of no field inside the conductor yields very tight bounds on what the photon's mass can be, for example. Maxwell's equations are different for a massive photon. See my answer to the question "How would charge be distributed in charged conductors if the Coulomb law was not $1/r^2$?" for further details of this.

  • $\begingroup$ Thank you very much, I appreciate your elaborated replay. $\endgroup$ – Danny Nov 10 '13 at 14:36

Metal being a crystal structure of atoms being locked in position,an electric field would have a net zero in relation to itself unless it is varied.


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