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I have some doubts about the electric field inside an ideal conductor (let's call it E). Precisely, I have read two different descriptions

1) On physics books I read that the electric field inside in a conductor in electrostatic equilibrium is equal to 0. The physical reason of this is the fact that all charges, at equilibrium, are distributed in the external surface of the conductor since only this distribution can reduce their repulsion forces. A mathematical view of this is given by the equation J = sigma * E. In fact, since sigma = infinite and J = 0 (since equilibrium means that charge do not move), necessarily we must have E = 0.

According to this explanation, E = 0 only in electrostatic equilibrium.

2) On Electromagnetic Fields books I read an explanation which is similar, but not identical. I read that, since J must have a finite value, and J = sigma * E and sigma = infinite, we get E = 0.

According to this explanation, there is not any mention of electrostatic equilibrium. It seems that inside a conductor in any condition we have E = 0. Also if there is a voltage source applied on it or something similar.

Now I have two questions:

  • Which is the correct description?

  • It is known that a metal is able to reflect EM waves. Is it due to the fact that E = 0 in its inner points?

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  • $\begingroup$ Kinka-Byo, the title of your question includes the term "ideal conductor" but your first question includes a statement about conductors in general and not just ideal conductors. Are you clear on the distinction between "conductor" and "ideal conductor"? $\endgroup$ Commented Sep 20, 2019 at 2:57
  • $\begingroup$ I think that a conductor has a finite (but not 0) conductivity, while an ideal conductor has an infinite conductivity, is it correct? So in practice is a conductor simply any material that is not a total dielectric ? $\endgroup$
    – Kinka-Byo
    Commented Sep 20, 2019 at 5:27
  • $\begingroup$ Kinka-Byo, it's important to keep in mind that an ideal conductor is un-physical (thus the adjective ideal) but is useful for simplifying calculations to get valid results (in the region of operation where the ideal conductor approximation is valid). For example, in a physical wire conductor, the mobile charge carriers are electrons while in an ideal conductor, we think only of mobile charge - the properties of the charge carrier have been abstracted away. There have been several recent questions here where the OP arrived at a contradiction pushing the ideal conductor approximation too far. $\endgroup$ Commented Sep 20, 2019 at 10:34
  • $\begingroup$ Also, don't forget that there are materials that are neither good conductors nor good insulators, e.g., semiconductors. $\endgroup$ Commented Sep 20, 2019 at 10:36

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The first description is true for all conductors while the second is only true for perfect conductors.

The reason the first works for all conductors is that in electrostatics, we get to say 'if $\textit{any}$ charge moves, our assumption of electrostatics is violated', so an electric field in even an imperfect conductor would violate the assumption and so is not allowed. On the other hand, if we allow a fully dynamical system, we can have moving charge in a conductor $\textit{if}$ it is not perfect. This should be obvious since we have charge moving through conductors all the time in the real world. So in this case, the argument only hold because the conductivity, $\sigma$, is infinite, which is only true for perfect conductors.

As for the second, the best answer I know to give is that we use boundary conditions to understand how the fields behave near surfaces, and when you carry out the calculations, you get reflection. If I recall correctly, $\textbf{E}=0$ is used to get the idealized perfect reflection in the idealized perfect conductor problem.

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  • $\begingroup$ Why are 1 and 2 true respectively fo conductors and perfect conductors? Where are there assumptions in those two descriptions? $\endgroup$
    – Kinka-Byo
    Commented Sep 20, 2019 at 1:13
  • $\begingroup$ @Kinka-Byo I added a section to answer your follow up question. $\endgroup$
    – Bronicki
    Commented Sep 20, 2019 at 2:33
  • $\begingroup$ Perfect, thank you very much $\endgroup$
    – Kinka-Byo
    Commented Sep 20, 2019 at 5:22

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