0
$\begingroup$

If we place a metallic conductor in a static, external electric field, the free electrons inside the conductor will move opposite to the external field leaving the immobile positive ions (nucleus+core electrons) where they are inside the conductor. This will cause an internal field to build up opposite to the external field, and in static equilibrium, they cancel out. This is how the net electric field inside a conductor vanishes, in static equilibrium. By Gauss' law, $\nabla\cdot{\vec E}=\rho/\varepsilon_0$, it further implies that the charge density $\rho({\vec r})$ also vanishes at every point inside the conductor.

But after equilibrium is established, let us consider that part of the conductor which contains only immobile positive ions (or certainly a large excess of it). If we now consider a small volume $\Delta V$ in that part of the conductor which contains predominantly immobile positive ions, $\Delta V$ will contain a nonzero positive charge $\Delta q$. When we take the limit $\Delta V\to 0$, we should get, $$\rho(\vec r)=\lim_{\Delta V\to 0}\frac{\Delta q}{\Delta V}\neq 0.$$ The figure below is a rough cartoon of what I mean (after static equilibrium is established).

enter image description here

What is wrong with the argument in the second paragraph?

$\endgroup$
16
  • $\begingroup$ The picture is misleading. $\endgroup$
    – my2cts
    Apr 10, 2022 at 7:34
  • $\begingroup$ @my2cts Why? By clue circles with a $`+'$, I represent immobile ions. I didn't show the bound/core electrons separately. $\endgroup$ Apr 10, 2022 at 7:37
  • $\begingroup$ You picture a solid made up from only positive ions. What does such an impossible solid have to do with a conductor? $\endgroup$
    – my2cts
    Apr 10, 2022 at 7:40
  • $\begingroup$ You (accurately) drew equal amounts of positive and negative charges on the surfaces. Why would you assume that the bulk would then be left with an excess of positive charge, when the overall conductor is neutral? $\endgroup$
    – Chris
    Apr 10, 2022 at 7:41
  • 1
    $\begingroup$ @Solidification Yes. Well, more realistically it's probably that all the electrons move over a bit, so some from one surface join the bulk and some from the bulk join the other surface. But the net result is that the bulk atoms are neutral on average. $\endgroup$
    – Chris
    Apr 10, 2022 at 8:19

3 Answers 3

2
$\begingroup$

The excess of positive ions is only on the surface, just like the excess of electrons on the other side is only on the surface. The inside of the conductor remains neutral.

$\endgroup$
3
  • 1
    $\begingroup$ I am not totally convinced. If we apply a strong enough electric field, it is possible that most of the free electrons are driven to one side/surface but leave the bulk of the conductor with bare immobile positive ions. The immobile positive ions of the bulk cannot move to one side in a crystalline conductor. $\endgroup$ Apr 10, 2022 at 7:04
  • 1
    $\begingroup$ @Solidification If there is an excess of positive ions in the bulk of the conductor, then there is an electric field in the conductor. If there is an electric field in the conductor, then the electrons will move until there isn't. If you apply an electric field that is so large that there are no electrons left in the conductor, what you have isn't a conductor anymore. It's separate piles of electrons and ions (both of which will react rather explosively). $\endgroup$
    – Chris
    Apr 10, 2022 at 7:07
  • $\begingroup$ If there are only positive ions left there is indeed a field. A field so strong that the solid undergoes Coulomb explosion. $\endgroup$
    – my2cts
    Apr 10, 2022 at 7:34
1
$\begingroup$

It is wrong to mix macroscopic and microscopic descriptions.

The statement that, inside a conductor, the net charge density is zero is valid only at the macroscopic level. There is a non-vanishing local density (positive at the nuclear positions and negative elsewhere due to the electronic density). In the presence of an external electric field, close to parts of the surface, there will be an excess of electrons. Close to other parts, a depletion, manifesting as negatively and positively charged regions on the surface.

At the macroscopic level, the excess and depletion regions are a few atomic layers wide. Therefore, at this level (or better, at the limit of a negligible volume of the surface layers as compared to the bulk volumes), they behave like a surface charge density, not as a volume charge density.

$\endgroup$
5
  • $\begingroup$ Consider the region $\Delta V$ shown in my diagram, and let us suppose that we do not take the limit $\Delta V\to $atomic volume, but a volume that contains many ions with positive charges. Still $\rho(\vec r)>0$. Why not? $\endgroup$ Apr 10, 2022 at 7:32
  • $\begingroup$ @Solidification Because it also contains many electrons with negative charges. Exactly as many, on average. $\endgroup$
    – Chris
    Apr 10, 2022 at 7:45
  • $\begingroup$ Dear @GiorgioP If no external field were applied from outside, the net charge inside $\Delta V$ would have been zero. This I agree. But when an external field is applied, the valence/free electrons have moved from the bulk to the surface, hence $\Delta V$ would contain lesser electrons (only core electrons will be there). So don't we expect that $\Delta V$ now has an unneutralized positive charge? $\endgroup$ Apr 10, 2022 at 7:53
  • $\begingroup$ @Solidification No. The excess electrons on the left side of the metal in your diagram correspond to an absence of exactly as many electrons on the right side. The bulk of the metal remains neutral, there aren't fewer electrons per atom. $\endgroup$
    – Puk
    Apr 10, 2022 at 8:01
  • $\begingroup$ @Solidification, you are still mixing macro- and microscopic views. If $\Delta V$ becomes comparable with the atomic interparticle distance, the net charge is never zero. Only with a volume larger than the typical interatomic distances, the overall charge will be nearly zero in the bulk. That's the length scale where it is possible to represent the average charge density as a macroscopically smooth function. In the same limit, at the surface, the charge density becomes a Dirac delta centered at the surface, or, in an equivalent way, a non-vanishing surface charge density. $\endgroup$ Apr 10, 2022 at 14:42
0
$\begingroup$

Think of it this way. Because of the external field all valence electrons move a little to one side. This causes excess negative charge on one surface and excess positive charge on the opposite. In the bulk the conductor remains neutral.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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