6
$\begingroup$

When a conductor is in electrostatic equilibrium, as far as I know, the electric field on the surface must be perpendicular to the conductor. This leads to a rather difficult problem for me, if the electrostatic force (as far as I know, this force includes the attraction of the ions + the repel of electrons) is acting on the electron's outward direction as shown, then there must be a force F (attraction) pulling it inward to satisfy the equilibrium condition and prevent electrons from leaving the conductive material. So what is this force really?

I am looking for a qualitative explanation (not a mathematical one) and on top of that. I hope someone can clarify this for me. Thank you!

enter image description here

$\endgroup$
4
  • 3
    $\begingroup$ The electrostatic force from the positively charged atomic nuclei? $\endgroup$
    – fishinear
    Commented May 9, 2023 at 14:54
  • 1
    $\begingroup$ Your intuition is almost correct: if the electric field is strong enough, the electrons will be emitted from the surface. This is called "field emission" and it happens when an electric field on the order of tens of MV/m is present. The reason why we can ordinarily treat a metal surface as a perfect "barrier" for electrons is exactly the fact that it is very hard (and unusual) to make such strong fields. Field emission in vacuum should, by the way, not be confused with gas discharges, which happen in much weaker electric fields. $\endgroup$ Commented May 9, 2023 at 16:03
  • $\begingroup$ @FlatterMann that should be an answer $\endgroup$
    – Dale
    Commented May 9, 2023 at 17:07
  • 1
    $\begingroup$ Electrons in conductors are not free, they are in quantum mechanical solutions of the lattice describing the conductor . see physics.stackexchange.com/questions/266114/… $\endgroup$
    – anna v
    Commented May 9, 2023 at 17:15

2 Answers 2

8
$\begingroup$

As pointed out in the comments, electrons are bound in metals both by electrostatic and quantum mechanical interactions. Your intuition is correct, though: if the electric field is strong enough, the electrons will be emitted from the surface. This is called "field emission" and it happens when an electric field on the order of tens of MV/m is present. The reason why we can ordinarily treat a metal surface as a perfect "barrier" for electrons is exactly the fact that it is very hard (and unusual) to make such strong fields. Field emission in vacuum should, by the way, not be confused with gas discharges, which happen in much weaker electric fields.

$\endgroup$
4
$\begingroup$

A conductor consists of positively charged ions, forming a lattice, and negatively charged electrons, which are free to move within a conductor. The net charge is zero, and so the net force is actually zero.

The problem is that, according to the Earnshow's theorem, an equilibrium of a system of classical point charges is always an unstable one - that is, we would expect a metal to gradually lose its electrons and become positively charged. The reason why this doesn't happen is the exchange interaction between electrons, which is a quantum mechanical effect. There is a detailed and pedagogical calculation in Fetter&Walecka's book Quantum Theory of Many-Particle Systems.

Note also that the fact that the force is perpendicular to the conductor does not necessarily mean that it is directed outwards - it could as well be directed towards the conductor. Moreover, the claim that the force is perpendicular is based on classical electrostatics, which assumes that any tangential force would make electrons move and hence be screened by redistribution of charge - in a way, this argument already assumes that metals are stable (and a few other things.)

$\endgroup$
8
  • $\begingroup$ But I really want to know what is the nature of the force F(Attract) that pulls electrons inward really? If it is indeed an electrostatic attraction, isn't that attraction included in the F(electrostatic) force caused by En ? $\endgroup$ Commented May 10, 2023 at 15:04
  • $\begingroup$ There is a force due to the ions, which pulls electrons in, and there is a force due to other electrons, which pulls them out. Since the conductor is electrically neutral, the two balance each other. Your question seems to suggest that, for some reason, there is a net force pulling electrons away... but it is not clear what this claim is based upon. Provide references and/or citations for clarity - otherwise it is hard to say more. $\endgroup$
    – Roger V.
    Commented May 10, 2023 at 15:09
  • $\begingroup$ the force pulling the electrons away, as far as I know, is caused by the external electric field and the electric field due to the induced charges present in the conductor. $\endgroup$ Commented May 10, 2023 at 15:16
  • $\begingroup$ So, you are dealing with a problem, where there is 1) an external field present, and 2) there is possibly an additional charge brought to the conductor - this is not at all obvious from your question. I seriously suggest editing it. But the answer is most likely exchange interaction - it can be expressed in terms of an effective potential, and hence as a force, but it is an essentially quantum effect (even calling it interaction is not quite correct, since it is just a consequence of indistinguishably of particles, while interactions correspond to actual forces.) $\endgroup$
    – Roger V.
    Commented May 10, 2023 at 15:19
  • 1
    $\begingroup$ > "we would expect a metal to gradually lose its electrons and become positively charged." That's not what Earnshaw's theorem implies, so why would we expect that? E. theorem is just about static configurations, it says nothing about systems where charges move. $\endgroup$ Commented May 28, 2023 at 12:30

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.