This question was asked to me by a 'backyard scientist' and I'm having a frustrating difficulty answering it satisfactorily.

As an example of a charged object to focus attention on, I'll use this bench-top Van de Graaff generator (VDG):

Van de Graaff schematic

The metal sphere is $0.25\:\mathrm{m}$ diameter, rated maximum voltage $375,000\:\mathrm{V}$ and maximum charge $5.2\:\mathrm{\mu C}$.

Based on this, I made a few 'back of an envelope' calculations and estimates for this VDG, at max. charge:

Number of electrons on the sphere $= \frac{5.2 \times 10^{-6}}{1.6\times 10^{-19}}=3.25\times 10^{13}$

Surface area of the sphere = $\pi D^2=\pi \times (0.25)^2=0.20\:\mathrm{m^2}$

Estimated distance between electrons (square packing) = $\sqrt\frac{0.2}{3.25\times 10^{13}}=7.8\times10^{-8}\:\mathrm{m}$

Repulsive force between two adjacent electrons (Coulomb's Law)$=9\times10^{9}\times\frac{(1.6\times 10^{-19})^2}{(7.8\times10^{-8})^2}=3.8\times10^{-14}\:\mathrm{N}$

Electron acceleration due to repulsive force: $F=ma$:


That's quite a whopper! Hard to see how these electrons would not fly away from each other (and reduce their potential energy).

I found this P.S answer to a very similar question but don't find it very satisfying.

What am I missing?

  • $\begingroup$ They're pulled back in by the atomic nuclei. $\endgroup$ – knzhou Dec 4 '18 at 14:49
  • $\begingroup$ Can you clarify why "attraction by protons" and "work function" from the other answer don't satisfy you? $\endgroup$ – Jasper Dec 4 '18 at 14:51
  • $\begingroup$ You're also forgetting that the force that an electron feels is due to the sum of the electric fields from all of the electrons, not just one. $\endgroup$ – probably_someone Dec 4 '18 at 14:55
  • $\begingroup$ @knzhou what?? electrons $\endgroup$ – Sourabh Dec 4 '18 at 15:03
  • $\begingroup$ @Jasper: for one, steel contains almost no protons. $\endgroup$ – Gert Dec 4 '18 at 15:04

What you are missing is the fact that each electron is affected not only by a single electron adjacent to it, but by all of the other electrons surrounding it as well.

The repulsive force of the electrons will cause them to spread evenly across the surface. Once this has happened, each electron will feel an approximately equal force from all of its neighbours, resulting in a net force close to zero.

Note: Electrons will obviously still feel some net force perpendicular to the surface, but unless the total charge is extremely high, this won't be anywhere near enough to strip the electrons from the material.

  • $\begingroup$ each electron will feel an approximately equal force from all of its neighbours, How can that be when some are close to a given electron and some very far away? $\endgroup$ – Gert Dec 4 '18 at 15:09
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    $\begingroup$ If that were true, that would happen in all directions, forming a gas of electrons, as they fell even less force from neighbours by escaping away. You're missing the key point: they are bounded to the atoms. $\endgroup$ – FGSUZ Dec 4 '18 at 15:17
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    $\begingroup$ @Katie: sorry, but this is not the explanation. If the reason would be the balance of the repulsive forces due to other electrons, it would not be understandable why electrons on a convex piece of metals are not ejected from the metal, since there would be a non equilibrated component of the force pushing them out of the metal. The real reason is in the presence of an energy barrier, due to quantum mechanics, which requires a finite energy to be overcome (see work function ( en.wikipedia.org/wiki/Work_function ). $\endgroup$ – GiorgioP Dec 4 '18 at 15:48
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    $\begingroup$ @GiorgioP: maybe you should formulate an answer to that effect? $\endgroup$ – Gert Dec 4 '18 at 17:51
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    $\begingroup$ @GiorgioP: I was taking it for granted that the electrons would all stay on the surface of the metal, as it seems obvious that the net force exerted perpendicular to the surface would not typically be enough to strip the electrons from the material. I was focusing on explaining why electrons confined to an enclosed surface don't fly around at crazy velocities on that surface, which is what it seemed like OP was actually asking about. $\endgroup$ – Katie Dec 5 '18 at 13:57

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