I have this thought experiment. Imagine you charge an electric ball with electrons - like the one in Van de Graaff generator. There's a hole in the ball.

Now if you shoot an electron in the hole, with enough force to overcome the repulsive magnetic field of the outer electrons, what does it do?

Will it just stick there like on the image?

enter image description here

I assume it can always exit the ball by quantum-tuneling out - that is by randomly "occurring" at the outer side and then leaving repulsed by the electric field of other electrons.

Maybe it can also somehow radiate itself away as elmag waves?

What would really happen?


Since this is a thought experiment, the material is perfectly even, we're in a perfect vacuum, and the hole is negligibly small relative to the size of the sphere, so we can ignore it for charge distribution.

But is the sphere conductive or insulating?

As you might know, for objects inside a spherical shell of constant density, the gravitational attraction cancels out. If your sphere is an insulator, then its excess electrons can't move around, and their ideally even distribution would mean that the same math applies. Once the free electrons enter the sphere, they will continue moving at constant velocity until they hit the other side of the sphere.

Whereas if your sphere is a conductor, then its electrons will move around in response to the presence of other charges, away from negative ones or towards positive ones, until their mutual repulsion balances out. I'm not prepared to do the math, but I'm pretty sure the net effect is electric force pushing outwards from the center of the sphere. The force is zero at the exact center, but it would be like trying to balance on the top of a frictionless hemisphere. The slightest perturbation sends you inexorably in motion.

Either way, it won't work as a trap for floating charged particles. They will almost certainly hit the walls.

  • 1
    $\begingroup$ If the charge distribution is indeed uniform on the shell and the small entry hole is negligible, then the force on the charged particle should be zero as you originally implied (see hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html), so the electron would continue at constant velocity inside the shell. But then you guessed that the force is only zero at the center and unstable. Those two pictures are not compatible, since if it were only zero at the center, it would experience a force everywhere else and would NOT go through at constant velocity. $\endgroup$ – C Perkins Apr 2 '18 at 2:58
  • $\begingroup$ @CPerkins, yes, my answer has two parts, one for an insulator and the other for a conductor. $\endgroup$ – Foo Bar Apr 2 '18 at 20:14

If you shoot an electron into the inside of a hollow metal sphere, it will eventually enter the metal and reside as a surface charge on the outside of the metal. You can charge a hollow sphere almost without limit by mechanically introducing electrons into the interior. This is the principle of the Faradays cup, which is also used in van de Graaff voltage generators. The limit is reached when the electric surface field reaches the breakdown field of air, or in vacuum when the Fowler-Nordheim field emission (tunnel) current becomes significant.

Thus you can definitely "trap" electrons (charge) in a conducting sphere, even when there is already significant charge present!

  • 1
    $\begingroup$ But a Faraday cup is usually grounded or held at some constant voltage relative to ground so that any extra charge generates a current. The current can be measured or is perhaps part of an automatic feedback circuit. Although such a configuration could accept charges without limit, the device itself does not retain charges. A floating (i.e. un-grounded) Faraday cup would not necessarily behave like a van der Graff generator... it would have no active mechanism and eventually charge up to repel electrons trying to enter the inner cavity. $\endgroup$ – C Perkins Mar 29 '18 at 5:58
  • $\begingroup$ Inside a van der Graff generator "sphere" there is no significant field as explained in other online resource and by other answers here. Earnshaw's theorem and the nature of the shape of electric fields (i.e. there cannot be an static electric "well" in all directions for a particle to fall into) are sufficient to explain why this trap cannot work. As for how and/or why an electron actually hitting the sphere would behave is definitely a quantum mechanical phenomenon separate from the idea of trapping the particle in the sphere interior. $\endgroup$ – C Perkins Mar 29 '18 at 6:06
  • $\begingroup$ @CPerkins - Think of the isolated metal sphere as a capacitor. You can transport charge (electrons) by mechanical means to the capacitor and it will distribute evenly on its surface. That is the principle of the van de Graaff high voltage generator. As long as no air conduction or tunnel field emission occurs, there will be no discharge current due to electrons leaving the metal sphere. Note: A Faraday cup is also a capacitor that is charged by mechanically moving charges into its interior. The charges will distribute on its surface just like in the case of the spherical capacitor. $\endgroup$ – freecharly Mar 29 '18 at 14:37
  • $\begingroup$ Okay, but unlike what is stated in the answer, such a device will not accept an indefinite amount of charge without some kind of effect. For the Faraday Cup, the charges would eventually build up until the external field would repel incoming electrons of a certain energy limit. A capacitor also charges only until its net charge repels further charged particles. As I already stated, I gather that the question is more about trapping particles inside the shell which just could not happen regardless of how the shell obtains its charge. $\endgroup$ – C Perkins Mar 29 '18 at 14:48
  • $\begingroup$ @C Perkins - I am sorry, but your reasoning is not correct. When you transport charge mechanically into the spherical shell, you have, of course, to overcome the repulsive electrostatic force, as in any capacitor you are charging. The charge (electrons) will then definitely be "trapped" on the spherical shell. I have also pointed out the limits to that. Quantum mechanics is only necessary to explain the Fowler-Nordheim tunnel emission. $\endgroup$ – freecharly Mar 29 '18 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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