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Aug 21 at 6:50 comment added dEmigOd @llama, this does not count for other charged particles, that can form further and further outer shells and can pull electrons in the opposite to repulsion direction.
Aug 20 at 22:37 vote accept vengaq
Aug 20 at 22:37
Aug 20 at 16:02 comment added llama @dEmigOd further, even if you did have an electron at r=0, v=0, that would still not be a bound system since it would be repelling the shell electrons (and they would be repelling each other). If you had a bunch of electrons in the centre, they would also be mutually repelling each other.
Aug 20 at 15:57 comment added llama @dEmigOd You could have a single electron at the centre of a spherically symmetric shell of electrons which would experience no net force, but this is only a stable equilibrium if the outer shell of electrons are fixed. Otherwise any movement from the exact centre will displace the outer shell asymmetrically.
Aug 20 at 13:48 comment added dEmigOd I'm not that good in symmetric arguments, but how do we know there does not exist some shape where you can place enough electrons inside a small radius, while none inner sphere electron experience a net force by some outer electron cloud?
Aug 19 at 22:49 comment added controlgroup You are correct, the gravity of the electrons is necessarily always outmatched by their electromagnetic repulsion, because gravity is much weaker than electromagnetism (or equivalently, the electron is more charged than it is heavy). In fact because they are so much more charged than they are massive, they can't even be brought together into a black hole, since such a Reissner-Nordstrom object would become a naked singularity.
Aug 19 at 22:35 comment added HiddenBabel I think OP already expected relativity to be required by mentioning mass-energy equivalence.
Aug 19 at 22:28 history answered David_h CC BY-SA 4.0