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Is there a point of balance where the gravitational pull of a sphere of electrons is equal to their electromagnetic repulsion?

That is to say, could it be possible to create stars that are made purely with electrons and that are stable and don't fly apart?

Qmechanic
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Nick Revenco
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  • https://en.wikipedia.org/wiki/White_dwarf – Nihar Karve Jul 25 '23 at 20:25
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    @NiharKarve, a White dwarf isn't just electrons - it's electrons and protons of an equal amount (so the entire star is not charged) and neutrons. The further collapse is prevented by the electrons' degeneracy pressure, but it's not made entirely of electrons. – Nadav Har'El Jul 25 '23 at 20:31
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    @NadavHar'El Sure, but it's a nice real life example of a related phenomenon which OP may not be aware of – Nihar Karve Jul 25 '23 at 20:38
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    If the electrons were somehow momentarily crammed into a space smaller than the Schwartzchild radius of the mass (e.g. by an array of multiple supernovas, in the style of a nuclear bomb with implosion-style initiator stage), the mass should form a highly charged black hole that would trap the electron matter inside. Certainly not an electron star though. – RC_23 Jul 25 '23 at 21:18
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    @RC_23 right. Also related to that is that when electrons are about a planck's length (around $10^{-35}$m) apart, their gravitational pull is actually stronger than the electric repelling force (gravity becomes stronger because the uncertainty principle gives the highly confined particles huge energies, which gravity pulls on just like it pulls mass). But at these densities, you basically get a black hole, not an electron star as you might envision one. – Nadav Har'El Jul 25 '23 at 22:17

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No, it's not possible:

The electric force between two electrons is is $2.4* 10^{43}$ times (trillion-trillion-trillion-trillion-trillion times!) stronger than the gravitational force between these two electrons. So if you add more and more electrons to the "electron star", the gravitational pull will indeed increase, but the electric repulsion between the electrons will also grow, and remain $2.4* 10^{43}$ stronger than the gravitational pull.

So a realistic star will need to be almost exactly neutral to have any chance of remaining intact. For example, the sun has around $10^{57}$ electrons and also $10^{57}$ protons and additional fewer neutrons (which, as their name suggests, are neutral).

Nadav Har'El
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    Good answer, but for the sake of accuracy: $2.4 \times 10^{43}$ works out to "only" be 24 million trillion trillion trillion. You've got a couple extra "trillions" in your parenthetic comment. – Michael Seifert Jul 25 '23 at 20:52
  • You're right, I think it was supposed to be approximately (but not exactly) the fifth power of billions, not trillions. – Nadav Har'El Jul 25 '23 at 22:12
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    a gram of fully ionized hydrogen gas at nominal density has enough electrostatic energy to unbind the Sun, gravitationally. – JEB Jul 26 '23 at 03:12
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    It's 24 million trillion trillion on the long scale (where a trillion is a million cubed); it's 24 million trillion trillion trillion on the short scale (where a trillion is a million squared), which has become far more popular in the English-speaking world. Yes, $n$-illion being $10^{6n}$ rather than $10^{3n+3}$ makes more sense, but it's convenient to have a name for each power of $10^3$. (The long scale calls $10^9$ milliard, but that means alternating suffixes.) You could also call $10^{42}$ a septillion on the long scale, or a tredecillion on the short scale. – J.G. Jul 26 '23 at 07:22