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The Coulomb's force is given by $$ F = {k q^2 \over r^2} $$ When $ r \rightarrow 0 $, $ F \rightarrow \infty $ Does this mean two electrons never touch each other?

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Define touch. Electrons are standing waves and are known to show interference patterns. I don't know what you mean by them 'touching'. –  Gerard Aug 4 '13 at 12:05
    
@user1305192 can there not be free electron in space at some given postion? –  hasExams Aug 5 '13 at 15:57
    
Yes, there can be an electron at a given position. But then you can't measure its momentum. So, for all intents and purposes, you don't know where it is. –  Gerard Aug 8 '13 at 15:04
    
Pauli Exclusion . . . –  Dimensio1n0 Aug 9 '13 at 11:55
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up vote 4 down vote accepted

Yes, two electrons or two charged particles are not allowed to overlap – the interaction energy goes like $1/r$ so it would diverge in the strict limit and the world doesn't have the infinite energy to do that. Two electrons are forbidden to overlap for another reason, the Pauli exclusion principle.

On the other hand, two particles of opposite signs attract by a force that goes to infinity in the $r\to 0$ limit, too. Nevertheless, even in this case, they won't end up overlapping due to the uncertainty principle of quantum mechanics.

For example, take the hydrogen atom (or any atom). The electron can't sit exactly at the nucleus even though it would save an infinite amount of energy. The reason is that a very low $r$ implies a very low $\Delta x$ of the same order and, therefore, a high $\Delta p \geq \hbar / \Delta x$. A high $\Delta p$ means a high average $\Delta p^2$, and therefore high kinetic energy, and this one actually wins if the proton-electron distance is too short. If the electrostatic force increased faster than $1/r^2$ at short distances, one could actually beat the kinetic energy and oppositely charged particles would choose to sit on top of each other.

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@ Lubos Motl what dou you mean by electrostatic force increased faster than 1/r^2 .please clarify on that. –  AaKASH Mar 31 '13 at 12:42
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