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According to coulomb law

$$ F = \frac{q_1q_2}{r^2} $$

I want to know what happens to force when $r=0$. If $F \to \infty$ then the charges can't be separated! But if an unlike charge of higher magnitude is placed beside any of $q_1$ or $q_2$ then it gets attracted. Can anyone clear me out?

sahin
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user3508453
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2 Answers2

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If r = 0 then you have a single charge, so the problem reduces to the electromagnetic self-force problem. A charge will interact with the electric field it is in, and that includes the field due to its own charge.As long as the charge is not accelerating, one can pretend as if there is no self-force, but for accelerating charges, the self-force will lead to the emission of electromagnetic radiation.

The rigorous treatment of the self-force was until recently an unsolved problem. It was only recently rigorously solved

Count Iblis
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In case the question concerned the case $r \rightarrow 0$, you would reach the situation where the charge (represented by a charged particle like electron, proton, positron) approaches the Coulomb field of the other particle and they would have a tendency to create a kind of a planetary system - but - quantum effects start to play a role here and those two charged particles create a bound quantum system that can pertain forever (like hydrogen atom) or explode (case of positron-electron).

The answer with $r=0$ (it becomes one single charge) is precise (+1), but I am afraid that you did not wanted to hear that.

jaromrax
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