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?

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    $\begingroup$ How would you get two different charges with $r=0$ in the first place? $\endgroup$ – ACuriousMind Apr 15 '15 at 18:34
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    $\begingroup$ Similar to this question about $F=GmM/r^2$ $\endgroup$ – Kyle Kanos Apr 15 '15 at 18:34
  • $\begingroup$ Quantum effects convert it into another story at r->0. I am having a feeling that your idea of Q3 (higher mag. charge) interacting with q1+q2 is not correct. $\endgroup$ – jaromrax Apr 15 '15 at 18:41
  • $\begingroup$ Why incorrect? Joromax $\endgroup$ – user3508453 Apr 15 '15 at 18:50
  • $\begingroup$ @user3508453 - first, I dont understand that part very much, so it may be my fault. But it sounds to me like you have (unlike) $q_1$ and $q_2$ already at $r=0$ and you put $Q_3$. Then it would be a problem to what is that one unlike to... To $q_{12}=q_1+q_2$? Maybe like this... $\endgroup$ – jaromrax Apr 15 '15 at 21:40

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

  • $\begingroup$ Any simple words? $\endgroup$ – user3508453 Apr 15 '15 at 19:07
  • $\begingroup$ @Count Iblis: As long as the charge is not accelerating, one can pretend as if there is no self-force- why did you use the word pretend? Is there actually any self-force acting on the charge when it is moving with uniform velocity? The Coulomb field blows up at $r=0$ but there can't be infinite force acting, isn't it? I think when the charge moves with uniform velocity, there is no force from its own field. What do you think, sir? $\endgroup$ – user36790 Jan 11 '16 at 6:57
  • $\begingroup$ I used to word "pretend" to mean that you can use the wrong model where self forces don't exist in general. Using that wrong model will still give you the correct answer when charges are not accelerating. The Coulomb force is a vector quantity and when the charge is at rest, it is clear by invoking spherical symmetry that the net force should be zero. If the charge is not at rest but moving at a uniform velocity then by Lorentz invariance, the self force cancels, as you can transform to the rest frame, evaluate the self force there and transform the result back to the original frame. $\endgroup$ – Count Iblis Jan 11 '16 at 17:20

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.


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