# Force when distance between charge is zero

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?

• How would you get two different charges with $r=0$ in the first place? – ACuriousMind Apr 15 '15 at 18:34
• Similar to this question about $F=GmM/r^2$ – Kyle Kanos Apr 15 '15 at 18:34
• 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. – jaromrax Apr 15 '15 at 18:41
• Why incorrect? Joromax – user3508453 Apr 15 '15 at 18:50
• @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... – jaromrax Apr 15 '15 at 21:40

• @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? – user36790 Jan 11 '16 at 6:57
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.