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If I had a fixed point-charge ($q_1$, +) and I placed another free-moving charge ($q_2$, -) some distance away, what is the potential energy between them?

The opposite charges attract drawing $q_2$ closer and, by Coulomb's Law, the force will increase as the they approach each other. So, the limit of the force as the separation goes to zero goes to infinity? Work and potential energy also go to infinity?

I hurt myself comparing it, mechanically, to a super compressed spring (I'm aware that is proportional force vs. the inverse square) and then thinking that two opposite point charges, regardless of their initial distance of separation, all have infinite potential energy?

It's been a long day... :)

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    $\begingroup$ Yes they approach infinity. What is wrong with that? $\endgroup$ – Aaron Stevens Sep 4 '19 at 23:04
  • $\begingroup$ The potential energy between them is $q_1q_2/d$ in Gaussian units. At finite separation $d$, it is finite. When the charges have opposite signs, it approaches negative infinity as the particles get closer and closer. Don’t think of it as a spring. $\endgroup$ – G. Smith Sep 4 '19 at 23:35
  • $\begingroup$ @G.Smith So, if someone were to ask, "at what minimum distance must you place an electron from a proton so that they collide with 'x' Joules of energy?" the answer would be "anywhere" assuming the electron/proton are point objects? A specific distance could only be determined knowing/given some final "collision separation distance" value? $\endgroup$ – Jon Sep 5 '19 at 4:30
  • $\begingroup$ That’s correct. $\endgroup$ – G. Smith Sep 5 '19 at 4:36
  • $\begingroup$ Classical physics doesn't deal well with this stuff. The best it can do is to say that finite particle size prevents the distance between particle centres from actually reaching zero. But that just sweeps the problem under the carpet. How does a single charged particle itself deal with the infinite field strength at its centre, and what stops it from being blown to pieces by self-repulsion? $\endgroup$ – PM 2Ring Sep 5 '19 at 5:52
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We know the force between two charges is always

$F=\frac{kq_1q_2}{r^2}$

The potential energy is simply related to force by $-\frac{dU}{dx}=F$ which means

$U=\frac{kq_1q_2}{r}+C$

where $C$ is any constant of choice. Now you may notice that the potential energy is indeed singular when $r=0$ so the difference in potential energy between any point and this one is indeed infinite. There is really no problem with this as when real particles get that close, other stronger nuclear forces repel them away. Alternatively, you may wonder if this law actually does hold all the way if those forces weren't there and I am not completely sure to the answer to that question but to my knowledge, it has held up to the smallest distances tested so far.

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  • $\begingroup$ Where does c comes,it is a definite integral $\endgroup$ – Yuvraj Singh... Sep 5 '19 at 4:38
  • $\begingroup$ The electron doesn't feel the strong nuclear force, and the weak force is insignificant in terms of attraction and repulsion, relative to the electromagnetic force. However, at close distances the Heisenberg uncertainty of position and momentum is important. $\endgroup$ – PM 2Ring Sep 5 '19 at 5:38
  • $\begingroup$ Even so, consider what happens when an electron and positron meet. They become temporarily bound into positronium, and then they annihilate, creating 2 or more photons. Note that the energy released in this process is equivalent to the rest mass + kinetic energy of the original electron and positron, it is not infinite, thank goodness. ;) $\endgroup$ – PM 2Ring Sep 5 '19 at 5:44

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