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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? $\endgroup$
    – BioPhysicist
    Commented Sep 4, 2019 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
    Commented Sep 4, 2019 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
    Commented Sep 5, 2019 at 4:30
  • $\begingroup$ That’s correct. $\endgroup$
    – G. Smith
    Commented Sep 5, 2019 at 4:36
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    $\begingroup$ @G.Smith Thanks for the replies, everyone; it's been quite some time since I've given it any real thought. A last clarification please, and I'll switch to the gravitational potential energy analog: $U(r) = -GMm/r$ is the stated formula for the potential energy and its value is "the energy available as object $m$ moves from infinity to final distance $r$ from $M$", correct? And not "the energy available as object $m$ collapses into point-object $M$"? Same idea for the $k(q_1)(q_2)/d$ formula? $\endgroup$
    – Jon
    Commented Sep 5, 2019 at 16:20

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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 $\endgroup$
    – आर्यभट्ट
    Commented Sep 5, 2019 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
    Commented Sep 5, 2019 at 5:38
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    $\begingroup$ 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
    Commented Sep 5, 2019 at 5:44

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