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Imagine there are two particles: A,B; of which, we observe A to be stationary, while B is moving towards A at velocity v. A,B has same mass; and they are charged both positive.

B has a potential wrt A. After some small time t, B's potential increases by an amount X; so its kinetic energy has been reduced by the same amount X. This amounts to B losing some velocity y, although it still has some velocity towards A. The kinetic to potential conversion means that B has its net mechanical energy unchanged till now.

Now, in that time, A has gained the amount y of velocity, away from B (conservation of momentum). That means A has gained some kinetic energy K. Of course, from the beginning time where A was stationary, and till now, A has continuously faced slow potential increase: each increases instantaneously dropped to add to its kinetic energy: so that now we have this kinetic energy K. Let, the potential energy of A wrt B was, at the beginning, P. Then the final potential of A ultimately is not lower than P. Then the total mechanical energy of A has actually increased till now.

So counting now, we have that the sum of energy of A,B has actually increased: in violation of conservation of energy.

I know, it's the little increases of potential of A, which I can't explain: why they are gained: before contributing to the kinetic energy of A. It seems that these potential increases suddenly popped out from nothing.

I have seen an example of ball falling from high to ground: there potential is exactly what is converting to kinetic energy. Everything was fine, until I thought about the circumstance I described.

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    $\begingroup$ There is no increase in the net energy of the two particles. Some of B's initial KE is converted into KE for A. By the way, you cannot start with two charged particles at rest in isolation without some external forces at play to begin with because the electric field from each acts over all distances (given enough time). $\endgroup$
    – honeste_vivere
    Commented Feb 5, 2016 at 14:04
  • $\begingroup$ When you say potential do you actually mean potential energy? In dealing with charged particles, it's important to realize there is a distinction between the two. Also, potential energy belongs to a system, not to individual particles. $\endgroup$
    – Bill N
    Commented Jan 2, 2018 at 4:11

3 Answers 3

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Potential and potential energy are defined for pairs of objects, not individual objects. It's meaningless to say "the potential energy of A". One must say "the potential energy of the system consisting of A and B". There is only one potential and potential energy in your problem.

Perhaps the confusion comes from the way potential is introduced in introductory textbooks. Typically we fix a positive charge at the origin, and calculate the work done to bring a charge in from an infinite distance. While doing this, your attention is focused on the moving charge. But what you've actually calculated is the potential energy of the system.

Potential energy is the energy of interaction of two objects. Two objects are required.

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There is no conflict here.

Let the two charged particles ($M,Q$) be the system with no external forces acting.

Momentum is conserved and so for all time

$M_BV = M_A V_{Af} + M_BV_{Bf}$

Energy is also conserved and so for all time

$\frac 12 M_B v_B^2 + \dfrac{kQ^2}{R_i} = \frac 12 M_B v_{Bf}^2 + \frac 12 M_A v_{Af}^2 + \dfrac{kQ^2}{R_f}$

The electric potential energy is stored in the system not in the individual particles.

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If A would gain kinetic energy, it would move far from B. As A would move more far, Potential Energy of B won't increase as distance had increased proportionally.

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