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this is my first question here, I'm sorry if I make any mistake. From what I read from my physics book, the expression for electric potential energy considering two point charges is: U=$\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}$. This, however, was proven when one charge was stationary, and the other one had free movement. If both charges are moving, then how do I consider this expression? U accounts for both charges? Meaning that in the expression $U_1+K_1=U_2+K_2$, in $K_1$ or $K_2$ I must take into account both kinetics energies? Or I just use the previous expression for each charge separately?

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  • $\begingroup$ Potential energy as a concept relies on the field being static, ie overcoming the E field from a stationary charge. With that being said, assigning PE for this situation doesn't make sense as the field is non conservative. To find the work needed to move a charge from infinity to the location, you need to manually compute the line integral for the specific situation. $\endgroup$ Jul 6, 2022 at 17:02
  • $\begingroup$ However I think you may be wanting, is the total field energy,$$ \iiint [\frac{1}{2}\epsilon_0 \vec{E}^2 + \frac{1}{2\mu_0}\vec{B}^2] dv$$ but this formula doesn't work for point charges, only distributions where $\rho$ is finite $\endgroup$ Jul 6, 2022 at 17:04
  • $\begingroup$ Actually you can use the general formula of total field energy for point charges, if you regularize it at small distances (eg assimilate the point by a sphere of small radius). The difference of regularized energy coincides with the difference in the usual Coulomb energy as the distance cutoff goes to $0$. More specifically, if the charges generate respective fields $E_1,E_2$, the cutoff will handle the constant self energy $\int E_i^2$ which cancels out in a difference, and the Coulomb energy is exactly $\epsilon_0\int E_1\cdot E_2$. $\endgroup$
    – LPZ
    Jul 6, 2022 at 20:43
  • $\begingroup$ @jensenpaull So since these point charges are moving, the E field varies with time and there's no guarantee that the force will be conservative? $\endgroup$
    – user336281
    Jul 14, 2022 at 20:16

1 Answer 1

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Electric potential energy of two charges is always (by definition)

$$ U = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r}. $$

In other words, potential energy does not depend on charges' velocities.

If you're asking whether the relation

$$ U + \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = const. $$

holds universally, then the answer is it doesn't. In EM theory there is no guarantee of conservation of kinetic plus potential energy, because part of the energy can move away in the form of radiation. There is still local energy conservation, but it cannot be expressed using charged particles' positions and velocities, one has to use EM energy density and EM energy flux density, depending on electric and magnetic fields.

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  • $\begingroup$ The potential, $\phi$, is one component of a relativistic four-vector. Therefor, "potential energy does not depend on charges' velocities." is not correct. $\endgroup$ Jul 7, 2022 at 18:03
  • $\begingroup$ @JerroldFranklin Relativistic four-vector is irrelevant in this question. Potential energy, by definition, is function of positions and not of velocities. In case of two charged point particles, it is given by the above formula. $\endgroup$ Jul 8, 2022 at 0:14
  • $\begingroup$ @JánLalinský So just to make sure I got it. The U expressions stands always, whether the charges are moving or not. However, K + U=const, not necessarily. If I still not take into account any energy loss due to radiation, might it be correct? $\endgroup$
    – user336281
    Jul 14, 2022 at 20:24
  • $\begingroup$ The expression for potential energy stands, but K+U=const. is not universally valid. It is valid when magnetic energy is negligible, which is only when the motions are slow enough. $\endgroup$ Jul 14, 2022 at 21:17

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