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By the work energy theorem we have that the total energy of a nonrelativistic point charge, $q_0$ of mass $m$, moving in an electric field $\mathbf{E}$ is

$ E = E_k + U_e = \frac{1}{2}mv^2 + q_0V \quad Eq.1 $

As I read in my text:

The electrical charge acquires potential energy. If the charge is released, work is done by the field and the charge accelerates. It means that its potential energy is converted into kinetic energy.

But if an accelerating charge emits electromagnetic radiation some electric potential energy, should be dissipated and could not be converted into kinetic energy. How can eq. 1 hold? The text mentions the dissipation of energy by accelerating charges only later in the course, and says nothing when treating energy conservation of moving charges in the electrostatic field.

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  • $\begingroup$ Eq 1 is not the work energy theorem $\endgroup$
    – Bob D
    Mar 1, 2023 at 11:59

2 Answers 2

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The conservation of energy in this problem is of course an approximation - like many things in physics (and one could even say that physics is only an approximation to the real world.) The energy dissipated by an accelerated charge can be calculated using the Larmor formula: $$ P=\frac{2}{3}\frac{q^2a^2}{c^3},$$ where $a$ is the charge acceleration. The speed-of-light entering the numerator hints that this is a relativistic corrections - something one is likely to ignore in problems, where the kinetic energy can be described by non-relativistic expression $\frac{mv^2}{2}$.

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  • $\begingroup$ If energy is not conserved, then it is not correct to consider the electrostatic field as a conservative field and introduce the electric potential $V$ $\endgroup$
    – gioretikto
    Mar 1, 2023 at 11:51
  • $\begingroup$ @gioretikto as already pointed in my answer - assuming that energy is conserved here is an approximation, which work very well. Like anything in science it is exact only up to certain precision. $\endgroup$
    – Roger V.
    Mar 1, 2023 at 11:54
  • $\begingroup$ @gioretikto e.g., $E=\sqrt{m^2c^4+p^2c^2}\approx mc^2+\frac{p^2}{2m}=mc^2+\frac{mv^2}{2}$ - so even without charge radiating, your expression for energy is inexact ("not correct" as you say.) $\endgroup$
    – Roger V.
    Mar 1, 2023 at 11:57
  • $\begingroup$ If I've understood correctly is the term $c^3$ in the denominator of Larmor formula which makes the approximation acceptable for non-relativistic velocities $\endgroup$
    – gioretikto
    Mar 1, 2023 at 12:02
  • $\begingroup$ @gioretikto yes, but the energy lost is very small - for typical potentials and accelerations. $\endgroup$
    – Roger V.
    Mar 1, 2023 at 12:06
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But if an accelerating charge emits electromagnetic radiation some electric potential energy, should be dissipated and could not be converted into kinetic energy. How can eq. 1 hold?

That equation holds for quasi-static rearrangements of charges in electrostatic field (no radiation, no magnetic field). As soon as charges move, one has to take into account magnetic energy as well, which eq. 1 does not do. And if charges accelerate, radiation may carry away EM energy away at the expense of local EM energy or other kind of energy, which is not taken into account by eq.1.

More general formulation of conservation of energy is the Law of local conservation of energy. In EM theory this is based on work-energy interpretation of the Poynting theorem: work of electric forces in a region plus increase of EM energy in that region equals influx of EM energy through boundary of the region from outside. This is strictly speaking not conservation in the region - energy in the region may change due to energy coming in, or leaving out, through the boundary.

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