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Let's use an electron and a 1V potential difference as a mode.

In school I learned that if the electron is at the negative end of the electric field, its potential energy is equal to the work that could be done on it by the electric field (analogous to standing on a building and having gravity-mediated potential energy). In this case the P.E. is 1 eV. Once the electron is moved across the 1V, that potential energy is converted into kinetic energy (just like the P.E. on top of the building being converted entirely to K.E. at the ground) and now the electron has K.E. of 1 eV. However, I just learned that accelerating charges emit EM radiation. If all the P.E. is converted into K.E., where does the energy for the radiation come from?

Thanks

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The final kinetic energy of the electron plus the radiated energy add up to the change in potential energy (1 eV in this case). So as a result of the radiation the final kinetic energy is less than 1 eV. In effect the radiation makes the electron seem slightly heavier so it accelerates slightly more slowly than it would otherwise do.

The radiated power is given by the Larmor formula:

$$ P = \frac{q^2a^2}{6\pi\varepsilon_0 c^3} $$

Suppose your electron travels 1cm so the field strength is $E = 100$ V/m. The force on the electron is $eE = 100e$ Newtons and the acceleration is therefore $a = 100e/m$ m/s$^2$. Substituting this into the equation for the power gives:

$$ P = \frac{10000 e^4}{6\pi\varepsilon_0 c^3 m^2} \approx 10^{-8} \text{eV/s} $$

The time taken by the electron to cross the 1cm gap is given by:

$$ t = \sqrt{\frac{2s}{a}} \approx 3.4 \times 10^{-8} \text{s} $$

So the total energy radiated is around $4 \times 10^{-16}$ eV, which is an immeasurably small fraction of the 1 eV potential energy change.

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  • $\begingroup$ OK perfect that makes sense. Thanks for the math. I'm assuming what they taught us omitted the effect of the radiation on the kinetic energy for simplicity, and because it's negligible. $\endgroup$ – Jory Mar 19 '16 at 17:16
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Energy is conserved. If the electron has radiated , its kinetic energy will be less to the amount radiated away. Here is an example.

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  • $\begingroup$ So are you saying it's possible for an electron to be accelerated and NOT emit radiation? I'm just trying to figure out whether what I learned in school is just a simplification to avoid confusion and illustrate the concept of PE and KE, or represents an actual physical possibility. I thought Maxwell's equations define radiation from accelerating charge, and that's the reason Rutherford's nuclear model was so surprising. $\endgroup$ – Jory Mar 19 '16 at 17:04
  • $\begingroup$ Accelerated charge will radiate if it is a macroscopiccharge distribution continuously, if it is an electron or a proton it will depend on the energies, and when decelerating but it will be in a quantized manner, consecutive photons leaving and to get detectable energy photons the energy of the electron should be relativistic $\endgroup$ – anna v Mar 19 '16 at 18:18
  • $\begingroup$ @Jory Sure, charges can accelerate without radiating. Consider a spherical shell of charge oscillating radially in and out, or look at Abbott and Griffiths' paper about it, a oaler you can find by searching "acceleration without radiation". $\endgroup$ – Timaeus Mar 20 '16 at 0:17
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To move a charge across a P.D , you need to do work on it , the work you do gets stored in it (in the form of potential energy if acceleration is negligible) , this energy is radiated.

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