Charge moved across a potential difference; where does the energy for emitted radiation come from? 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
 A: Energy is conserved. If the electron has radiated , its kinetic energy will be less to the amount radiated away. Here is an example.
A: 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.
A: 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.
