3
$\begingroup$

Generally we equate change in potential energy to change in kinetic energy but in case of a charged particle like electron this is inconsistent. Consider a case: An electron(of charge e)from rest is accelerated in potential difference V then the final kinetic energy is given by KE=eV But we know that accelerating charges radiate energy then why are we not taking into consideration this form of energy when charges accelerate as in simple cases like flow current in wires etc.

$\endgroup$
1
  • $\begingroup$ Not sure what (if any) question you are asking - but conservation of energy is valid, even for electrons. It's just that "kinetic" and "potential" are not the only two forms of energy that need to be considered. You need the sum over all forms of energy... $\endgroup$
    – Floris
    Feb 16, 2016 at 17:58

3 Answers 3

2
$\begingroup$

Although these unbound accelerating electrons do radiate relative to their kinetic energy it is very small in a lot of cases and so can be neglected. For example if you connect a wire to the input of an oscilloscope, the display of the oscilloscope will, amongst other waveform, show a signal at the frequency of the mains power supply. Relative to the energies transmitted by the mains supply the energy of these em waves is very small.

Of course there are examples of the amount of em radiation emitted being significant. Antennas and X-ray tubes are examples of this.

$\endgroup$
2
$\begingroup$

Conservation of energy is valid, even for electrons.

But when you look at electrons "accelerating in wires", it is worth calculating the relative energy lost due to radiation, to understand whether it's a term worth worrying about.

The Larmor Equation describes the power radiated by a charge $q$ accelerating with acceleration $a$ - it shows that power goes with acceleration squared:

$$P = \frac23 \frac{q^2a^2}{4\pi\epsilon_0 c^3}$$

Let's take an electron that is oscillating at 1 MHz - what fraction of the energy does it lose due to radiation in each cycle?

If the amplitude is $A$ and the frequency $\omega$, then acceleration $a = -\omega^2 A\cos\omega t$, and the mean value of $a^2$ over one cycle is $\frac12A^2\omega^4$. It follows that the energy lost per cycle is

$$E = P\frac{2\pi}{\omega} = \frac16 \frac{q^2 A^2 \omega^3}{\epsilon_0 c^3}$$

For such motion, the maximum kinetic energy is $KE=\frac12 m v^2 = \frac14 m \omega^2 A^2$. The ratio (fraction lost per cycle) is:

$$\frac{E}{KE} = \frac23\frac{q_e^2\omega}{m_e\epsilon_0 c^3}$$

for the numbers above, this evaluates to $7.5\cdot 10^{-16}$ - meaning it would take a really, really long time for these electrons to lose a significant fraction of their power.

Note that the fraction scales with frequency - so as frequency goes up, the fractional energy dissipated per cycle also increases (and the fractional energy per unit time goes as $\omega^2$, so even faster).

What this tells us is that it is hard to elicit bremsstrahlung from accelerating charges; and the fraction of energy used is small. For many problems where you are concerned with the dynamics and equations of motion of an electron, ignoring this effect will make an insignificant difference to the calculation.

$\endgroup$
2
$\begingroup$

Farcher and Floris have fine answers to this question.

I did want to point out that potential is not potential energy per charge. That kind of fiction can approximately hold in some situations, particularly for static charges and charges that are moved by forces that produce very very small accelerations.

It's like if you assembled some static charges to a particular locations and spent a huge amount of time assembling them. If you accelerated them slowly enough then the energy required to assemble them is really close to the electrostatic potential times the charge. But the actual energy required will depend on exactly how slowly you assembled them.

So you were lied to whenever someone said potential is potential energy per charge. Really, energy is exchanged between electromagnetic fields and charges.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.