Electromagnetic waves > Properties of emission and reflection of electromagnetic waves > Problem 12.1.13

Statement

$12.1.13.$

A charged flat capacitor is swung by moving it parallel to the plates. As the os-
cillation frequency ν increases, the average intensity I of the electromagnetic
waves emitted by the capacitor first increases, then decreases to zero, then
increases again, and so on. At what frequencies does the capacitor not emit
energy? Estimate the frequency at which the 1st and $kth$ emission maxima
are observed.

Solution

The capacitor has two parallel plates with opposite charges $\pm Q$ and area S. When oscillating along the x-axis (parallel to the plates) with velocity $v(t)$, each plate is equivalent to a surface current:

$K_+ = \sigma v(t)\, \hat{x}, \qquad K_- = -\sigma v(t)\, \hat{x}, \qquad \sigma = \frac{Q}{S}$

These time-varying currents generate electromagnetic fields that propagate: the capacitor becomes a transmitting antenna.

Each plate emits electromagnetic waves. For a distant observer in the direction perpendicular to the plates (z-axis), the waves emitted by the front and back plates travel different paths. The path difference is the separation between the plates, d.

If the wavelength is
$\lambda = c/\nu$

the phase difference between the two waves at the observation point is:

$\Delta\phi = \frac{2\pi}{\lambda}\, d = \frac{2\pi\nu}{c}\, d$

Constructive interference (intensity maximum):
It occurs when the waves arrive in phase:
$ \Delta\phi = 2\pi k \quad\Rightarrow\quad \frac{2\pi\nu d}{c} = 2\pi k \quad\Rightarrow\quad$

$ \boxed{\nu_k' = \frac{c}{d}\, k},\quad k = 1,2,3,\dots$

These are the frequencies of the emission maxima.
The first maximum $(k = 1) is \nu_1' = c/d$

Destructive interference (zero intensity):
It occurs when the waves arrive in opposite phase:
$ \Delta\phi = (2k+1)\pi \quad\Rightarrow\quad \frac{2\pi\nu d}{c} = (2k+1)\pi \quad\Rightarrow\quad$
$ \boxed{\nu_k'' = \frac{c}{d}\left(k + \frac{1}{2}\right)},\quad k = 0,1,2,\dots$

At these frequencies, the contributions from the two plates cancel out and the capacitor emits no energy (in that main direction). That is why the average intensity drops to zero.

As the frequency \nu increases from zero, the phase difference \Delta\phi grows. For very low frequencies,

$\Delta\phi \approx 0$
and there is reinforcement. When reaching
$\Delta\phi = \pi $(first cancellation,$ k = 0)$
the intensity goes to zero. Then, upon reaching $ \Delta\phi = 2\pi (k = 1)$,
it becomes maximum again, and so on periodically. The intensity oscillates between high values and zeros, giving rise to the observed alternation.

Answer

$\boxed{\nu_k' = \frac{c}{d}\, k},\quad k = 1,2,3,\dots$

$\boxed{\nu_k'' = \frac{c}{d}\left(k + \frac{1}{2}\right)},\quad k = 0,1,2,\dots$