Solution

For a charged plate moving with velocity v, the radiation electric field at a distance x is

$E_{\text{rad}} = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$

where E is the field inside the capacitor and c is the speed of light.
So we start by parts

a) Parallel‑plate capacitor with constant acceleration a

Data:

· Plate separation: d
· Internal field: E
· Acceleration: a (parallel to the plates)
· Observation point A at distance x from the capacitor.

Solution:

1. Contribution from the upper plate (distance x):

   E_1 = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right).

2. Contribution from the lower plate (distance x + d, opposite charge)
   $E_2 = \frac{v}{2c}\,E\!\left(t - \frac{x+d}{c}\right)$

The sign change is because the lower plate has charge -Q, which reverses the direction of the emitted field

$v(t) = a t. The retarded times are t - x/c and t - (x+d)/c$

$E_1 = -\frac{a}{2c}\,E\left(t - \frac{x}{c}\right), \qquad$

$E_2 = \frac{a}{2c}\,E\left(t - \frac{x+d}{c}\right)$

$E_{\text{rad}} = E_1 + E_2 = -\frac{aE}{2c}\left[\left(t - \frac{x}{c}\right) - \left(t - \frac{x+d}{c}\right)\right]$

Simplifying:

$E_{\text{rad}} = -\frac{aE}{2c}\left( \frac{d}{c} \right) = -\frac{a d E}{2c^2}$

$\boxed{E_{\text{rad}} = \frac{a d E}{2c^2}}$

b) Plate with sinusoidal current

Data:
Linear current density on the plate: amplitude$i_0$, sinusoidal variation.
Find the amplitude of the electric field of the emitted wave, in SI and in CGS.

Solution:

1. Relation between surface current and emitted field:

   A plate with surface current$K (A/m)$

emits a plane wave whose electric field in the perpendicular direction is:
$E_{\text{rad}} = \frac{\mu_0 c}{2}\,K$
The amplitude of the surface current is $i_0.$
2. Expression in SI:

$E_{\text{rad}} = \frac{\mu_0 c}{2}\,i_0$ Using $\mu_0 c = \frac{1}${c\varepsilon_0}$

it can also be written as:

$\boxed{E_{\text{rad}} = \frac{i_0}{2c\varepsilon_0}} \quad (\text{SI})$
3. Expression in CGS:
In the Gaussian system, the same relation is:
$\boxed{E_{\text{rad}} = \frac{2\pi i_0}{c}} \quad (\text{CGS})$

c) Reflection coefficient of a thin conducting film

Data:

· Film thickness: x
· Density of conduction electrons: $n_e$
· Electron charge: e, mass:$m_e$
· Frequency of the incident wave:

$\nu (\omega = 2\pi\nu)$

Perpendicular incident wave:

$E = E_0 \sin\omega t$

Solution:

  From the equation of motion

$m_e \dot{v} = e E_0 \sin\omega t: v(t) = -\frac{e E_0}{m_e \omega} \cos\omega t$

  Current density: $ j = n_e e v$.
  Thickness x → surface current (per unit width): $K = j x$
  

Amplitude of this current

$K_0 = n_e e \cdot \frac{e E_0}{m_e \omega} \cdot x = \frac{n_e e^2 x}{m_e \omega} E_0$

  As in part (b):

$E_{\text{rad}} = \frac{\mu_0 c}{2} K_0 = \frac{\mu_0 c}{2} \cdot \frac{n_e e^2 x}{m_e \omega} E_0$.

Substituting

$\mu_0 c = 1/(c\varepsilon_0) and \omega = 2\pi\nu$:
$E_{\text{rad}} = \frac{n_e e^2 x}{4\pi m_e \nu \varepsilon_0 c}\,E_0$

$k = \left( \frac{E_{\text{rad}}}{E_0} \right)^2$
$\boxed{k = \left( \frac{n_e e^2 x}{4\pi m_e \nu \varepsilon_0 c} \right)^2}$