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+### Statement
+
+$12.1.15.$ [Insert the problem statement]
+
+### 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$
+ $\boxed{k = \left( \frac{n_e e^2 x}{4\pi m_e \nu \varepsilon_0 c} \right)^2}$
+
+#### Answer
+
+[Insert a concise answer or boxed result]