Solution

Let's prove the formula using the basic laws of electrodynamics.

1. The static electric field of an infinite charged plate is directed along the normal to it (along the x-axis) and is equal to $E = \frac{\sigma}{2\varepsilon_0}$.

2. When the plate begins to move parallel to itself with a velocity $v(t)$, it creates a surface current $K(t) = \sigma v(t)$.

3. According to Ampere's law, this surface current generates a magnetic field near the plate (as $x \to 0$) equal to:
   $$B(0, t) = \frac{\mu_0 K(t)}{2} = \frac{\mu_0 \sigma v(t)}{2}$$
   By multiplying the numerator and denominator by $\varepsilon_0$ and considering that $c^2 = \frac{1}{\mu_0 \varepsilon_0}$, and $E = \frac{\sigma}{2\varepsilon_0}$, we obtain the boundary condition for the magnetic field at the plate surface:
   $$B(0, t) = \mu_0 \varepsilon_0 v(t) \frac{\sigma}{2\varepsilon_0} = \frac{v(t)}{c^2} E$$

4. Any change in this magnetic field (caused by a change in velocity) detaches from the plate and propagates through space as a plane electromagnetic wave at the speed of light $c$. Consequently, the field at a distance $x$ at time $t$ exactly repeats the field at the plate, but delayed by a time $\Delta t = x/c$:
   $$B(x, t) = B(0, t - x/c) = \frac{v(t - x/c)}{c^2} E$$

5. In a traveling plane electromagnetic wave, the transverse electric field strength $E_{\text{rad}}$ (which is directed parallel to the plate) and the magnetic induction $B$ are strictly related by $E_{\text{rad}} = c B$.

6. Substituting the derived expression for $B(x, t)$ into this relation, we arrive at the required formula:
   $$E_{\text{rad}}(t, x) = c \cdot \frac{v(t - x/c)}{c^2} E = \frac{v(t - x/c)}{c} E$$

Application to graphs (b) and (c):
The derived formula is universal because the wave equation is linear, and the boundary condition $B(0,t)$ directly tracks any function $v(t)$.

• For graph (b): the velocity drops abruptly from $v_0$ to $0$ at time $t_0$. In space, this will look like a wavefront behind which (closer to the plate) the transverse field $E_{\text{rad}}$ is zero, and ahead of which it is $\frac{v_0}{c}E$. This termination front moves away from the plate at speed $c$.
• For graph (c): the velocity increases abruptly from $0$ to $v_0$ at time $t_0$. This generates a wavefront behind which a uniform transverse field $E_{\text{rad}} = \frac{v_0}{c}E$ is established, propagating behind the front at speed $c$.
  In both cases, the formula $\frac{v_{t-x/c}}{c}E$ provides the exact instantaneous value of the radiated field at any point in space.