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+### Statement
+
+$12.1.28.$ [Insert the problem statement]
+
+### Solution
+
+$12.1.28$ A plane electromagnetic wave falls on a metal wall moving at velocity $v$ perpendicular
+to its surface. Electric field strength of wave $E$. What pressure in
+SI and GHS does the wave exert on the wall?
+
+Using the Lorentz transformation for electric and magnetic fields and the relation between these fields $E = B c$, we can calculate the electric field in
+the moving frame:
+
+\begin{equation}
+ E_1 = \frac{E - (-v) B}{\sqrt{1-v^2/c^2}} = \frac{E - (-v) E/c}{\sqrt{1-v^2/c^2}} = E \sqrt{\frac{1+v/c}{1-v/c}}
+\end{equation}
+
+The pressure on the moving wall is:
+
+\begin{equation}
+ \mathcal{P} = 2 \frac{I}{c} = 2 \frac{c \epsilon_0 E_1^2}{c} = 2 \epsilon_0 E_1^2 = 2 \epsilon_0 E^2 \frac{1+v/c}{1-v/c}
+\end{equation}
+
+The factor of two is because there is a pressure due to the incoming flow and another due to the outgoing flow.
+
+Also, the result in the $GHS$ system of units is:
+
+\begin{equation}
+ \mathcal{P} = \frac{1}{8 \pi} \epsilon_0 E^2 \frac{1+v/c}{1-v/c}
+\end{equation}
+
+\begin{equation}
+ \mathcal{P} = 2 \epsilon_0 E^2 \frac{1+v/c}{1-v/c}
+\end{equation}
+
+#### Answer
+
+[Insert a concise answer or boxed result]