Electromagnetic waves > Properties of emission and reflection of electromagnetic waves > Problem 12.1.22

Statement

$12.1.22$ The amplitude of the electric field strength of a plane sinusoidal wave is equal
to $E_0$. What is the average pressure exerted by this wave on a flat metal wall
when it is normally incident on it?

Solution

The energy per unit area per unit time striking the metal wall is equal to the Poynting vector:

\begin{equation}
\vec{S} = c^2 \epsilon_0 (\vec{E} \times \vec{B})
\end{equation}

For this case:

\begin{equation}
\vec{E} = \hat{y} E_0 \cos(\omega t) \quad \text{and} \quad \vec{B} = \hat{z} \frac{E_0}{c} \cos(\omega t)
\end{equation}
The pressure exerted by the incoming flow on the wall is:

\begin{equation}
P = \frac{\left\langle S\right\rangle_T }{c} = c \epsilon_0 \left\langle \vec{E} \times \vec{B}\right\rangle_T = \frac{\epsilon_0 E_0^2}{2}
\end{equation}
In this equation we have taken the time average of the cross product.
This is half of the answer, because there is a pressure exerted by the incoming flow and another equal pressure from the outgoing flow; our answer must be twice that for this reason:

\begin{equation}
P_{\text{total}} = \epsilon_0 E_0^2
\end{equation}

Answer

\begin{equation}
P_{\text{total}} = \epsilon_0 E_0^2
\end{equation}