Show that the average value of radiant flux density $S$ over a single period $T$ is given by $S = \frac{1}{2c\mu_0}E_0^2$.

(N/A) The radiant flux density (Poynting vector) is given by $S = \frac{1}{\mu_0}(\vec{E} \times \vec{B})$.
Since $B = \frac{E}{c}$,the magnitude is $S = \frac{EB}{\mu_0} = \frac{E^2}{c\mu_0}$.
For an electromagnetic wave,$E = E_0 \cos(kx - \omega t)$.
Substituting this into the expression for $S$,we get $S = \frac{E_0^2 \cos^2(kx - \omega t)}{c\mu_0}$.
The average value of $\cos^2(\theta)$ over a full period $T$ is $\frac{1}{T} \int_0^T \cos^2(\omega t) dt = \frac{1}{2}$.
Therefore,the average radiant flux density $\langle S \rangle$ is $\langle S \rangle = \frac{E_0^2}{c\mu_0} \times \frac{1}{2} = \frac{E_0^2}{2c\mu_0}$.