#### Show that the average value of radiant flux density S over a single period T is given by$S= \frac{(\frac{1}{2}c\mu _{0})}{E{_{0}}^{2}}$

We know that radiant flux density is:

$\overrightarrow{S}=\frac{1}{\mu _{0}}(\overrightarrow{E}\times \overrightarrow{B})=C^{2}\epsilon _{0} (\overrightarrow{E}\times \overrightarrow{B})$              $[as, c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}]$

Let, along the X-axis, electromagnetic waves are propagating. If along the Y-axis, the electric field vector of electromagnetic waves is there. In the same way, along the z-axis, magnetic field vectors will be there. Hence,

$E=E_{0}\cos (kx-\omega t)$

$B=B_{0}\cos (kx-\omega t)$

$E\times B=\left ( E_{0}B_{0} \right )\cos^{2} (kx-\omega t)$

$S=C^{2}\varepsilon _{0}\left ( E\times B \right )=C^{2}\varepsilon _{0}\left ( E_{0}B_{0} \right )\cos ^{2}\left ( kx-\omega t \right )$

The average value of the radiant flux density is

$S_{av}=C^{2}\varepsilon _{0}\left | E_{0}\times B_{0} \right |\frac{1}{T}\int_{0}^{T} \cos ^{2}\left ( kx-\omega t \right )dt=c^{2}\varepsilon _{0}E_{0}B_{0}\left ( \frac{1}{T} \right )\left ( \frac{T}{2} \right )=\left ( c^{2}\frac{\varepsilon _{0}E_{0}}{2} \right ) \left ( \frac{E_{0}}{c} \right ) \; \; \; \; \;$

$S_{av}=\left ( \frac{\varepsilon _{0}c E{_{0}}^{2}}{2} \right )\; \; \; \; \; \; \; \left [ as, c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}} \right ]$

Therefore,

$S_{av}=\left ( \frac{E{_{0}}^{2}}{2c\mu _{0}} \right )$