**1 Answer**

written 6.9 years ago by |

Poynting theorem generally provides the direction of EM wave and the power associated with it. That can be obtained using cross product of $\bar{E}$ and \bar{H} hence,

$\bar{P}= \bar{E} × \bar{H}=|\bar{E} | a_x ×|\bar{H} | a_y \\ \bar{P}= |\bar{E} ||\bar{H} | (a_x × a_y ) \\ \bar{P}=|P| \overline{a_z} w/m^2 \\ |\bar{E} | a_x$

Where $\bar{P}$ is called pointing vector which shows the direction of EM wave and the power associated with it

$\bar{P}= \bar{E} × \bar{H}$

Taking divergence on both sides

$∇.\bar{P}= ∇.(\bar{E} × \bar{H}) \\ ∇.P ̅= \bar{H} (∇×\bar{E} )- \bar{E} (∇×\bar{H}) \\ ∇.\bar{P}= \bar{H} \dfrac{-μ d\bar{H}}{dt}- \bar{E} \bigg(σ\bar{E} +E \dfrac{d\bar{E}}{dt} \\ ∇.\bar{P}= - μ H \dfrac{(d\bar{H}}{dt}- \bar{E} σ \bar{E}-E \dfrac{d\bar{E}}{dt} \\ ∇.\bar{P}=\dfrac{- μ}{2} \dfrac{d}{dt} |\bar{H} |- σ |\bar{E} |^2-\dfrac{E}{2} \dfrac{d}{dt} |\bar{E} |^2$

In integrated format-

$∮_s\bar{P} .d\bar{s}= -∮_u \bigg[σ |\bar{E} |^2+ \dfrac{E}{2} \dfrac{d}{dt} |\bar{E} |^2 v+ \dfrac{μ }{2} \dfrac{d}{dt} |\bar{H} |^2 \bigg]$