 Solution

Use the Maxwell curl equationsDot the first with $\mathbf{E}/{\mu }_0$ and the second with $\mathbf{B}/{\mu }_0$, then combine them. The identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot \nabla \times \mathbf{E}-\mathbf{E}\cdot \nabla \times \mathbf{B}$ givesThis proves the Poynting theorem $W_t+\nabla \cdot \mathbf{S}+\mathbf{J}\cdot \mathbf{E}=0$, with the stated energy density and Poynting vector. The last term is work transferred from the electromagnetic field to matter.

For the specified vacuum wave, substitution into the same curl equations gives $k{E}_0=\omega B_0$ and $k{B}_0={ϵ}_0{\mu }_0\omega E_0$. Thus ${\omega }^2=k^2/{ϵ}_0{\mu }_0$, and for propagation along positive $z$,Putting $\vartheta =kz-\omega t$,The equal electric and magnetic contributions account for the factor in $W$. Energy propagates at speed $c$, whose square is $1/({ϵ}_0{\mu }_0)$; the reciprocal product is not itself the speed.