Waveguide wave equation

Question

Consider a wave guide built from two parallel slabs at $x=0, x=L$. between the slabs there is a dielectric material with $\mu=\epsilon=1$ and electric conductivity with a frequency of $\sigma(\omega)=\frac{\omega^2}{4\pi \omega_0}$. an electric field that moves to the $\hat z$ direction and linearly polarized at the $\hat y$ enters the waveguide.

my prof. used the following wave equation:

$\nabla^2 E=\frac{1}{c^2}\frac{\partial^2E }{\partial t^2}+\frac{4\pi\sigma(\omega)}{c^2}\frac{\partial E }{\partial t}$

I don't understand why this equation is used and I did not find any info about it.

Answer 1

I think that you are working in cgs/Gaussian units with $\vec \nabla E = -\dfrac 1 c \dfrac{\partial \vec B}{\partial t}$ and $\vec \nabla \times \vec B = \dfrac 1 c \dfrac{\partial \vec E}{\partial t}+ \dfrac{4\pi}{c}\vec J$ as two of Maxwell's equations?

With no losses $\vec J = 0$ but in this case you must use Ohm's law $\vec J=\sigma \vec E$ to get the result that you require when deriving the lossy wave equation.