# Waveguide wave equation

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.

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.