Home exercise 5.5.1: A wave in a lossy medium

With $\varphi =\omega t-\beta z$ ($\partial \varphi /\partial z=-\beta$),

$E_x$ does not depend on $x$ or $y$, so ${\nabla }^2{E}_x={\partial }^2{E}_x/\partial z^2$ (and $\nabla \cdot E=0$).

After division by $E_0{e}^{-\alpha z}$,

which holds if and only if both brackets vanish:

Put $\alpha =\omega \mu \sigma /(2\beta )$ into the first equation: ${\beta }^4-{\omega }^2\mu \epsilon {\beta }^2-(\omega \mu \sigma /2{)}^2=0$, a quadratic in ${\beta }^2$ whose positive root gives

$\omega =4\pi \times 10^{8}rad/s$, $\omega \sqrt{\mu \epsilon }=\omega \sqrt{16}/c=16.755rad/m$ and $\sigma /(\omega \epsilon )=0.01125$:

${\beta }^2-{\alpha }^2=280.74={\omega }^2\mu \epsilon$ and $2\alpha \beta =3.158=\omega \mu \sigma$. For $\sigma /(\omega \epsilon )\ll 1$ the square root is $1+\frac{1}{2}(\sigma /\omega \epsilon {)}^2$ and $\alpha \to \frac{\sigma }{2}\sqrt{\mu /\epsilon }$, the $\alpha$ of Problem 5.5.4. Here the two agree to four digits ($0.094246$ against $0.094248Np/m$).