Problem 5.6.7: Frequency and conductivity from k

The phasor has the form $E_0{e}^{-jkx}$ with $k=50(2-j)=100-j50=\beta -j\alpha$, so $\beta =100rad/m$ and $\alpha =50Np/m$: a wave going to $+x$ that decays as $e^{-50x}$. Known: $\mu ={\mu }_0$, $\epsilon =3{\epsilon }_0$. Unknown: $\omega$ and $\sigma$.

The region is source-free, so ${\nabla }^{2}E+k^{2}E=0$. With $E=10e^{-50(1+2j)x}{a}_y$,

The real part is ${\omega }^2{\mu }_0(3{\epsilon }_0)=3{\omega }^2/c^2$:

The imaginary part:

$(100-j50{)}^2=10^4-2500-j10^4=7500-j10^4$. The loss tangent is $\tan \delta =\sigma /\omega \epsilon =10000/7500=4/3$ ($\delta =53.1^{\circ }$): conduction and displacement currents are of the same size, and the medium is neither a good dielectric nor a good conductor. The amplitude $10$ of the phasor plays no part: it cancels in the Helmholtz equation.