Exercise 5.2.2: Conductor or dielectric?

$J=\sigma E_0\cos \omega t$ and $\partial D/\partial t=-\epsilon \omega E_0\sin \omega t$: they are in quadrature, with the amplitude ratio

$\sigma =\omega \epsilon$ at $f_c=\sigma /(2\pi \epsilon )=0.05/(2\pi \times 4\times 8.842\times 10^{-12})=2.25\times 10^8\mathrm{Hz}=225\mathrm{MHz}$.

The ratio is $f_c/f$: at $1\mathrm{MHz}$ it is $225$ (the soil is a conductor, drop $\partial D/\partial t$), at $1\mathrm{GHz}$ it is $0.225$ (the soil is a lossy dielectric, the conduction current is a $22\%$ correction that cannot be dropped, but the displacement current dominates).

Units: $\sigma /(\omega \epsilon )$ is $(S/m)/({\mathrm{s}}^{-1}\cdot F/m)=\mathrm{S}\cdot \mathrm{\Omega }=1$, a pure number. This is why ground-penetrating radar works at hundreds of MHz and above, while low-frequency fields are shielded by moist ground, and why the same material needs two different descriptions in the two bands.