Whether a medium behaves as a conductor or a dielectric is not a fixed property — it depends on frequency, and the deciding quantity is the loss tangent.

\(\tan\delta=\dfrac{\sigma}{\omega\varepsilon}=\dfrac{\text{conduction current}}{\text{displacement current}}\)

Step 1 — compute the denominator.

\(\omega=2\pi f=2\pi\left(10^{10}\right)=6.283\times10^{10}\ \text{rad/s}\)

\(\varepsilon=\varepsilon_{r}\varepsilon_{0}=10\times8.854\times10^{-12}=8.854\times10^{-11}\ \text{F/m}\)

\(\omega\varepsilon=6.283\times10^{10}\times8.854\times10^{-11}=5.563\)

Step 2 — form the ratio.

\(\tan\delta=\dfrac{10^{-2}}{5.563}=1.8\times10^{-3}\)

Step 3 — classify.

At \(1.8\times10^{-3}\) the displacement current outweighs the conduction current by more than five hundred to one, so the earth behaves as a good dielectric — option 3.

The frequency dependence is the real lesson. Since \(\tan\delta\propto1/f\), the same soil that is a good dielectric at 10 GHz is a good conductor at low frequencies. Repeating the calculation at 1 MHz gives

\(\tan\delta=\dfrac{10^{-2}}{2\pi\left(10^{6}\right)\left(8.854\times10^{-11}\right)}=18\)

— firmly conducting. This is exactly why ground-wave propagation works at medium-wave broadcast frequencies, where the earth acts as a conductor and guides the wave along its surface, while at microwave frequencies the ground behaves as a lossy dielectric that reflects and refracts instead.

Note what does not enter the calculation : the permeability \(\mu_{r}=2\) is a distractor. It affects the wave impedance and velocity but not the conductor/dielectric classification, which depends only on \(\sigma\), \(\varepsilon\) and \(\omega\).

Hence, at 10 GHz the earth behaves like a good dielectric.