Question

A plane electromagnetic wave of angular frequency ω propagates in a poorly conducting medium of conductivity σ and relative permittivity ɛ . The ratio of conduction current density and displacement current density in the medium is

• A. $\frac{\sigma}{\omega \epsilon \epsilon_0}$
• B. $\frac{\omega \epsilon \epsilon_0}{\sigma}$
• C. $\frac{\sigma}{\omega \epsilon}$
• D. $\frac{\omega \epsilon}{\sigma}$

Answer 1

Answer: (C)

$\frac{\sigma}{\omega \epsilon}$

Answer 2

The conduction current density is given by $\vec{J}_c = \sigma \vec{E}$. The displacement current density is given by $\vec{J}_d = \epsilon \frac{\partial \vec{E}}{\partial t}$. For a plane electromagnetic wave, the electric field varies sinusoidally with time, $\vec{E} = \vec{E}_0 \cos(\omega t - kz)$. The time rate of change of the electric field is $\frac{\partial \vec{E}}{\partial t} = -\omega \vec{E}_0 \sin(\omega t - kz)$. The amplitude of the conduction current density is $|\vec{J}_c|_{amp} = \sigma E_0$. The amplitude of the displacement current density is $|\vec{J}_d|_{amp} = \epsilon \omega E_0$. The ratio of the magnitudes of the conduction current density to the displacement current density is $\frac{|\vec{J}_c|_{amp}}{|\vec{J}_d|_{amp}} = \frac{\sigma E_0}{\epsilon \omega E_0} = \frac{\sigma}{\omega \epsilon}$. The question states "relative permittivity ɛ", which implies the absolute permittivity of the medium is $\epsilon_{abs} = \epsilon \epsilon_0$. However, in many contexts, especially in simpler problems or when $\epsilon_0$ is implicitly assumed or not relevant to the ratio's structure, $\epsilon$ is treated as the absolute permittivity. Given the options, it is most likely that $\epsilon$ in the question refers to the absolute permittivity of the medium, leading to the ratio $\frac{\sigma}{\omega \epsilon}$.