# ratio between conduction current and displacement current

First, recall that Maxwell displacement current for a plane wave is $$\vec j_D = \epsilon \partial_t \vec E = \epsilon \partial_t (\vec E_o cos(\vec k \cdot \vec r - \omega t)) = \epsilon \omega \vec E$$ and then we know that the conduction current $\vec J$ is related to E from Ohm's law $$\vec j_C = \sigma \vec E$$

The ratio between these two currents, $$Q = \frac{\vec j_D}{\vec j_C} = \frac{\epsilon \omega}{\sigma} = \frac{k_R}{k_I}$$

incidentally, this is also related to the ration between the real and complex parts of the wavevector $\vec k$ of a plane wave propagating through a conducting medium. the algebra in its derivation is quite tedious so i'll forgo putting it here.

So we know that when Q approaches infinity, $k_R$ >> $k_I$, we have a bad conductor. The wave does not get dissipated, it just passes through.

when Q approaches zero, $k_I$ >> $k_R$, we have a good conductor. There is a finite skin depth.

Those two statements above I'm not really 100% confident if they're correct. some other sources say that a good conductor has Q = 1.

So what does Q = 1 imply, or what are the properties of a medium with Q = 1?