From "Schaum's Electromagnetics crash course", 2003.
Solved Problem 4.1:
In a Material for which:
$$\begin{aligned}\sigma &= 5~~~[\text{seimen/meter}] \\ \epsilon &= 1~~~[\text{farad/meter}]\end{aligned}$$
the electric field intensity is:
$$E = 250 \sin(10^{10}t)~~~[\text{volt/meter}]$$
Find:
- conduction current density $J_c$
- displacement current density $J_i$
- frequency $\omega$ at which they will have equal magnitudes.
I'm ok with answer for #1 and #2:
$$J_c = 1250 \sin(10^{10}t)~~~[\text{coulomb/meter}^2]$$
$$J_D = 22.1 \cos(10^{10}t)~~~[\text{coulomb/meter}^2]$$
its #3 that makes no sense…. it seems to me that $J_c$ and $J_d$ already have the same frequency of $10^{10}~\text{rad}$?
However solution says:
that $i_c = i_d$ when $\sigma = \omega \epsilon$
thus,
$\omega = \frac{5.0}{8.854\times10^-12} = 5.65 \times 10^11~\text{rad/s} = 89.9~GHz$
I don't get it… where does this formula $\sigma = \omega \epsilon$ come from? There's nothing in the book prior to this page that even mentions it.
Best Answer
Displacement current density is $\epsilon \partial {\bf E}/\partial t$, whilst conduction current density is $\sigma {\bf E}$.
If ${\bf E} = {\bf E_0} \sin \omega t$, then the displacement current density depends on frequency, whilst the conduction current density does not.
The two are out of phase, but have equal amplitudes when $\sigma =\omega \epsilon$ as can be seen by simply differentiating the electric field with respect to time.