# Ampere's law: frequency where conduction current equals displacement current

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:

1. conduction current density $$J_c$$
2. displacement current density $$J_i$$
3. 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.

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.
• Kind of makes sense.. so the trick is to replace $10^{10}$ with $\omega$ in the sinusoid for E then work the problem...