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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.

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1 Answer 1

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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.

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  • $\begingroup$ Kind of makes sense.. so the trick is to replace $10^{10}$ with $\omega$ in the sinusoid for E then work the problem... $\endgroup$
    – pico
    Commented Dec 31, 2019 at 17:14

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