5.2.3 Conduction or displacement? The crossover frequency

In a material medium both terms of Eq. (5.2.1) are present at once. A medium with conductivity $σ$ and permittivity $ε$ carrying the field $E = E_{0} cos ω t$ has the conduction current density $J = σ E$ and the displacement current density $ε \partial E / \partial t$ :

\nabla \times H = conduction σ E_{0} cos ω t displacement - ε ω E_{0} sin ω t . (5.2.9)

The two terms are a quarter period apart in time, and their amplitudes are in the ratio

\frac{∣ J ∣}{∣ \partial D / \partial t ∣} = \frac{σ}{ω ε} = tan δ, (5.2.10)

a number called the loss tangent of the medium. The ratio falls with frequency: every medium behaves as a conductor at low enough frequencies and as a dielectric at high enough ones. The two terms are equal at the crossover frequency

f_{c} = \frac{σ}{2 π ε}, (5.2.11)

above which the displacement current dominates. Table 5.2.1 and Figure 5.2.4 give three examples, computed with $ε_{0} = 1 0^{- 9} / (36 π) F/m$ .

Table 5.2.1. Three media and their crossover frequencies $f_{c} = σ / (2 π ε_{r} ε_{0})$ .

Medium	$σ$	$ε_{r}$	$f_{c}$ (where $σ = ω ε$ )
copper	$5.8 \times 1 0^{7} S/m$	$1$	$1.0 \times 1 0^{18} Hz$
sea water	$4 S/m$	$81$	$0.89 GHz$
wet soil	$1 0^{- 2} S/m$	$30$	$6.0 MHz$

The loss tangent \sigma/(\omega\varepsilon) of copper, sea water and wet soil against frequency. Each line crosses the dashed line \tan\delta=1 at the crossover frequency of Table 5.2.1. Above the line the medium is a conductor and the conduction current dominates, below it the medium is a dielectric and the displacement current dominates. — **Figure 5.2.4.** The loss tangent $σ / (ω ε)$ of copper, sea water and wet soil against frequency. Each line crosses the dashed line $tan δ = 1$ at the crossover frequency of Table 5.2.1. Above the line the medium is a conductor and the conduction current dominates, below it the medium is a dielectric and the displacement current dominates.

Drag the frequency marker and read off where each medium changes from conductor to dielectric.

InteractiveLoading simulation…

Reading the graph

Copper is a conductor right through the visible band, which is why the displacement current may be dropped inside metals at every frequency an electrical engineer meets. Sea water stops being a conductor near $1 GHz$ . This is why communication with submarines uses very low frequencies (VLF, $3$ to $30 kHz$ ), where the water is a good conductor and the depth to which a wave penetrates is metres rather than millimetres, while a microwave signal is absorbed within a few centimetres. Exercise 5.2.2 works out damp soil, which changes character between $1 MHz$ and $1 GHz$ .

Quick check

Is sea water ( $σ = 4 S/m$ , $ε_{r} = 81$ ) a conductor or a dielectric at $1 kHz$ ?

$σ / (ω ε) = f_{c} / f = 0.89 \times 1 0^{9} /1 0^{3} \approx 9 \times 1 0^{5}$ : the conduction current exceeds the displacement current by almost six orders of magnitude, and sea water is a conductor. (The same ratio at $10 GHz$ is $0.09$ , a dielectric with losses.)