Irodov 3.350: Two Currents Out of Step – Amit & Deepali Sinha Foundation

Problem Irodov 3.350

A parallel-plate capacitor has round disc plates a distance $d$ apart, and the gap is filled with a uniform, weakly conducting medium of conductivity $\sigma$ and permittivity $\varepsilon$ . An alternating voltage $V = V_m\cos\omega t$ is applied across it. Ignoring edge effects, find the magnetic field strength between the plates at a distance $r$ from their common axis.

The disc capacitor edge on: the medium of permittivity ε and conductivity σ fills the gap d, and the leads carry the voltage V = V_mω t across the plates. — The disc capacitor edge on: the medium of permittivity $\varepsilon$ and conductivity $\sigma$ fills the gap $d$ , and the leads carry the voltage $V = V_m\cos\omega t$ across the plates.

Given

$\sigma$
$\varepsilon$
gap $d$
$V = V_m\cos\omega t$
distance $r$ from the axis
edge effects neglected

Find

H(r, t)

between the plates

Solution

Step 1The field and the two current densities

With the edge effects neglected the field between the plates is uniform and normal to them, $E = V/d$ . The medium carries a conduction current by Ohm’s law, and the changing field adds a displacement current; both are normal to the plates. Each is measured along the direction the field has when $V > 0$ :

j = \sigma E = \frac{\sigma V_m}{d}\cos\omega t \tag{1}

Differentiating $D = \varepsilon_0\varepsilon E$ in time:

j_d = \varepsilon_0\varepsilon\,\frac{\partial E}{\partial t} = -\frac{\varepsilon_0\varepsilon\omega V_m}{d}\sin\omega t \tag{2}

Step 2The circulation of H round a coaxial circle

The capacitor at an instant chosen with the left plate positive and charging (plate charges not drawn): the conduction current j and the displacement current j_d both run across the gap, and H circulates round the coaxial circle of radius r. — The capacitor at an instant chosen with the left plate positive and charging (plate charges not drawn): the conduction current $\vect j$ and the displacement current $\vect j_d$ both run across the gap, and $\vect H$ circulates round the coaxial circle of radius $r$ .

Take a circle of radius $r$ about the axis, in a plane between the plates, spanned by the flat disc it bounds. Both current densities are uniform over that disc, so the circulation theorem gives

H\cdot 2\pi r = \pi r^2\,(j + j_d)

Dividing by $2\pi r$ and substituting (1) and (2):

H = \frac{r}{2}\,(j + j_d) = \frac{rV_m}{2d}\left(\sigma\cos\omega t - \varepsilon_0\varepsilon\omega\sin\omega t\right) \tag{3}

Here $H$ is the component along the circle taken counterclockwise about the field direction at $V > 0$ (right-hand rule).

Step 3One oscillation with a phase lead

Two oscillations a quarter period apart add into one. Writing $\sigma = \sqrt{\sigma^2 + (\varepsilon_0\varepsilon\omega)^2}\,\cos\alpha$ and $\varepsilon_0\varepsilon\omega = \sqrt{\sigma^2 + (\varepsilon_0\varepsilon\omega)^2}\,\sin\alpha$ , the bracket in (3) is a cosine of a sum:

\sigma\cos\omega t - \varepsilon_0\varepsilon\omega\sin\omega t = \sqrt{\sigma^2 + (\varepsilon_0\varepsilon\omega)^2}\,\cos(\omega t + \alpha)

\begin{gathered} H = H_m\cos(\omega t + \alpha)\\ H_m = \frac{rV_m}{2d}\sqrt{\sigma^2 + (\varepsilon_0\varepsilon\omega)^2}\\ \tan\alpha = \frac{\varepsilon_0\varepsilon\omega}{\sigma} \end{gathered}

(4)

The field grows linearly with $r$ and leads the voltage by the phase $\alpha$ .

Check

Limits. A perfect insulator ( $\sigma = 0$ ) gives $\alpha = 90^\circ$ and $H_m = \varepsilon_0\varepsilon\omega V_m r/(2d)$ , the field of a pure displacement current in an ordinary capacitor. Direct current ( $\omega = 0$ ) gives $\alpha = 0$ and $H = \sigma V r/(2d)$ , the field inside a uniform conductor carrying the current density $\sigma V/d$ , growing as $r$ just like the field inside a straight wire. Units: $\sigma V_m r/d$ is $(\text{S/m})\cdot\text{V}\cdot\text{m}/\text{m} = \text{A/m}$ .

Remark

The ratio $\varepsilon_0\varepsilon\omega/\sigma$ compares the displacement current with the conduction current, and it is $\omega$ times the medium’s charge relaxation time $\varepsilon_0\varepsilon/\sigma$ . A medium counts as a conductor at frequencies well below $\sigma/(\varepsilon_0\varepsilon)$ and as a dielectric well above it; for sea water ( $\sigma \approx 4$ S/m, $\varepsilon \approx 81$ ) the crossover lies near $6\cdot 10^{9}$ s $^{-1}$ .