Irodov 3.348: A Leak That Makes No Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.348

A parallel-plate capacitor has two disc-shaped plates, and the space between them is filled with a uniform, weakly conducting medium. The capacitor is charged and then disconnected from the source, so it discharges through the medium. Ignoring edge effects, show that there is no magnetic field between the plates.

The disc capacitor edge on, its gap filled with the uniform poorly conducting medium, of permittivity ε and resistivity ρ; it is disconnected from any source. — The disc capacitor edge on, its gap filled with the uniform poorly conducting medium, of permittivity $\varepsilon$ and resistivity $\rho$ ; it is disconnected from any source.

Given

disc plates
uniform poorly conducting medium in the gap
capacitor isolated after charging
edge effects neglected

Find

show that

\vect H = 0

between the plates

Solution

Step 1The conduction current between the plates

Let the plates have area $S$ and carry the charges $\pm q$ at some instant, and let the medium have permittivity $\varepsilon$ and resistivity $\rho$ . With the edge effects neglected the field between the plates is uniform and normal to them, with $\varepsilon_0\varepsilon E = q/S$ by Gauss’s law. Ohm’s law then gives the conduction current density, running from the positive plate to the negative one:

j = \frac{E}{\rho} = \frac{q}{\varepsilon_0\varepsilon\rho\, S} \tag{1}

The capacitor is disconnected, so this current is the only way charge leaves the positive plate, and it does so at the rate

\frac{dq}{dt} = -jS \tag{2}

Step 2The displacement current cancels it

The displacement current density is the rate of change of the displacement $D$ , which by Gauss’s law equals $q/S$ , directed along the field when $D$ grows. Differentiating and using (2):

j_d = \frac{\partial D}{\partial t} = \frac{1}{S}\,\frac{dq}{dt} = \frac{-jS}{S}

Here $j_d$ is measured along the field, the direction of $\vect j$ , so as vectors

\vect j + \vect j_d = 0 \tag{3}

at every point of the gap: the field weakens exactly as fast as the leak drains the plates.

The disconnected capacitor discharging through its medium: conduction current j from the left plate (taken positive), displacement current j_d opposite to it, and a coaxial circle of radius r in the gap. — The disconnected capacitor discharging through its medium: conduction current $\vect j$ from the left plate (taken positive), displacement current $\vect j_d$ opposite to it, and a coaxial circle of radius $r$ in the gap.

Step 3The circulation of H is zero round every coaxial circle

Take a circle of radius $r$ about the plates’ axis, in a plane between the plates and parallel to them, and the flat disc it bounds as the spanning surface. The circulation theorem for $\vect H$ counts the conduction and displacement currents through that disc, and by (3) their sum is zero at every point of it:

\oint \vect H\cdot d\vect l = \int (\vect j + \vect j_d)\cdot d\vect S = 0

With $H_\varphi$ the same all round the circle, the circulation is $2\pi r H_\varphi$ , which must vanish for every $r$ :

\vect H = 0\ \text{between the plates}

(4)

Check

Compare a capacitor charged through leads by a current $I$ : there $\vect j + \vect j_d$ through the gap is $I/S$ , the continuation of the lead current, and the same circle gives $H = Ir/(2\pi R^2)$ inside plates of radius $R$ , not zero. Here the leads are gone and the total current through any surface is zero. As a limit, a perfect insulator ( $\rho \to \infty$ ) makes $j$ and $j_d$ both vanish by (1), and a static charged capacitor has no magnetic field either, as (4) says.

Remark

A current flowing with no magnetic field looks paradoxical only if the displacement current is forgotten: the circulation law with conduction current alone would demand $H = jr/2$ round the axis. Maxwell’s term is what removes it, and the cancellation (3) is nothing more than the conservation of charge read at each point.