Irodov 3.347: Leak and Displacement Cancel Exactly – Amit & Deepali Sinha Foundation

Problem Irodov 3.347

Two concentric metal spheres have the space between them filled with a uniform, weakly conducting medium of resistivity $\rho$ and permittivity $\varepsilon$ . At the instant $t = 0$ the inner sphere is given some charge, which then leaks through the medium. Find:

how the displacement current density and the conduction current density are related at any point of the medium at the same instant;
the displacement current through any closed surface that lies wholly in the medium and encloses the inner sphere, at the instant when the inner sphere’s charge is $q$ .

The two concentric spheres in section, with the poorly conducting medium of permittivity ε and resistivity ρ between them; the inner sphere carries the charge q. — The two concentric spheres in section, with the poorly conducting medium of permittivity $\varepsilon$ and resistivity $\rho$ between them; the inner sphere carries the charge $q$ .

Given

medium: resistivity $\rho$ , permittivity $\varepsilon$ , uniform
inner sphere’s charge $q$ at the instant considered
no free charge inside the medium (assumed)

Find

(a) the relation between

\vect j_d

and

\vect j

; (b) the displacement current

I_d

through a closed surface enclosing the inner sphere

Solution

Step 1The field at any instant

On a concentric sphere of radius $r$ in the medium, Gauss’s law for $\vect D$ gives, with $\vect e_r$ the outward unit vector and $q$ the inner sphere’s charge at that instant,

\vect D = \frac{q}{4\pi r^2}\,\vect e_r \tag{1}

Step 2The conduction current drains the inner sphere

In the medium Ohm’s law gives the conduction current density from the field $\vect E = \vect D/(\varepsilon_0\varepsilon)$ . Using (1):

\vect j = \frac{\vect E}{\rho} = \frac{q}{4\pi\varepsilon_0\varepsilon\rho\, r^2}\,\vect e_r \tag{2}

This current carries charge outward through every concentric sphere, so the inner sphere’s charge falls at the rate of the total current through the sphere of radius $r$ :

\frac{dq}{dt} = -4\pi r^2 j \tag{3}

The radial conduction current in the medium and a concentric sphere of radius r. At a point of it the conduction current density j points outward and the displacement current density j_d inward. — The radial conduction current in the medium and a concentric sphere of radius $r$ . At a point of it the conduction current density $\vect j$ points outward and the displacement current density $\vect j_d$ inward.

Step 3(a) The two current densities at a point

The displacement current density is the rate of change of $\vect D$ at a fixed point. Differentiating (1) with $r$ held fixed, then using (3):

\vect j_d = \frac{\partial\vect D}{\partial t} = \frac{1}{4\pi r^2}\,\frac{dq}{dt}\,\vect e_r = \frac{-4\pi r^2 j}{4\pi r^2}\,\vect e_r

\vect j_d = -\vect j

(4)

At every point of the medium, at every instant, the displacement current is as large as the conduction current and opposite to it.

Step 4(b) The displacement current through a closed surface

By (4) the total current density $\vect j + \vect j_d$ is zero everywhere in the medium, so its flux through any closed surface lying in the medium vanishes, whatever the surface’s shape:

I_d = \oint \vect j_d\cdot d\vect S = -\oint \vect j\cdot d\vect S

The conduction current through a surface enclosing the inner sphere is the same through every such surface; through the concentric sphere it is $4\pi r^2 j$ , which by (2) is

\oint \vect j\cdot d\vect S = 4\pi r^2\cdot\frac{q}{4\pi\varepsilon_0\varepsilon\rho\, r^2}

With the outward normal, the displacement current is minus this:

I_d = -\oint \vect j\cdot d\vect S = -\frac{q}{\varepsilon_0\varepsilon\rho}

|I_d| = \frac{q}{\varepsilon_0\varepsilon\rho}

(5)

The displacement current flows inward through the surface, against the conduction current, for $q > 0$ ; for $q < 0$ both reverse.

Check

The charge’s history agrees. Putting $4\pi r^2 j = q/(\varepsilon_0\varepsilon\rho)$ into (3) gives $dq/dt = -q/(\varepsilon_0\varepsilon\rho)$ , so $q = q_0\,e^{-t/\varepsilon_0\varepsilon\rho}$ : the charge relaxes with the time $\varepsilon_0\varepsilon\rho$ , which has the unit $(\text{F/m})\cdot\Omega\cdot\text{m} = \text{s}$ . Then $|I_d| = |dq/dt|$ , the rate at which the enclosed charge changes, exactly as Gauss’s law requires of the flux of $\partial\vect D/\partial t$ .

Remark

Because the total current $\vect j + \vect j_d$ vanishes everywhere, this leaking capacitor has no magnetic field at all, even though a current flows: by symmetry a radial current could only make a radial $\vect H$ , and $\nabla\cdot\vect B = 0$ forbids that. Without the displacement current the circulation law would demand an $\vect H$ round any loop threaded by the radial current; $\vect j_d = -\vect j$ is exactly what reconciles it with the symmetry.