Irodov 3.203: A Leak Set by the Medium Alone – Amit & Deepali Sinha Foundation

Problem Irodov 3.203

A spherical capacitor has electrodes of radii $a$ and $b$ , with $a < b$ , and the space between them is filled with a uniform material of permittivity $\varepsilon$ and resistivity $\rho$ . The capacitor starts uncharged. At the moment $t = 0$ the inner electrode is given a charge $q_0$ . Find:

how the charge on the inner electrode varies with time
the heat released while the charge spreads

The spherical capacitor in section: the inner electrode of radius a with the charge q_0, the medium of permittivity ε and resistivity ρ, and the outer electrode of radius b. — The spherical capacitor in section: the inner electrode of radius $a$ with the charge $q_0$ , the medium of permittivity $\varepsilon$ and resistivity $\rho$ , and the outer electrode of radius $b$ .

Given

radii $a < b$
permittivity $\varepsilon$
resistivity $\rho$
charge $q_0$ on the inner electrode at $t = 0$

Find

(a) the charge

q(t)

on the inner electrode, (b) the heat

Q

released

Solution

Step 1The current that leaks through the medium

Let the inner electrode hold the charge $q$ at some moment. By symmetry the field in the medium is radial and the same all round, so Gauss’s theorem for $\vect D$ over a sphere of radius $r$ between the electrodes gives

\varepsilon_0\varepsilon E\cdot 4\pi r^2 = q

The medium obeys Ohm’s law in its local form, so the current density is $E/\rho$ , outward, and the total current through the sphere is

\begin{aligned} I &= \frac{E}{\rho}\cdot 4\pi r^2\\ &= \frac{q}{4\pi\varepsilon_0\varepsilon r^2\rho}\cdot 4\pi r^2 \end{aligned}

I = \frac{q}{\varepsilon_0\varepsilon\rho}

(1)

The leakage current: radial streamlines through the medium from the inner electrode to the outer one, drawn for q_0 > 0. — The leakage current: radial streamlines through the medium from the inner electrode to the outer one, drawn for $q_0 > 0$ .

Step 2(a) The charge decays exponentially

The current (1) is the rate at which the inner electrode loses charge:

\begin{aligned} \frac{dq}{dt} &= -\frac{q}{\varepsilon_0\varepsilon\rho}\\ \ln\frac{q}{q_0} &= -\frac{t}{\varepsilon_0\varepsilon\rho} \end{aligned}

q(t) = q_0\,e^{-t/\varepsilon_0\varepsilon\rho}

(2)

The relaxation time $\varepsilon_0\varepsilon\rho$ is a property of the material alone: the radii $a$ and $b$ do not enter.

Step 3(b) The heat, from the Joule power

The voltage between the electrodes is the integral of the field from $a$ to $b$ :

U = \int_a^b \frac{q\,dr}{4\pi\varepsilon_0\varepsilon r^2} = \frac{q}{4\pi\varepsilon_0\varepsilon}\left(\frac{1}{a} - \frac{1}{b}\right) \tag{3}

The Joule power released in the medium is the current times this voltage, with (1) and (3):

P = IU = \frac{q^2}{4\pi\varepsilon_0^2\varepsilon^2\rho}\left(\frac{1}{a} - \frac{1}{b}\right)

Integrating over the whole spreading with (2):

\begin{aligned} Q &= \frac{1}{4\pi\varepsilon_0^2\varepsilon^2\rho}\left(\frac{1}{a} - \frac{1}{b}\right)\int_0^\infty q_0^2\,e^{-2t/\varepsilon_0\varepsilon\rho}\,dt\\ &= \frac{1}{4\pi\varepsilon_0^2\varepsilon^2\rho}\left(\frac{1}{a} - \frac{1}{b}\right)\frac{q_0^2\,\varepsilon_0\varepsilon\rho}{2} \end{aligned}

Q = \frac{q_0^2}{8\pi\varepsilon_0\varepsilon}\left(\frac{1}{a} - \frac{1}{b}\right)

(4)

Check

Energy balance. At $t = 0$ the field in the gap stores $q_0^2/2C$ , with the capacitance $C = 4\pi\varepsilon_0\varepsilon ab/(b - a)$ of the spherical capacitor, and this equals (4). At the end the gap is field-free. Outside, the outer electrode’s outer face carries $q_0$ all the time (its total charge and the charge on its inner face change together), so the field beyond $b$ never changes: the heat is exactly the energy that was in the gap. Dimensions: $\varepsilon_0\rho$ is (F/m)( $\Omega$ m), that is F $\,\Omega$ , a time.

Remark

The relaxation time $\varepsilon_0\varepsilon\rho$ is the same for any shape of electrodes in a uniform medium, because $\vect j = \vect E/\rho$ and $\vect D = \varepsilon_0\varepsilon\vect E$ are proportional everywhere: the current leaving any conductor is its flux of $\vect D$ divided by $\varepsilon_0\varepsilon\rho$ , that is its charge divided by the same time. Any leaky capacitor therefore has $RC = \varepsilon_0\varepsilon\rho$ . For a metal this time is about $10^{-19}$ s, which is why a conductor carries no volume charge in practice.