Irodov 3.171: The Filler Sets the Leak Time – Amit & Deepali Sinha Foundation

Problem Irodov 3.171

A capacitor is filled with a dielectric of permittivity $\varepsilon = 2.1$ . Once charged, it loses half of its charge in a time $\tau = 3.0$ min. Assuming that the charge leaks only through the dielectric filling, find the resistivity of the dielectric.

A capacitor filled with the dielectric of permittivity ε and resistivity ρ, drawn as a parallel-plate capacitor; the plates may have any shape. — A capacitor filled with the dielectric of permittivity $\varepsilon$ and resistivity $\rho$ , drawn as a parallel-plate capacitor; the plates may have any shape.

Given

$\varepsilon = 2.1$
half the charge lost in $\tau = 3.0$ min
leakage only through the filler
uniform filler holding all of the field (assumed)

Find

the resistivity

\rho

of the filler

Solution

Step 1The leakage current is fixed by the charge

Let the positive plate carry the charge $q$ , and take a closed surface $\Sigma$ just around it, inside the filler. The filler is uniform and holds all of the plate’s field, the edge field outside it neglected (assumed). Gauss’s theorem for $\vect D = \varepsilon_0\varepsilon\vect E$ gives

\oint_\Sigma \vect E\cdot d\vect S = \frac{q}{\varepsilon_0\varepsilon}

The leakage current leaving the plate through $\Sigma$ follows from Ohm’s law in local form, $\vect j = \vect E/\rho$ , with the same uniform $\rho$ everywhere on $\Sigma$ :

I = \oint_\Sigma \vect j\cdot d\vect S = \frac{1}{\rho}\oint_\Sigma \vect E\cdot d\vect S = \frac{q}{\varepsilon_0\varepsilon\rho} \tag{1}

Step 2The charge decays exponentially

The capacitor discharging through its filler: the conduction current j runs from the positive plate to the negative one, and between the streamlines the displacement current j_d points the other way. — The capacitor discharging through its filler: the conduction current $\vect j$ runs from the positive plate to the negative one, and between the streamlines the displacement current $\vect j_d$ points the other way.

The leakage current (1) is the rate at which the plate loses charge:

\frac{dq}{dt} = -\frac{q}{\varepsilon_0\varepsilon\rho}

Integrating from the charge $q_0$ at $t = 0$ :

q(t) = q_0\,e^{-t/\varepsilon_0\varepsilon\rho} \tag{2}

Step 3The resistivity from the half-time

Half the charge is left at $t = \tau$ . Setting $q(\tau) = q_0/2$ in (2) and taking the logarithm:

\frac{\tau}{\varepsilon_0\varepsilon\rho} = \ln 2

Numerically, with $\tau = 180$ s:

\rho = \frac{180~\text{s}}{8.85\times 10^{-12}~\text{F/m}\times 2.1\times 0.693}

\rho = \frac{\tau}{\varepsilon_0\varepsilon\ln 2} \approx 1.4\times 10^{13}~\Omega\cdot\text{m}

(3)

Check

Units: $\varepsilon_0$ in F/m times $\Omega\cdot$ m gives $\Omega\cdot$ F, which is seconds, so $\tau/\varepsilon_0$ is indeed a resistivity. A parallel-plate check: with area $S$ and gap $d$ , $C = \varepsilon_0\varepsilon S/d$ and the leakage resistance $\rho d/S$ multiply to $\varepsilon_0\varepsilon\rho$ , the decay time of (2). The value, of order $10^{13}~\Omega\cdot$ m, is that of a good insulating plastic.

Remark

Inside the filler the conduction current and the displacement current cancel exactly: the total current $\vect j + \partial\vect D/\partial t$ vanishes at every point, since (1) and (2) make $j = -\dot D$ . Where the filler occupies every place the field reaches (a spherical capacitor filled up to its outer electrode, say), the leaking capacitor therefore produces no magnetic field at all, however large its leakage current.