Irodov 3.156: A Leaky Capacitor Times Itself – Amit & Deepali Sinha Foundation

Problem Irodov 3.156

Two conducting concentric spheres of radii $a$ and $b$ , with $a < b$ , have the space between them filled with a uniform, poorly conducting medium; the capacitance of the system is $C$ . The spheres are charged and then disconnected from the voltage source, after which the potential difference between them falls by a factor $\eta$ in a time $\Delta t$ . Find the resistivity of the medium.

The two concentric spheres of radii a and b, in a section through the centre, with the poorly conducting medium between them. — The two concentric spheres of radii $a$ and $b$ , in a section through the centre, with the poorly conducting medium between them.

Given

radii $a < b$
capacitance $C$
voltage falls $\eta$ -fold in $\Delta t$
spheres disconnected

Find

the resistivity

\rho

of the medium

Solution

Step 1The resistance of the medium

Let a current $I$ cross the medium from the inner sphere to the outer one. The spheres are equipotentials, so by symmetry the current is radial and the same in every direction, and all of $I$ crosses each concentric sphere of radius $r$ :

j\cdot 4\pi r^2 = I

By the local form of Ohm’s law $E = \rho j$ , and the voltage is its integral across the gap:

V = \int_a^b \frac{\rho I}{4\pi r^2}\,dr = \frac{\rho I}{4\pi}\left(\frac{1}{a} - \frac{1}{b}\right)

So the medium is a resistor between the spheres:

R = \frac{V}{I} = \frac{\rho\,(b - a)}{4\pi ab} \tag{1}

Step 2The discharge through the medium

Once the spheres are disconnected, the only way the charge $q = CV$ of the inner sphere can leave is through the medium, as the current $I = V/R$ . The charge lost per unit time is that current:

The discharge with the inner sphere positive: the current carries its charge q radially through the medium to the outer sphere. — The discharge with the inner sphere positive: the current carries its charge $q$ radially through the medium to the outer sphere.

-C\,\frac{dV}{dt} = \frac{V}{R}

Separating the variables and integrating from the start, where the voltage is $V_0$ , gives an exponential decay with the time constant $RC$ :

\int_{V_0}^{V}\frac{dV'}{V'} = -\frac{1}{RC}\int_0^t dt'

V = V_0\,e^{-t/RC} \tag{2}

Step 3The time constant from the measured fall

After $\Delta t$ the voltage is $V_0/\eta$ . Putting this into (2) and taking logarithms:

\begin{aligned} \frac{1}{\eta} &= e^{-\Delta t/RC}\\ RC &= \frac{\Delta t}{\ln\eta} \end{aligned}

Substituting $R$ from (1) and solving for $\rho$ :

\frac{\rho\,(b - a)}{4\pi ab}\,C = \frac{\Delta t}{\ln\eta}

\rho = \frac{4\pi ab\,\Delta t}{(b - a)\,C\ln\eta}

(3)

Check

Units: $ab/(b - a)$ is a length, and a farad is a second per ohm, so $\Delta t/C$ is in ohms and the whole is ohm metres, a resistivity. Limits: if the voltage does not fall at all, $\eta \to 1$ , then $\ln\eta \to 0$ and $\rho \to \infty$ , a perfect insulator; a faster fall (larger $\eta$ in the same $\Delta t$ ) means a smaller $\rho$ , as it should.

Remark

If the medium has permittivity $\varepsilon$ , the capacitance of the spheres is $C = 4\pi\varepsilon_0\varepsilon ab/(b - a)$ , the reciprocal of the geometric factor that multiplies $\rho$ in (1), so the time constant $RC = \varepsilon_0\varepsilon\rho$ does not depend on the electrodes at all. It is the time constant with which a free charge placed inside any uniform conductor drains away to its surfaces; the given $C$ in (3) is a way of supplying $\varepsilon$ without naming it.