Irodov 3.161: One Flux Gives Charge and Current – Amit & Deepali Sinha Foundation

Problem Irodov 3.161

Two conductors of any shape are embedded in an infinite, uniform, weakly conducting medium of resistivity $\rho$ and permittivity $\varepsilon$ . Let $R$ be the resistance of the medium between the conductors and $C$ their mutual capacitance with the medium present. Find the product $RC$ .

Two conductors 1 and 2 of arbitrary shape in the medium of resistivity ρ and permittivity ε. — Two conductors $1$ and $2$ of arbitrary shape in the medium of resistivity $\rho$ and permittivity $\varepsilon$ .

Given

resistivity $\rho$
permittivity $\varepsilon$
conductors of any shape
medium infinite and uniform

Find

the product

RC

Solution

Step 1The same field serves both questions

Keep a voltage $V$ between the conductors. In the steady state $\nabla\cdot\vect j = 0$ , and with $\vect j = \vect E/\rho$ and $\rho$ uniform the potential obeys Laplace’s equation in the medium, with each conductor an equipotential. In the purely electrostatic problem of the same two conductors at the same voltage in the dielectric $\varepsilon$ , the potential obeys the same equation with the same boundary values. By uniqueness the two fields are one and the same field $\vect E$ .

Step 2Charge and current from one surface integral

A closed surface S wrapping conductor 1, lying wholly in the medium. The flux of E through it fixes both the charge on the conductor and the current it sends out. — A closed surface $S$ wrapping conductor $1$ , lying wholly in the medium. The flux of $\vect E$ through it fixes both the charge on the conductor and the current it sends out.

Wrap conductor 1 in a closed surface $S$ lying wholly in the medium. Gauss’s theorem for $\vect D = \varepsilon_0\varepsilon\vect E$ gives the free charge $q$ on conductor 1, and the outward flux of $\vect j = \vect E/\rho$ gives the current $I$ that leaves it:

\begin{gathered} q = \oint_S \vect D\cdot d\vect S = \varepsilon_0\varepsilon\oint_S \vect E\cdot d\vect S \\ I = \oint_S \vect j\cdot d\vect S = \frac{1}{\rho}\oint_S \vect E\cdot d\vect S \end{gathered}

Dividing one by the other, the unknown flux of $\vect E$ , which holds all the geometry, cancels:

\frac{q}{I} = \varepsilon_0\varepsilon\rho \tag{1}

Step 3The product

The capacitance is $C = q/V$ and the resistance is $R = V/I$ , so their product is $q/I$ , given by (1):

RC = \frac{V}{I}\cdot\frac{q}{V} = \varepsilon_0\varepsilon\rho

(2)

Check

Two parallel plates of area $A$ and gap $d$ filled with the medium have

\begin{gathered} R = \frac{\rho d}{A}\\ C = \frac{\varepsilon_0\varepsilon A}{d} \end{gathered}

whose product is $\varepsilon_0\varepsilon\rho$ . Dimensions: (F/m)( $\Omega\cdot$ m) is F $\cdot\Omega$ , a time, as $RC$ is.

Remark

The product is the relaxation time of the medium: a charge placed inside a uniform conductor decays as $e^{-t/\varepsilon_0\varepsilon\rho}$ wherever it sits. So a resistance between electrodes of any shape can be found from their capacitance alone, which is how earthing resistances and leakage through cable insulation are estimated.