Irodov 3.157: Two Spreading Resistances in Series – Amit & Deepali Sinha Foundation

Problem Irodov 3.157

Two identical metal balls, each of radius $a$ , are embedded in a uniform, weakly conducting medium of resistivity $\rho$ . The balls are very far apart compared with their radius. What is the resistance of the medium between the two balls?

The two metal balls of radius a in the weakly conducting medium of resistivity ρ (drawn schematically: the separation is much larger than a). — The two metal balls of radius $a$ in the weakly conducting medium of resistivity $\rho$ (drawn schematically: the separation is much larger than $a$ ).

Given

ball radius $a$
resistivity $\rho$
separation $\gg a$
medium fills all space (assumed)

Find

the resistance

R

between the balls

Solution

Step 1The steady current is an electrostatics problem

Let a current $I$ enter the medium through ball 1 and leave it through ball 2, and take the medium to fill all space. In a steady state the charge in any volume of the medium is constant, so $\nabla\cdot\vect j = 0$ , and the field is still conservative. With Ohm’s law in local form, $\vect j = \vect E/\rho$ and $\rho$ uniform, the potential obeys Laplace’s equation in the medium:

\nabla\cdot\vect j = \frac{1}{\rho}\,\nabla\cdot\vect E = -\frac{1}{\rho}\,\nabla^2\varphi = 0

The metal balls are equipotentials, so this is the same boundary problem as two charged balls in vacuum, and solutions superpose.

Step 2The potential of a ball that injects a current

I

Alone in the medium, a ball feeding the current $I$ spreads it evenly in all directions, so by symmetry $\vect j$ is radial with the same magnitude over each concentric sphere of radius $r$ :

j\cdot 4\pi r^2 = I

The field is $E = \rho j$ , and integrating it in from infinity gives the potential at distance $r$ :

\varphi(r) = \int_r^{\infty}\frac{\rho I}{4\pi r'^2}\,dr' = \frac{\rho I}{4\pi r} \tag{1}

Step 3The voltage between the balls

Current lines in the medium, from the ball the current enters by (+) to the ball it leaves by (-), with equipotentials; the centers are a distance l apart. Close to either ball the lines are radial. — Current lines in the medium, from the ball the current enters by ( $+$ ) to the ball it leaves by ( $-$ ), with equipotentials; the centers are a distance $l$ apart. Close to either ball the lines are radial.

Ball 2 withdraws the current $I$ , a source of strength $-I$ . Call the distance between the centers $l$ . On ball 1 the potential is its own term (1) at $r = a$ plus the term of ball 2 at $r = l$ , and the same holds for ball 2 with the signs reversed:

\begin{gathered} \varphi_1 = \frac{\rho I}{4\pi}\left(\frac{1}{a} - \frac{1}{l}\right) \\ \varphi_2 = -\frac{\rho I}{4\pi}\left(\frac{1}{a} - \frac{1}{l}\right) \end{gathered}

Their difference is the voltage, and since $l \gg a$ the $1/l$ terms drop out:

V = \varphi_1 - \varphi_2 = \frac{\rho I}{2\pi}\left(\frac{1}{a} - \frac{1}{l}\right) \approx \frac{\rho I}{2\pi a}

Step 4The resistance

The resistance is $R = V/I$ , and $I$ cancels:

R = \frac{\rho}{2\pi a}

(2)

Check

Dimensions: resistivity is $\Omega\cdot$ m, so $\rho/a$ is in ohms. Each ball alone contributes the spreading resistance $\rho/(4\pi a)$ , the resistance of a spherical shell from $r = a$ to infinity,

\int_a^{\infty}\frac{\rho\,dr}{4\pi r^2} = \frac{\rho}{4\pi a}

and two of them in series give $\rho/(2\pi a)$ , the boxed result.

Remark

The distance between the balls drops out: almost all of the resistance sits within a few radii of each ball, where the current is crowded into a small cross-section, while the wide region between them, however long, adds almost nothing. That is why an earthing electrode is judged by its size and the soil next to it, not by how far away the other electrode is.