Irodov 3.155: The Resistance Lives Near the Ball – Amit & Deepali Sinha Foundation

Problem Irodov 3.155

A metal ball of radius $a$ sits at the centre of a thin concentric metal shell of radius $b$ . The space between the two electrodes is filled with a uniform, poorly conducting medium of resistivity $\rho$ . Find the resistance of the gap between the electrodes, and discuss the result in the limit $b \to \infty$ .

The ball of radius a and the concentric shell of radius b, in a section through the centre, with the medium of resistivity ρ between them. — The ball of radius $a$ and the concentric shell of radius $b$ , in a section through the centre, with the medium of resistivity $\rho$ between them.

Given

radii $a < b$
resistivity $\rho$
metal electrodes

Find

the resistance

R

of the gap; its limit as

b \to \infty

Solution

Step 1The current density at radius

r

Let a steady current $I$ flow from the ball to the shell. Both electrodes are metal, so each is an equipotential, and by symmetry the current in the medium is radial and the same in every direction. All of $I$ crosses every concentric sphere of radius $r$ between the electrodes:

j\cdot 4\pi r^2 = I

By the local form of Ohm’s law the field in the medium is $\rho$ times the current density:

E(r) = \rho j = \frac{\rho I}{4\pi r^2} \tag{1}

The steady current in the medium with the ball at the higher potential: radial lines of j in a section through the centre. — The steady current in the medium with the ball at the higher potential: radial lines of $\vect{j}$ in a section through the centre.

Step 2The voltage and the resistance

The voltage between the electrodes is the line integral of (1) along a radius:

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) = \frac{\rho I}{4\pi}\,\frac{b - a}{ab}

The current $I$ cancels in $V/I$ :

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

(2)

Step 3The limit of a distant shell

Written as a difference of two terms, (2) keeps $b$ only in the second:

R = \frac{\rho}{4\pi}\left(\frac{1}{a} - \frac{1}{b}\right)

As $b \to \infty$ that term vanishes and the resistance stays finite:

R_\infty = \frac{\rho}{4\pi a}

(3)

Pushing the outer electrode to infinity adds an unlimited amount of medium, yet the resistance stays bounded. The current spreads over spheres whose area grows as $r^2$ , so the layer between $r$ and $r + dr$ adds only $\rho\,dr/4\pi r^2$ , and these contributions die off fast enough to sum to a finite value. Half of it lies within $r = 2a$ of the centre and $90$ percent within $r = 10a$ .

Check

Dimensions: $\rho/a$ is ohm metres over metres, a resistance. Thin gap: with $b = a + \delta$ , $\delta \ll a$ , (2) gives $\rho\,\delta/(4\pi a^2)$ , which is $\rho \times$ thickness $/$ area for a layer of area $4\pi a^2$ . And (2) grows steadily with $b$ toward (3) from below, as adding medium in series must.

Remark

This is the physics of an earthing electrode. Buried deep in soil of resistivity $\rho$ , a metal sphere of radius $a$ has the spreading resistance $\rho/4\pi a$ to the whole Earth, set by the electrode and not by the distance to wherever the current returns. A larger electrode, or soil made more conductive right around it, is what lowers the earthing resistance.