Irodov 3.158: A Ball, a Plane and Its Mirror – Amit & Deepali Sinha Foundation

Problem Irodov 3.158

A metal ball of radius $a$ sits at a distance $l$ from an infinite, perfectly conducting plane, with $a \ll l$ . The space around the ball is filled with a uniform, weakly conducting medium of resistivity $\rho$ .

A potential difference $V$ is kept between the ball and the plane. Find the current density at the plane as a function of the distance $r$ from the ball.
Find the resistance of the medium between the ball and the plane.

The metal ball of radius a, its center a distance l from the ideally conducting plane, in the medium of resistivity ρ (drawn schematically: the problem has a l). — The metal ball of radius $a$ , its center a distance $l$ from the ideally conducting plane, in the medium of resistivity $\rho$ (drawn schematically: the problem has $a \ll l$ ).

Given

ball radius $a$
distance $l \gg a$
resistivity $\rho$
voltage $V$

Find

(a)

j(r)

at the plane, (b) the resistance

R

Solution

Step 1The plane is replaced by a mirror source

Let the ball feed a current $I$ into the medium. In the steady state $\nabla\cdot\vect j = 0$ and $\vect j = \vect E/\rho$ with uniform $\rho$ , so the potential obeys Laplace’s equation in the medium, exactly as in electrostatics. A point source of current $I$ in an unbounded medium gives, by symmetry, a radial current spread evenly over each sphere of radius $r$ about it:

\begin{gathered} j\cdot 4\pi r^2 = I\\ \varphi = \frac{\rho I}{4\pi r} \end{gathered} \tag{1}

The ideally conducting plane is an equipotential, which we call zero. A source $I$ at the ball’s center and a sink $-I$ at its mirror point, a distance $l$ behind the plane, make the whole plane $\varphi = 0$ , so in the medium this pair is the field.

Step 2The current fed by the voltage

V

On the ball the image is a distance $2l$ away. Its potential, the ball’s own term at $r = a$ plus the image’s at $2l$ , is the voltage $V$ against the plane:

V = \frac{\rho I}{4\pi}\left(\frac{1}{a} - \frac{1}{2l}\right) \approx \frac{\rho I}{4\pi a}

The image term is smaller by the factor $a/2l$ and is dropped for $a \ll l$ . Solving for the current:

I = \frac{4\pi a V}{\rho} \tag{2}

Step 3The current density at the plane

Current lines in the medium. They leave the ball and end on the plane at right angles; at a point P of the plane, a distance r from the ball's center, j is along the normal (for V > 0). — Current lines in the medium. They leave the ball and end on the plane at right angles; at a point $P$ of the plane, a distance $r$ from the ball’s center, $\vect j$ is along the normal (for $V > 0$ ).

A point $P$ of the plane is the same distance $r$ from the source and from the image. Each contributes a current density of magnitude $I/(4\pi r^2)$ by (1), the source’s pointing away from the ball and the image’s toward the mirror point. Along the plane they cancel; along the normal each has the fraction $l/r$ of its magnitude, and they add:

j = 2\cdot\frac{I}{4\pi r^2}\cdot\frac{l}{r} = \frac{I\,l}{2\pi r^3}

Substituting the current (2):

j = \frac{2 a l V}{\rho\, r^3}

(3)

The current enters the plane along its normal, falling off as $1/r^3$ with the distance from the ball.

Step 4The resistance

The resistance is the voltage over the current, from (2):

R = \frac{V}{I} = \frac{\rho}{4\pi a}

(4)

Check

The current density integrates to the whole current. With $r^2 = l^2 + s^2$ for a point a distance $s$ from the foot of the perpendicular, the current through the plane is

\begin{aligned} \int_0^{\infty}\frac{2alV}{\rho\,(l^2 + s^2)^{3/2}}\,2\pi s\,ds &= \frac{4\pi a l V}{\rho}\left[-\frac{1}{(l^2 + s^2)^{1/2}}\right]_0^{\infty}\\ &= \frac{4\pi a l V}{\rho}\cdot\frac{1}{l} = \frac{4\pi a V}{\rho} \end{aligned}

which is $I$ in (2). Dimensions: $alV/(\rho r^3)$ is V/( $\Omega\cdot$ m $^2$ ), that is A/m $^2$ .

Remark

The resistance is half that of two distant balls, $\rho/(2\pi a)$ : the plane is a perfect electrode, so only one spreading resistance, the one around the ball, remains. To this order it does not depend on $l$ , but the distribution over the plane does, and its width is set by $l$ , not by $a$ .