Irodov 3.111: A Ball Barely Notices the Plane – Amit & Deepali Sinha Foundation

Problem Irodov 3.111

A metal ball of radius $a$ has its center a distance $l$ from an infinite conducting plane, with $l \gg a$ . Find the capacitance of the system formed by the ball and the plane.

The metal ball of radius a with its center a distance l from the conducting plane (drawn schematically: the problem has l a). — The metal ball of radius $a$ with its center a distance $l$ from the conducting plane (drawn schematically: the problem has $l \gg a$ ).

Given

ball radius $a$
center-to-plane distance $l$
$l \gg a$

Find

the capacitance

C

of ball and plane

Solution

Step 1Replace the plane by an image

Put the charge $q$ on the ball and $-q$ on the plane. Since $l \gg a$ , the field of the plane’s induced charge is weak and nearly uniform across the ball and shifts its charge only slightly, so the charge stays almost uniform over the ball and acts outside it like a point charge $q$ at its center. The field in the half-space of the ball is then that of $q$ and its mirror image $-q$ a distance $l$ behind the plane, which makes the plane an equipotential at zero.

The ball replaced by its charge q at the center, and the plane by the image -q at the mirror point a distance l below the surface; the lines that reach the plane meet it at right angles, and those leaving upward return to it outside the frame. — The ball replaced by its charge $q$ at the center, and the plane by the image $-q$ at the mirror point a distance $l$ below the surface; the lines that reach the plane meet it at right angles, and those leaving upward return to it outside the frame.

Step 2The voltage between ball and plane

The plane is at potential zero. The ball’s potential is the potential of its own uniform charge, $q/(4\pi\varepsilon_0 a)$ , plus that of the image, which is $2l$ from the ball’s center. Averaged over the ball’s surface the image contributes exactly its value at the center, so

V = \frac{q}{4\pi\varepsilon_0 a} - \frac{q}{4\pi\varepsilon_0\cdot 2l} = \frac{q}{4\pi\varepsilon_0 a}\left(1 - \frac{a}{2l}\right)

Step 3The capacitance

The capacitance is $C = q/V$ , with $q$ cancelling:

C = \frac{4\pi\varepsilon_0 a}{1 - a/2l}

For $l \gg a$ the correction $a/2l$ is negligible:

C \approx 4\pi\varepsilon_0 a

(1)

To first order in the small ratio the plane raises the capacitance slightly, $C \approx 4\pi\varepsilon_0 a\,(1 + a/2l)$ .

Check

Limit: as $l \to \infty$ the plane disappears to infinity, the image’s term vanishes, and $C$ becomes $4\pi\varepsilon_0 a$ , the capacitance of an isolated ball, whose partner conductor is the one at infinity. Sign of the correction: the image lowers the ball’s potential for a given charge, so the capacitance must rise as the ball approaches the plane, as the factor $1/(1 - a/2l)$ does.

Remark

For a small conductor far from everything else, “capacitance” and “capacitance to the ground” are the same number: the surroundings enter only at relative order $a/l$ . That is why an isolated sphere’s $4\pi\varepsilon_0 a$ , about $0.7$ mF for the Earth, is a meaningful quantity at all, even though no conductor is truly alone.