Irodov 3.129: The Image Sets the Energy – Amit & Deepali Sinha Foundation

Problem Irodov 3.129

A point charge $q$ is at a distance $l$ from an infinite conducting plane. Find the interaction energy of the charge with the charges it induces on the plane.

The point charge q at distance l from the conducting plane. — The point charge $q$ at distance $l$ from the conducting plane.

Given

point charge $q$
distance $l$ from the plane

Find

the interaction energy

W

q

with the induced charge

Solution

Step 1The induced charge replaced by an image

The plane is an equipotential, at zero potential like the far field. Above it, the field of the induced surface charge is the field of an image charge $q' = -q$ at the mirror point, a distance $l$ behind the plane: $q$ and $q'$ together give zero potential everywhere on the plane, and the solution in the region above it is unique.

The lines of q end on the negative charge induced on the plane; the image q' (dashed) inside the metal reproduces the induced charge's field above the plane. — The lines of $q$ end on the negative charge induced on the plane; the image $q'$ (dashed) inside the metal reproduces the induced charge’s field above the plane.

At the position of $q$ , a distance $2l$ from the image, the induced charge therefore produces the potential

\varphi' = \frac{1}{4\pi\varepsilon_0}\frac{q'}{2l} = -\frac{q}{8\pi\varepsilon_0 l} \tag{1}

Step 2The interaction energy

The interaction energy of a charge with a given set of other charges is the charge times the potential those others produce at its position. Substituting (1):

W = q\varphi' = q\left(-\frac{q}{8\pi\varepsilon_0 l}\right)

W = -\frac{q^2}{8\pi\varepsilon_0 l}

(2)

Check

Dimensions: $q^2/(\varepsilon_0 l)$ is an energy. Sign: the induced charge is of the opposite sign and attracts $q$ , so the interaction energy is negative, and it grows in magnitude without bound as $l \to 0$ , as for two unlike charges brought together. Independent route through the surface charge: on the plane at distance $\rho$ from the foot of the perpendicular, $\sigma = -ql/[2\pi(l^2 + \rho^2)^{3/2}]$ , and

\begin{aligned} W &= \int_0^\infty \frac{q\,\sigma\,2\pi\rho\,d\rho}{4\pi\varepsilon_0\sqrt{l^2 + \rho^2}} = -\frac{q^2 l}{4\pi\varepsilon_0}\int_0^\infty\frac{\rho\,d\rho}{(l^2 + \rho^2)^2}\\ &= -\frac{q^2 l}{4\pi\varepsilon_0}\cdot\frac{1}{2l^2}, \end{aligned}

the same result.

Remark

This $W$ is not the work needed to pull the charge away. The induced charge is not a fixed set of charges: it rearranges as $q$ moves, and it has a positive self energy of its own, $+q^2/(16\pi\varepsilon_0 l)$ . Apart from the constant self energy of $q$ , the system’s energy is therefore $W/2 = -q^2/(16\pi\varepsilon_0 l)$ , and the work to remove $q$ slowly to infinity is only half of $\lvert W\rvert$ .