Irodov 3.55: Half the Energy of the Pair – Amit & Deepali Sinha Foundation

Problem Irodov 3.55

A point charge $q$ is at a distance $l$ from an infinite conducting plane. How much work must be done to move the charge slowly to a very large distance from 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 work

A

to remove the charge slowly to infinity

Solution

Step 1The attraction at any height

On its way out the charge passes every height $y$ from $l$ to infinity. At height $y$ the plane, an equipotential at zero, acts on it as the image $q' = -q$ at the mirror point, a distance $2y$ from the charge.

The charge at an intermediate height y: its field lines end on the induced negative charge of the plane, and the image q' (dashed) inside the metal reproduces that field above it. F is the attraction. — The charge at an intermediate height $y$ : its field lines end on the induced negative charge of the plane, and the image $q'$ (dashed) inside the metal reproduces that field above it. $\vect F$ is the attraction.

Coulomb’s law with that separation gives the attraction toward the plane.

F(y) = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{(2y)^2} = \frac{q^2}{16\pi\varepsilon_0 y^2}

(1)

Step 2The work against the attraction

Moving slowly means the applied force just balances $F(y)$ at every height, with no kinetic energy gained, so the work done is the integral of (1) over the path from $y = l$ to infinity.

\begin{aligned} A &= \int_l^{\infty} F(y)\,dy = \frac{q^2}{16\pi\varepsilon_0}\int_l^{\infty}\frac{dy}{y^2} \\ &= \frac{q^2}{16\pi\varepsilon_0}\left[-\frac{1}{y}\right]_l^{\infty} \end{aligned}

The bracket is $1/l$ .

A = \frac{q^2}{16\pi\varepsilon_0 l}

(2)

Check

Units: $q^2/(\varepsilon_0 l)$ is charge times potential, an energy. An independent route through field energy: the field of $q$ and its image is mirror symmetric about the plane, and the real field is that pattern above the plane and nothing below it. Apart from the self energy of $q$ , which never changes, the real field therefore stores half of the pair’s interaction energy, and the work must supply that half:

A = \frac{1}{2}\cdot\frac{1}{4\pi\varepsilon_0}\frac{q^2}{2l} = \frac{q^2}{16\pi\varepsilon_0 l}

in agreement with (2). Treating the image as a fixed charge would miss the factor one half and give twice the answer.

Remark

This image attraction is part of what holds electrons in a metal. For an electron, $q = e$ , and the work to escape from a height $y = 0.1$ nm above the surface amounts to several electronvolts, the scale of a metal’s work function:

\frac{e^2}{16\pi\varepsilon_0 y} = \frac{1}{4}\cdot\frac{1.44\ \text{eV\,nm}}{0.1\ \text{nm}} \approx 3.6\ \text{eV}

Closer in the formula diverges, which only says that the image picture fails at atomic distances, where the surface is no longer a smooth plane.