Irodov 3.60: A Line Charge and Its Mirror – Amit & Deepali Sinha Foundation

Problem Irodov 3.60

A thin, infinitely long thread with charge $\lambda$ per unit length runs parallel to an infinite conducting plane, at a distance $l$ from it. Find:

the magnitude of the force on each unit length of the thread;
the induced surface charge density $\sigma(x)$ on the plane, where $x$ is the distance of a point of the plane from the plane that contains the thread and is perpendicular to the conducting surface.

The thread λ, seen end on, at distance l from the conducting plane, and a point of the plane at distance x from the foot of the perpendicular through the thread. — The thread $\lambda$ , seen end on, at distance $l$ from the conducting plane, and a point of the plane at distance $x$ from the foot of the perpendicular through the thread.

Given

charge per unit length $\lambda$
thread parallel to the plane
distance $l$

Find

(a) the force

F_1

per unit length · (b) the density

\sigma(x)

Solution

Step 1The image thread and the field of a thread

The plane is an equipotential, so above it the field is that of the thread and its mirror image, a thread $-\lambda$ at depth $l$ , parallel to it: every point of the plane is equally far from the two. The field of each thread is the standard one; by symmetry it is radial and the same all round, so Gauss’s law on a coaxial cylinder of radius $s$ and length $L$ gives it.

E \cdot 2\pi s L = \frac{\lambda L}{\varepsilon_0} \implies E(s) = \frac{\lambda}{2\pi\varepsilon_0 s} \tag{1}

Step 2(a) The pull of the image

The real thread lies in the field of the image, at the distance $2l$ from it.

The image thread λ' (dashed) at the mirror position inside the metal and the force F on the real thread, toward the plane. — The image thread $\lambda'$ (dashed) at the mirror position inside the metal and the force $\vect F$ on the real thread, toward the plane.

On a unit length of the thread, carrying the charge $\lambda$ , the image’s field (1) exerts the force

F_1 = \lambda\, E(2l) = \lambda\cdot\frac{\lambda}{2\pi\varepsilon_0\cdot 2l}

F_1 = \dfrac{\lambda^2}{4\pi\varepsilon_0 l}

(2)

The force is an attraction toward the plane, whatever the sign of $\lambda$ .

Step 3(b) The field at a point of the plane

A point of the plane at distance $x$ from the foot of the perpendicular is the same distance $\rho = \sqrt{l^2 + x^2}$ from both threads, so the two fields there have the same size $E(\rho)$ .

At a point of the plane a distance x from the foot of the perpendicular: the field E_1 of the thread, E_2 of its image λ' (dashed), and their sum E normal to the plane. The glyphs show the induced charge, densest under the thread. — At a point of the plane a distance $x$ from the foot of the perpendicular: the field $\vect E_1$ of the thread, $\vect E_2$ of its image $\lambda'$ (dashed), and their sum $\vect E$ normal to the plane. The glyphs show the induced charge, densest under the thread.

Their components along the plane cancel, and their components normal to it add, each a fraction $l/\rho$ of $E(\rho)$ . For $\lambda > 0$ the sum points into the metal.

E = 2\cdot\frac{\lambda}{2\pi\varepsilon_0 \rho}\cdot\frac{l}{\rho} = \frac{\lambda l}{\pi\varepsilon_0 (l^2 + x^2)} \tag{3}

Step 4(b) The surface density

At a conductor’s surface $\sigma = \varepsilon_0 E_n$ , with $E_n$ the field’s component along the outward normal, from the metal into the vacuum. The field (3) points into the metal, so $E_n = -E$ .

\sigma = \varepsilon_0 E_n = -\varepsilon_0\,\frac{\lambda l}{\pi\varepsilon_0 (l^2 + x^2)}

\sigma(x) = -\dfrac{\lambda l}{\pi (l^2 + x^2)}

(4)

The induced charge has the sign opposite to $\lambda$ ; its magnitude peaks under the thread at $\lambda/(\pi l)$ and halves at $x = l$ .

Check

Every field line of the thread ends on the plane, so a strip of the plane of unit length along the thread must carry the total $-\lambda$ . Integrating (4) across the plane:

\int_{-\infty}^{\infty} \sigma\,dx = -\frac{\lambda l}{\pi}\int_{-\infty}^{\infty}\frac{dx}{l^2 + x^2} = -\frac{\lambda l}{\pi}\cdot\frac{1}{l}\left[\arctan\frac{x}{l}\right]_{-\infty}^{\infty} = -\frac{\lambda l}{\pi}\cdot\frac{\pi}{l} = -\lambda

as required. Units: $\lambda/l$ is a charge per area.

Remark

Half the induced charge lies within $|x| < l$ , because $\arctan 1 = \pi/4$ is half of $\pi/2$ . The density falls off as $1/x^2$ , one power more slowly than under a point charge, where it goes as $1/r^3$ : the field of a line charge spreads in two dimensions only. For the same reason the force (2) falls only as $1/l$ , where a point charge’s image force falls as $1/l^2$ .