Irodov 3.58: A Dipole Facing Its Twin – Amit & Deepali Sinha Foundation

Problem Irodov 3.58

A point electric dipole of moment $\vect{p}$ is at a distance $l$ from an infinite conducting plane, with $\vect{p}$ perpendicular to the plane. Find the magnitude of the force on the dipole.

The point dipole p, normal to the conducting plane, at distance l from it. — The point dipole $\vect p$ , normal to the conducting plane, at distance $l$ from it.

Given

dipole moment $\vect p$
$\vect p$ normal to the plane
distance $l$

Find

the force

F

on the dipole

Solution

Step 1The image of a normal dipole is an equal, parallel dipole

Represent the dipole as a charge $q$ at height $l + a/2$ above the plane and $-q$ at height $l - a/2$ , with $p = qa$ and $a \to 0$ . The plane, an equipotential, is replaced by the mirror images: $-q$ at depth $l + a/2$ and $+q$ at depth $l - a/2$ . Measured upward, the image pair runs from its $-q$ to its $+q$ , so its moment is

p' = q\left[\left(-l + \frac{a}{2}\right) - \left(-l - \frac{a}{2}\right)\right] = qa = p

The dipole as the pair q, -q and its images q'_1, q'_2 (dashed) in the metal. The image moment p ' (dashed) equals p and points the same way, a distance 2l below it on the same axis. — The dipole as the pair $q$ , $-q$ and its images $q'_1$ , $q'_2$ (dashed) in the metal. The image moment $\vect p\,'$ (dashed) equals $\vect p$ and points the same way, a distance $2l$ below it on the same axis.

The image is a dipole equal and parallel to $\vect p$ , on the same axis, a distance $2l$ away. Reversing $\vect p$ reverses both charges and their images, so the image stays parallel to the dipole either way.

Step 2The image’s field along the common axis

On the axis of a dipole the fields of its two charges are collinear, and at a distance $s$ from the dipole, far compared with $a$ , they add to the standard axial field along the moment.

E(s) = \frac{1}{4\pi\varepsilon_0}\frac{2p}{s^3} \tag{1}

Put the $y$ axis along $\vect p$ , normal to the plane, with the plane at $y = 0$ . The image sits at $y = -l$ , so a point of the axis at height $y$ is at $s = y + l$ from it.

E_y(y) = \frac{1}{4\pi\varepsilon_0}\frac{2p}{(y + l)^3}

Step 3The force from the field’s gradient

A dipole along $y$ in a field that varies along $y$ feels $F_y = p\, \partial E_y/\partial y$ , the difference of the fields on its two charges. The gradient is taken with the image held where it is: the force is that of the induced charge as it stands.

\frac{\partial E_y}{\partial y} = -\frac{1}{4\pi\varepsilon_0}\frac{6p}{(y + l)^4}

At the dipole, $y = l$ , the distance to the image is $2l$ .

F_y = p\left(-\frac{1}{4\pi\varepsilon_0}\frac{6p}{(2l)^4}\right) = -\frac{6p^2}{64\pi\varepsilon_0 l^4}

F = \dfrac{3p^2}{32\pi\varepsilon_0 l^4}

(2)

The minus sign of $F_y$ says the dipole is attracted toward the plane.

Check

The charge picture gives the same force directly. On $+q$ (height $l + a/2$ ) the like image at distance $2l$ pushes up and the unlike one at $2l + a$ pulls down; on $-q$ (height $l - a/2$ ) the unlike image at $2l - a$ pulls down and the like one at $2l$ pushes up. With $(2l \pm a)^{-2} \approx (4l^2)^{-1}(1 \mp a/l + 3a^2/4l^2)$ :

F_y = \frac{q^2}{4\pi\varepsilon_0}\left[\frac{2}{4l^2} - \frac{1}{(2l + a)^2} - \frac{1}{(2l - a)^2}\right] \approx -\frac{q^2}{4\pi\varepsilon_0}\frac{3a^2}{8l^4} = -\frac{3p^2}{32\pi\varepsilon_0 l^4}

in agreement with (2); the first-order terms cancel, which is why the force starts at $a^2$ , that is at $p^2$ .

Remark

The force falls as $1/l^4$ , two powers faster than a point charge’s image force, because the dipole and its image are both neutral. Integrating (2) from $l$ to infinity gives the work to pull the dipole away,

A = \int_l^\infty \frac{3p^2}{32\pi\varepsilon_0 y^4}\,dy = \frac{p^2}{32\pi\varepsilon_0 l^3}

which is half the magnitude of the interaction energy of two coaxial dipoles $2l$ apart,

|W_{\text{pair}}| = p\,E(2l) = \frac{1}{4\pi\varepsilon_0}\frac{2p^2}{(2l)^3} = \frac{p^2}{16\pi\varepsilon_0 l^3}

the same factor one half as for a point charge and its image, for the same reason, the real field filling only half of the mirror-symmetric pattern.