Irodov 3.39: Two Charges Shrink to One Moment – Amit & Deepali Sinha Foundation

Problem Irodov 3.39

A dipole of electric moment $\vect{p}$ sits at the origin. A point $S$ has position vector $\vect{r}$ , of length $r$ , making the angle $\theta$ with the direction of $\vect{p}$ . Show that the potential of the dipole’s field at $S$ can be written as $\varphi = \vect{p}\cdot\vect{r}/(4\pi\varepsilon_0 r^3)$ . Then, from this expression, find the magnitude of the electric field at $S$ as a function of $r$ and $\theta$ .

The dipole p at the origin, pointing up the z axis, and the point S at position r, its polar angle θ measured from the z axis. — The dipole $\vect{p}$ at the origin, pointing up the $z$ axis, and the point $S$ at position $\vect{r}$ , its polar angle $\theta$ measured from the $z$ axis.

Given

dipole moment $\vect{p}$ along $z$ at the origin
$S$ at $\vect{r}$ , angle $\theta$ to $\vect{p}$

Find

show

\varphi = \vect{p}\cdot\vect{r}/(4\pi\varepsilon_0 r^3)

E(r, \theta)

Solution

Step 1The dipole as a pair of charges

A dipole is a charge $+q$ and a charge $-q$ a small distance $l$ apart, with moment $\vect{p} = q\vect{l}$ , where $\vect{l}$ runs from $-q$ to $+q$ . Put the pair on the $z$ axis, $+q$ at $z = l/2$ and $-q$ at $z = -l/2$ , and call their distances from $S$ $r_+$ and $r_-$ . The potentials of the two charges add:

\varphi = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r_+} - \frac{1}{r_-}\right) = \frac{q}{4\pi\varepsilon_0}\,\frac{r_- - r_+}{r_+ r_-} \tag{1}

The dipole as the pair +q, -q on the z axis (the separation drawn much larger than it is) and the distances r_+ and r_- from each charge to S. — The dipole as the pair $+q$ , $-q$ on the $z$ axis (the separation drawn much larger than it is) and the distances $r_+$ and $r_-$ from each charge to $S$ .

Step 2Far from the pair

At $S$ , a distance $r \gg l$ away, the lines from the two charges are nearly parallel to $\vect{r}$ . The $-q$ charge is then farther by the projection of $\vect{l}$ on that direction, and the product of the two distances is $r^2$ to first order in $l$ :

\begin{aligned} r_- - r_+ &\approx l\cos\theta\\ r_+ r_- &\approx r^2 \end{aligned}

Putting both into (1), and using $\vect{p}\cdot\vect{r} = q l r\cos\theta$ :

\varphi = \frac{q l\cos\theta}{4\pi\varepsilon_0 r^2} = \frac{q l r\cos\theta}{4\pi\varepsilon_0 r^3}

\varphi = \frac{\vect{p}\cdot\vect{r}}{4\pi\varepsilon_0 r^3}

(2)

Step 3The field along r and across it

The field is minus the gradient of (2), written as $\varphi = p\cos\theta/(4\pi\varepsilon_0 r^2)$ . In polar coordinates in the plane of $\vect{p}$ and $\vect{r}$ , its radial part is $-\partial\varphi/\partial r$ and its part along the direction of growing $\theta$ is $-(1/r)\,\partial\varphi/\partial\theta$ :

E_r = -\frac{\partial\varphi}{\partial r} = \frac{p}{4\pi\varepsilon_0}\,\frac{2\cos\theta}{r^3} \tag{3}

E_\theta = -\frac{1}{r}\frac{\partial\varphi}{\partial\theta} = \frac{p}{4\pi\varepsilon_0}\,\frac{\sin\theta}{r^3} \tag{4}

The field at S split along r (E_r) and along the direction of growing θ (E_θ). — The field at $S$ split along $\vect{r}$ ( $E_r$ ) and along the direction of growing $\theta$ ( $E_\theta$ ).

Step 4The magnitude

The two components are perpendicular, so the magnitude is the root of the sum of their squares. With (3) and (4):

\begin{aligned} E &= \sqrt{E_r^2 + E_\theta^2} = \frac{p}{4\pi\varepsilon_0 r^3}\sqrt{4\cos^2\theta + \sin^2\theta} \end{aligned}

Writing $\sin^2\theta = 1 - \cos^2\theta$ under the root:

E = \frac{p}{4\pi\varepsilon_0 r^3}\sqrt{1 + 3\cos^2\theta}

(5)

Check

On the axis, $\theta = 0$ , (5) gives $2p/(4\pi\varepsilon_0 r^3)$ . The pair gives the same directly: the two Coulomb fields on the axis differ by

\frac{q}{4\pi\varepsilon_0}\left[\frac{1}{(r - l/2)^2} - \frac{1}{(r + l/2)^2}\right] \approx \frac{q}{4\pi\varepsilon_0}\,\frac{2l}{r^3}

On the equator, $\theta = 90^\circ$ , (5) gives $p/(4\pi\varepsilon_0 r^3)$ , half the axial value, and there (3) vanishes, so the field is all $E_\theta$ , antiparallel to $\vect{p}$ .

Remark

The pair potential differs from (2) only by terms of relative order $(l/r)^2$ , not $l/r$ . Expanding each $1/r_\pm$ in powers of $l/r$ , the even powers come with the same sign for both charges and cancel in the difference: the pair, antisymmetric about its midpoint, has no quadrupole term, and its next correction is the octupole. So the point dipole is a better description of a symmetric pair than its first-order derivation suggests.