Irodov 3.43: Two Rings Make an Axial Dipole – Amit & Deepali Sinha Foundation

Problem Irodov 3.43

Two thin wire rings of the same radius $R$ share an axis and are a small distance $l$ apart ( $l \ll R$ ). They carry charges $q$ and $-q$ . The $x$ axis runs along the common axis, pointing from the ring with $-q$ toward the ring with $q$ , with its origin $O$ midway between them. Find the potential and the field on the axis as functions of $x$ , sketch both in one drawing, and find their form for $|x| \gg R$ .

The two rings of radius R on the common x axis, +q above and -q below (their separation l drawn much larger than it is), and the origin O midway between them. — The two rings of radius $R$ on the common $x$ axis, $+q$ above and $-q$ below (their separation $l$ drawn much larger than it is), and the origin $O$ midway between them.

Given

rings of radius $R$
charges $q$ and $-q$
separation $l \ll R$
$x$ along the axis from $O$

Find

\varphi(x)

and

E_x(x)

, their plots, and their form at

|x| \gg R

Solution

Step 1The potential of one ring on its axis

Every element of a ring of charge $Q$ is the same distance $\sqrt{R^2 + s^2}$ from a point of the axis a distance $s$ from the ring’s plane, so the potentials of the elements simply add to

\varphi_1(s) = \frac{Q}{4\pi\varepsilon_0\sqrt{R^2 + s^2}} \tag{1}

Step 2The pair, as a difference

The ring with $q$ sits at $x = l/2$ and the ring with $-q$ at $x = -l/2$ . Write (1) as $Q f(s)/(4\pi\varepsilon_0)$ , with

f(s) = \frac{1}{\sqrt{R^2 + s^2}}

Then the point $x$ of the axis has the potential

\varphi(x) = \frac{q}{4\pi\varepsilon_0}\left[f\!\left(x - \frac{l}{2}\right) - f\!\left(x + \frac{l}{2}\right)\right]

For $l \ll R$ the bracket is $-l$ times the slope of $f$ at $x$ :

f\!\left(x - \frac{l}{2}\right) - f\!\left(x + \frac{l}{2}\right) \approx -l\,\frac{df}{dx} = \frac{l\,x}{(R^2 + x^2)^{3/2}}

\varphi = \frac{ql}{4\pi\varepsilon_0}\,\frac{x}{(R^2 + x^2)^{3/2}}

(2)

Step 3The field on the axis

By symmetry the field on the axis lies along it, and its projection is $E_x = -d\varphi/dx$ . Differentiating (2) as a product:

\begin{aligned} \frac{d}{dx}\,\frac{x}{(R^2 + x^2)^{3/2}} &= \frac{1}{(R^2 + x^2)^{3/2}} - \frac{3x^2}{(R^2 + x^2)^{5/2}}\\ &= \frac{R^2 - 2x^2}{(R^2 + x^2)^{5/2}} \end{aligned}

E_x = -\frac{ql}{4\pi\varepsilon_0}\,\frac{R^2 - 2x^2}{(R^2 + x^2)^{5/2}}

(3)

Step 4The shapes of the two curves

The potential (2) is odd in $x$ , zero at $O$ , and has its extrema where its slope, and so $E_x$ , vanishes:

\begin{aligned} R^2 - 2x^2 &= 0\\ x &= \pm\frac{R}{\sqrt{2}} \end{aligned}

The field (3) is even. At $O$ it is negative, pointing from the positive ring to the negative one, with the value

E_x(0) = -\frac{ql}{4\pi\varepsilon_0 R^3}

It changes sign at $x = \pm R/\sqrt{2}$ and is positive beyond, so outside the pair the field on the axis points away from it on the positive side and toward it on the negative side.

The field E_x (crimson, top) and the potential φ (teal, bottom) along the axis. The extrema of φ at x = R/sqrt(2) line up with the zeros of E_x. — The field $E_x$ (crimson, top) and the potential $\varphi$ (teal, bottom) along the axis. The extrema of $\varphi$ at $x = \pm R/\sqrt{2}$ line up with the zeros of $E_x$ .

Step 5Far along the axis

For $|x| \gg R$ the $R^2$ is negligible beside $x^2$ in (2) and (3), and $(x^2)^{1/2} = |x|$ :

\begin{gathered}\varphi \approx \frac{ql}{4\pi\varepsilon_0}\,\frac{x}{|x|^3} = \pm\frac{ql}{4\pi\varepsilon_0 x^2}\\ E_x \approx \frac{ql}{2\pi\varepsilon_0 |x|^3}\end{gathered}

(4)

The potential takes the sign of $x$ . These are the axial potential and field of a point dipole $p = ql$ pointing along $+x$ .

Check

At $O$ , a distance $l/2$ from each ring’s plane, each ring alone gives the axial field

\frac{q\,(l/2)}{4\pi\varepsilon_0(R^2 + l^2/4)^{3/2}}

Both point along $-x$ , away from the positive ring above and toward the negative one below, so they add to $ql/(4\pi\varepsilon_0 R^3)$ along $-x$ for $l \ll R$ , which is $E_x(0)$ found from (3). Also, the potential returns to zero at both ends of the axis, so the area under $E_x$ over the whole axis must vanish: the negative dip between $\pm R/\sqrt{2}$ is balanced by the two positive wings, as the plot shows.

Remark

Near the pair, for $|x| < R/\sqrt{2}$ , the axial field points from $+q$ to $-q$ , as in a capacitor; farther out it points the other way, along the moment $ql$ , as a dipole’s does. A single ring has its strongest axial field at the same distance $R/\sqrt{2}$ , and that is no coincidence: the pair’s potential is $-l$ times the slope of one ring’s potential, that is, $l$ times one ring’s axial field (with $q$ for $Q$ ), so the extrema of $\varphi$ , where $E_x$ of the pair vanishes, sit exactly where the single ring’s field is strongest.