Irodov 3.30: Two Rings, Two Distances – Amit & Deepali Sinha Foundation

Problem Irodov 3.30

Two thin wire rings of the same radius $R$ share a common axis, and their centres are a distance $a$ apart. One ring carries charge $q$ and the other $-q$ . Find the potential difference between the two centres.

The two coaxial rings of radius R carrying q and -q, with centres O_1 and O_2 a distance a apart on the common axis. — The two coaxial rings of radius $R$ carrying $q$ and $-q$ , with centres $O_1$ and $O_2$ a distance $a$ apart on the common axis.

Given

radius $R$ each
charges $q$ and $-q$
centres $a$ apart

Find

\varphi_1 - \varphi_2

, between the centres

O_1

and

O_2

Solution

Step 1The potential of one ring at two points

Potential is a scalar, so a ring’s potential at a point is the sum of $dq/(4\pi\varepsilon_0 r)$ over its elements. At its own centre every element is at distance $R$ ; at the other ring’s centre every element is at the same distance, the hypotenuse of $R$ and $a$ .

Every element of either ring is a distance (R^2 + a^2)^1/2 from the other ring's centre; one such line is drawn from ring 2 to O_1. — Every element of either ring is a distance $(R^2 + a^2)^{1/2}$ from the other ring’s centre; one such line is drawn from ring 2 to $O_1$ .

For a ring of charge $Q$ the two values are therefore

\begin{aligned} \varphi_{\text{own centre}} &= \frac{Q}{4\pi\varepsilon_0 R}\\ \varphi_{\text{other centre}} &= \frac{Q}{4\pi\varepsilon_0\sqrt{R^2 + a^2}} \end{aligned}

Step 2The potentials at the two centres

At $O_1$ the ring $q$ contributes from distance $R$ and the ring $-q$ from distance $\sqrt{R^2+a^2}$ ; at $O_2$ the roles swap. Adding the two rings at each centre:

\begin{aligned} \varphi_1 &= \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{R} - \frac{1}{\sqrt{R^2+a^2}}\right)\\ \varphi_2 &= \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{\sqrt{R^2+a^2}} - \frac{1}{R}\right) \end{aligned} \tag{1}

Step 3The difference

Subtracting the two lines of (1), the difference is twice $\varphi_1$ :

\varphi_1 - \varphi_2 = \frac{2q}{4\pi\varepsilon_0}\left(\frac{1}{R} - \frac{1}{\sqrt{R^2+a^2}}\right)

Taking $1/R$ out of the bracket:

\varphi_1 - \varphi_2 = \frac{q}{2\pi\varepsilon_0 R}\left(1 - \frac{1}{\sqrt{1 + (a/R)^2}}\right)

(2)

For $q > 0$ the centre of the positive ring is at the higher potential.

Check

Two limits of (2). With $a \to 0$ the rings coincide, their charges cancel everywhere, and the bracket $1 - 1$ gives zero, as it must. With $a \gg R$ the bracket tends to $1$ and the difference to $q/(2\pi\varepsilon_0R)$ : each ring then sits far from the other, so its centre has only its own potential $\pm q/(4\pi\varepsilon_0R)$ , and the two differ by twice that.

Remark

The field on the axis of a ring is zero at its centre, yet the potential difference between the centres is not zero: along the axis from $O_1$ to $O_2$ the field of the pair points from the positive ring toward the negative one everywhere between them, and its line integral is what (2) measures. For $a \ll R$ the result is about $qa^2/(4\pi\varepsilon_0R^3)$ , quadratic in the separation, because the field each ring makes near its centre itself grows only linearly with distance along the axis.