Irodov 3.5: Extra Tension from a Central Charge – Amit & Deepali Sinha Foundation

Problem Irodov 3.5

A thin wire ring of radius $r$ carries a charge $q$ . A point charge $q_0$ is then placed at the centre of the ring. By how much does the tension in the wire increase?

The thin ring of radius r with charge q spread along it, and the point charge q_0 at its centre, both drawn positive. — The thin ring of radius $r$ with charge $q$ spread along it, and the point charge $q_0$ at its centre, both drawn positive.

Given

ring radius $r$ , charge $q$
point charge $q_0$ at the centre

Find

the increase

\Delta T

of the tension in the wire

Solution

Step 1The push of the central charge on half the ring

By symmetry the charge stays spread evenly round the ring, with density $q/(2\pi r)$ per unit length. Cut the ring along a diameter and consider the upper half. An element at angle $\varphi$ from the diameter, of length $r\,d\varphi$ , carries the charge

dq = \frac{q}{2\pi r}\,r\,d\varphi = \frac{q}{2\pi}\,d\varphi

The upper half of the ring cut free. Each element at angle φ is pushed radially outward by dF from q_0; the wire below pulls on the two cut ends with the extra tension , tangent to the ring there. — The upper half of the ring cut free. Each element at angle $\varphi$ is pushed radially outward by $d\vect{F}$ from $q_0$ ; the wire below pulls on the two cut ends with the extra tension $\Delta\vect{T}$ , tangent to the ring there.

Every element is at the distance $r$ from $q_0$ , so Coulomb’s law gives the same radial push on each:

dF = \frac{1}{4\pi\varepsilon_0}\,\frac{q_0\,dq}{r^2} = \frac{q q_0}{8\pi^2\varepsilon_0 r^2}\,d\varphi \tag{1}

The elements at $\varphi$ and $\pi - \varphi$ are mirror images, so the components of their pushes along the diameter cancel; the components perpendicular to it, $dF\sin\varphi$ , add. Integrating (1) over the half-ring:

\begin{aligned} F &= \frac{q q_0}{8\pi^2\varepsilon_0 r^2}\int_0^{\pi}\sin\varphi\,d\varphi\\ &= \frac{q q_0}{8\pi^2\varepsilon_0 r^2}\Big[-\cos\varphi\Big]_0^{\pi} = \frac{q q_0}{4\pi^2\varepsilon_0 r^2} \end{aligned}

Step 2Balance against the tension at the cuts

Before $q_0$ arrived the half-ring was already in equilibrium. The new push $F$ must therefore be balanced by the increase $\Delta T$ of the tension, which pulls on each cut end along the wire, that is along the tangent, perpendicular to the diameter. The two ends together give $2\Delta T$ :

2\,\Delta T = \frac{q q_0}{4\pi^2\varepsilon_0 r^2}

\Delta T = \frac{q q_0}{8\pi^2\varepsilon_0 r^2}

(2)

Check

The same result from a single element: an element subtending $d\varphi$ , held by the tension $T$ at both ends, feels a net inward pull $T\,d\varphi$ from the two slightly inclined tensions, so an outward radial force $f\,r\,d\varphi$ (with $f$ the force per unit length) raises the tension by $fr$ . Here $f\,r\,d\varphi$ is (1), so the increase is

\Delta T = \frac{dF}{d\varphi} = \frac{q q_0}{8\pi^2\varepsilon_0 r^2}

in agreement with (2). Units: $q q_0/(\varepsilon_0 r^2)$ is a force, as Coulomb’s law shows.

Remark

For $q_0$ of the sign opposite to $q$ the formula gives a negative $\Delta T$ : the central charge pulls the wire inward and slackens it. The problem asks only for the increment because the tension the ring’s own charge produces is not finite for an ideally thin wire: neighbouring elements repel each other so strongly that the self tension grows without bound, though only logarithmically, as the wire’s thickness goes to zero. The increment, set by a charge a full radius away, does not care about the thickness at all.