Irodov 3.11: The Thread Samples the Ring's Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.11

A thin wire ring of radius $R$ carries a total charge $q$ . A very long, uniformly charged thread with charge $\lambda$ per unit length lies along the ring’s axis, starting at the center of the ring and running away from it on one side. Find the force with which the ring and the thread act on each other.

The ring of radius R with charge q, and the thread of density λ starting at the ring's center and running along its axis; drawn for q, λ > 0. — The ring of radius $R$ with charge $q$ , and the thread of density $\lambda$ starting at the ring’s center and running along its axis; drawn for $q, \lambda > 0$ .

Given

ring: radius $R$ , charge $q$
thread along the axis, density $\lambda$ , from the center to infinity

Find

the force

F

between the ring and the thread

Solution

Step 1The ring’s field on its axis

Measure $x$ along the axis from the center, on the thread’s side. Every element $dq$ of the ring is the same distance from a point $P$ of the axis:

r = \sqrt{x^2 + R^2} \tag{1}

One element dq of the ring, its distance r to a point P of the thread, its field d E there and the axial component dE_x. — One element $dq$ of the ring, its distance $r$ to a point $P$ of the thread, its field $d\vect E$ there and the axial component $dE_x$ .

The element’s field at $P$ has magnitude $dq/(4\pi\varepsilon_0 r^2)$ along the line from the element, and its axial component carries the cosine $x/r$ of the angle between that line and the axis:

dE_x = \frac{1}{4\pi\varepsilon_0}\,\frac{dq}{r^2}\cdot\frac{x}{r}

All elements share $r$ and $x$ , so summing over the ring adds up its charge. With (1):

E_x = \frac{x}{4\pi\varepsilon_0 r^3}\oint dq = \frac{qx}{4\pi\varepsilon_0 r^3}

E_x(x) = \frac{1}{4\pi\varepsilon_0}\,\frac{qx}{(x^2+R^2)^{3/2}}

(2)

Step 2Sum the force over the thread

A piece $dx$ of the thread at distance $x$ carries charge $\lambda\,dx$ , and by (2) the ring pushes it along the axis with

dF = \lambda\,dx\,E_x(x) = \frac{q\lambda}{4\pi\varepsilon_0}\,\frac{x\,dx}{(x^2+R^2)^{3/2}}

Every piece is pushed the same way, so the forces add. The thread runs from the center, $x = 0$ , to infinity:

\begin{aligned} F &= \frac{q\lambda}{4\pi\varepsilon_0}\int_0^{\infty}\frac{x\,dx}{(x^2+R^2)^{3/2}}\\ &= \frac{q\lambda}{4\pi\varepsilon_0}\left[-\frac{1}{\sqrt{x^2+R^2}}\right]_0^{\infty} = \frac{q\lambda}{4\pi\varepsilon_0}\left(0 + \frac{1}{R}\right) \end{aligned}

F = \frac{q\lambda}{4\pi\varepsilon_0 R}

(3)

For $q\lambda > 0$ the thread is pushed along the axis away from the ring, and by Newton’s third law the ring is pushed the opposite way with the same force; for $q\lambda < 0$ they attract.

Check

Units: $\lambda R$ is a charge, so $q\lambda/(\varepsilon_0 R) = q\,(\lambda R)/(\varepsilon_0R^2)$ has the form of Coulomb’s $q_1q_2/(\varepsilon_0 r^2)$ , a force. Limit $R \to 0$ : the ring shrinks to a point charge $q$ sitting at the thread’s end, and (3) grows without bound; directly, that point charge would push the thread with $\int_0^\infty q\lambda\,dx/(4\pi\varepsilon_0x^2)$ , which diverges at the end $x = 0$ , so the two agree.

Remark

Although the thread is infinite, the force is finite, because the ring’s field falls as $1/x^2$ far away. Most of it comes from close in: the integral up to $x$ is $1 - R/\sqrt{x^2+R^2}$ of the total, so the stretch $0 < x < \sqrt{3}\,R$ already supplies half the force, even though the field itself is zero at the center.