Irodov 3.19: Half a Thread Through a Ring – Amit & Deepali Sinha Foundation

Problem Irodov 3.19

A very long, uniformly charged straight thread lies along the axis of a circle of radius $R$ , with one of its ends at the centre of the circle, so that the thread runs off to one side of the circle’s plane. The thread carries charge $\lambda$ per unit length. Find the flux of the electric field through the area of the circle.

The circle of radius R with centre O, and the charged thread along its axis, starting at O and running off to one side (drawn for λ > 0). — The circle of radius $R$ with centre $O$ , and the charged thread along its axis, starting at $O$ and running off to one side (drawn for $\lambda > 0$ ).

Given

$R$
$\lambda$
thread from the centre along the axis, one side only

Find

the flux

\Phi

\vect E

through the circle

Solution

Step 1The normal field at a point of the circle

A plane through the axis, the circle seen edge-on as the dashed line. An element dq of the thread at distance z from the circle's plane, and its field d E at the point P at distance r from O, with its component dE_n normal to the circle. — A plane through the axis, the circle seen edge-on as the dashed line. An element $dq$ of the thread at distance $z$ from the circle’s plane, and its field $d\vect E$ at the point $P$ at distance $r$ from $O$ , with its component $dE_n$ normal to the circle.

Only the component of $\vect E$ normal to the circle’s plane carries flux. Take a point $P$ of the circle at distance $r$ from the centre, and an element $dq = \lambda\,dz$ of the thread at distance $z$ from the plane. It is at distance $\sqrt{z^2 + r^2}$ from $P$ , and the normal (axial) part of its field is the fraction $z/\sqrt{z^2 + r^2}$ of the whole:

dE_n = \frac{\lambda\,dz}{4\pi\varepsilon_0 (z^2 + r^2)}\cdot\frac{z}{\sqrt{z^2 + r^2}} = \frac{\lambda}{4\pi\varepsilon_0}\,\frac{z\,dz}{(z^2 + r^2)^{3/2}} \tag{1}

Every element pushes the field away from the thread side, so all the normal parts point the same way and simply add. Integrating (1) along the whole thread:

E_n(r) = \frac{\lambda}{4\pi\varepsilon_0}\int_0^\infty \frac{z\,dz}{(z^2 + r^2)^{3/2}} = \frac{\lambda}{4\pi\varepsilon_0}\left[-\frac{1}{\sqrt{z^2 + r^2}}\right]_0^\infty

The upper limit gives zero and the lower one $1/r$ :

E_n(r) = \frac{\lambda}{4\pi\varepsilon_0 r} \tag{2}

Step 2Add up over the circle

The normal field depends only on $r$ , so split the circle into rings of radius $r$ and area $2\pi r\,dr$ and sum (2) over them:

\Phi = \int_0^R E_n(r)\,2\pi r\,dr = \int_0^R \frac{\lambda}{4\pi\varepsilon_0 r}\,2\pi r\,dr = \frac{\lambda}{2\varepsilon_0}\int_0^R dr

|\Phi| = \frac{\lambda R}{2\varepsilon_0}

(3)

For $\lambda > 0$ the flux goes through the circle away from the side the thread is on.

Check

The same result follows from solid angles. A charge $dq$ on the axis at distance $z$ sends through the circle the fraction $\Omega/(4\pi)$ of its flux $dq/\varepsilon_0$ , with the cone’s solid angle $\Omega = 2\pi\bigl(1 - z/\sqrt{z^2 + R^2}\bigr)$ :

\Phi = \frac{\lambda}{2\varepsilon_0}\int_0^\infty\Bigl(1 - \frac{z}{\sqrt{z^2 + R^2}}\Bigr)dz = \frac{\lambda}{2\varepsilon_0}\Bigl[z - \sqrt{z^2 + R^2}\Bigr]_0^\infty = \frac{\lambda R}{2\varepsilon_0}

since $z - \sqrt{z^2 + R^2}$ tends to $0$ at infinity and equals $-R$ at $z = 0$ .

Remark

The flux is finite although the thread is infinitely long: elements far along the thread see the circle under a solid angle that falls as $1/z^2$ , so their contributions die out fast enough. Equivalently, $\lambda R/(2\varepsilon_0)$ is the flux that a length $R/2$ of thread would send through a closed surface: only the part of the thread within a few radii of the circle really matters.