Irodov 3.20: Two Charges Share One Window – Amit & Deepali Sinha Foundation

Problem Irodov 3.20

Point charges $q$ and $-q$ are a distance $2l$ apart. A circle of radius $R$ lies in the plane midway between them, perpendicular to the line joining them and centred on that line, as in the figure. Find the flux of the electric field through the circle.

The charges q and -q on their axis, each a distance l from the centre of the circle of radius R, which stands in the mid-plane, perpendicular to the axis. — The charges $q$ and $-q$ on their axis, each a distance $l$ from the centre of the circle of radius $R$ , which stands in the mid-plane, perpendicular to the axis.

Given

$q$ and $-q$ , $2l$ apart
circle of radius $R$ in the mid-plane

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 edge-on as the dashed line. At its point P, a distance r from the axis, both charges are a distance s away; their fields E_1 and E_2 add up to E along the axis, normal to the circle. — A plane through the axis, the circle edge-on as the dashed line. At its point $P$ , a distance $r$ from the axis, both charges are a distance $s$ away; their fields $\vect E_1$ and $\vect E_2$ add up to $\vect E$ along the axis, normal to the circle.

Take a point $P$ of the circle at distance $r$ from the axis. Both charges are at the same distance from it:

s = \sqrt{l^2 + r^2}

The field of $q$ points away from $q$ , the field of $-q$ toward $-q$ : both have a component along the axis in the same direction, from $q$ toward $-q$ , while their components across the axis are equal and opposite. Each axial component is the fraction $l/s$ of a Coulomb field, so the field normal to the circle is

E_n = 2\cdot\frac{q}{4\pi\varepsilon_0 s^2}\cdot\frac{l}{s} = \frac{q\,l}{2\pi\varepsilon_0\,(l^2 + r^2)^{3/2}} \tag{1}

Step 2Add up over the circle

Split the circle into rings of radius $r$ and area $2\pi r\,dr$ , on each of which (1) is constant:

\Phi = \int_0^R E_n\,2\pi r\,dr = \frac{q\,l}{\varepsilon_0}\int_0^R \frac{r\,dr}{(l^2 + r^2)^{3/2}} = \frac{q\,l}{\varepsilon_0}\left[-\frac{1}{\sqrt{l^2 + r^2}}\right]_0^R

Evaluating at the limits:

\Phi = \frac{q\,l}{\varepsilon_0}\left(\frac{1}{l} - \frac{1}{\sqrt{l^2 + R^2}}\right)

|\Phi| = \frac{q}{\varepsilon_0}\left(1 - \frac{l}{\sqrt{l^2 + R^2}}\right)

(2)

For $q > 0$ the flux crosses the circle in the direction from $q$ toward $-q$ .

Check

For $R \to \infty$ the circle becomes the whole mid-plane and the flux tends to $q/\varepsilon_0$ : half of the total flux $q/\varepsilon_0$ of $q$ crosses the plane, and the field of $-q$ points toward $-q$ , so it crosses the plane in the same sense and adds the other half. For $R \ll l$ the bracket is about $R^2/(2l^2)$ , and the flux is the nearly uniform field $2q/(4\pi\varepsilon_0 l^2)$ at the centre times the area $\pi R^2$ , as it should be.

Remark

Each charge alone sends through the circle the fraction $\Omega/(4\pi)$ of its flux, with $\Omega = 2\pi(1 - l/\sqrt{l^2 + R^2})$ the solid angle under which it sees the circle; the answer is just twice that, because the two contributions have the same sense. A circle in the mid-plane of two equal charges of the same sign would carry zero flux.