Irodov 3.44: Field Leaking Through a Hole – Amit & Deepali Sinha Foundation

Problem Irodov 3.44

Two infinite parallel planes a distance $l$ apart carry uniform surface charges of density $\sigma$ and $-\sigma$ . Each plane has a round hole of radius $R$ , the two holes on a common axis, with $l \ll R$ . The $x$ axis runs along that common axis, pointing from the plane with $-\sigma$ toward the plane with $\sigma$ , with its origin $O$ at the center of the holes, midway between the planes. Find the potential $\varphi$ and the projection $E_x$ of the field on the axis as functions of $x$ , and sketch $\varphi(x)$ .

Pieces of the two planes, +σ above and -σ below (their separation l drawn much larger than it is), the coaxial holes of radius R, and the x axis through their centers with O midway. — Pieces of the two planes, $+\sigma$ above and $-\sigma$ below (their separation $l$ drawn much larger than it is), the coaxial holes of radius $R$ , and the $x$ axis through their centers with $O$ midway.

Given

planes $\sigma$ and $-\sigma$ , a distance $l$ apart
coaxial holes of radius $R$
$l \ll R$

Find

\varphi(x)

and

E_x(x)

on the axis · the plot of

\varphi(x)

Solution

Step 1Cut the planes into ring pairs

A ring of charge $Q$ and radius $\rho$ gives, at a point of its axis a distance $s$ from its plane, the potential $Q/(4\pi\varepsilon_0\sqrt{\rho^2 + s^2})$ , because every element of it is at the same distance. Cut each plane into thin annuli of radius $\rho$ and width $d\rho$ , from the rim of the hole outward. The annulus on the positive plane carries

dq = \sigma\cdot 2\pi\rho\,d\rho \tag{1}

and the one facing it on the negative plane carries $-dq$ .

One pair of annuli of radius ρ (sky), carrying dq on the upper, positive plane and -dq on the lower one, and a point P of the axis at the coordinate x. — One pair of annuli of radius $\rho$ (sky), carrying $dq$ on the upper, positive plane and $-dq$ on the lower one, and a point $P$ of the axis at the coordinate $x$ .

Step 2The potential of one ring pair

The $dq$ ring sits at $x = l/2$ and the $-dq$ ring at $x = -l/2$ . Their potentials at $P$ differ only through the small shift $l$ , so their sum is $-l$ times the derivative, with respect to $x$ , of one ring’s potential at $x$ . Every ring is at least $R$ from the axis and $l \ll R$ , so this holds at every point of the axis:

d\varphi = -l\,\frac{\partial}{\partial x}\,\frac{dq}{4\pi\varepsilon_0\sqrt{\rho^2 + x^2}} = \frac{dq\,l\,x}{4\pi\varepsilon_0(\rho^2 + x^2)^{3/2}}

With (1):

d\varphi = \frac{\sigma l x}{2\varepsilon_0}\,\frac{\rho\,d\rho}{(\rho^2 + x^2)^{3/2}} \tag{2}

Step 3Adding all the ring pairs

The planes start at the rim of the hole and go on for ever, so (2) is integrated from $\rho = R$ to infinity:

\int_R^\infty\frac{\rho\,d\rho}{(\rho^2 + x^2)^{3/2}} = \left[-\frac{1}{\sqrt{\rho^2 + x^2}}\right]_R^\infty = \frac{1}{\sqrt{R^2 + x^2}}

\varphi = \frac{\sigma l}{2\varepsilon_0}\,\frac{x}{\sqrt{x^2 + R^2}}

(3)

Step 4The field on the axis

By symmetry the field on the axis lies along it, and $E_x = -d\varphi/dx$ . Differentiating (3):

\begin{aligned} \frac{d}{dx}\,\frac{x}{\sqrt{x^2 + R^2}} &= \frac{1}{\sqrt{x^2 + R^2}} - \frac{x^2}{(x^2 + R^2)^{3/2}}\\ &= \frac{R^2}{(x^2 + R^2)^{3/2}} \end{aligned}

E_x = -\frac{\sigma l R^2}{2\varepsilon_0(x^2 + R^2)^{3/2}}

(4)

The field on the axis points along $-x$ everywhere, from the positive plane’s side toward the negative plane’s side, strongest at $O$ . The potential rises steadily through zero at $O$ and levels off at $\pm\sigma l/(2\varepsilon_0)$ far away on either side.

The field E_x (crimson, top) and the potential φ (teal, bottom) along the axis: φ rises through O and levels off on both sides; E_x is negative everywhere, strongest at O. — The field $E_x$ (crimson, top) and the potential $\varphi$ (teal, bottom) along the axis: $\varphi$ rises through $O$ and levels off on both sides; $E_x$ is negative everywhere, strongest at $O$ .

Check

Far from the hole on both sides the potentials differ by

\varphi(+\infty) - \varphi(-\infty) = \frac{\sigma l}{\varepsilon_0}

which is the jump across the double layer without any hole: the field $\sigma/\varepsilon_0$ between the planes times the gap $l$ . The integral of $E_x$ along the whole axis gives the same, since the integral of $R^2/(x^2 + R^2)^{3/2}$ over all $x$ is 2. Units: $\sigma l/\varepsilon_0$ is a field times a length, a potential.

Remark

Far along the axis, $|x| \gg R$ , the field is $E_x \approx -\sigma l R^2/(2\varepsilon_0 |x|^3)$ , the axial field of a point dipole of moment $\sigma l\cdot\pi R^2$ pointing along $-x$ . That is the hole: the complete double layer has no field outside the gap, so the hole acts as the missing disc of moment $\sigma l\cdot\pi R^2$ with its sign reversed, and the field leaks through it from the positive side round to the negative.