Irodov 3.45: The Field Outside a Capacitor – Amit & Deepali Sinha Foundation

Problem Irodov 3.45

A capacitor is made of two thin round parallel plates, each of radius $R$ , a distance $l$ apart ( $l \ll R$ ), charged uniformly with surface densities $\sigma$ and $-\sigma$ . On the axis of the capacitor, outside it, at a distance $x \gg l$ from the plates, find the potential of the electric field and the magnitude of the field as functions of $x$ . Then find their form for $x \gg R$ .

The capacitor cut through its axis: the plates of radius R seen edge-on, σ on the right and -σ on the left (the gap l drawn much larger than it is), and a point P of the axis at the distance x from them. — The capacitor cut through its axis: the plates of radius $R$ seen edge-on, $\sigma$ on the right and $-\sigma$ on the left (the gap $l$ drawn much larger than it is), and a point $P$ of the axis at the distance $x$ from them.

Given

round plates of radius $R$
charges $\sigma$ and $-\sigma$
gap $l \ll R$
$x \gg l$

Find

\varphi(x)

and

E(x)

on the axis; their form for

x \gg R

Solution

Step 1Cut the plates 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 plate into thin annuli of radius $\rho$ and width $d\rho$ . The annulus on the positive plate carries

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

and the one facing it on the negative plate carries $-dq$ . Measure $x$ along the axis from the middle of the gap, positive on the side of the positive plate; for $x \gg l$ this is the distance from the plates.

One pair of annuli of radius ρ, cut by the page: dq on the positive plate, -dq on the negative one, and the point P on the axis. — One pair of annuli of radius $\rho$ , cut by the page: $dq$ on the positive plate, $-dq$ on the negative one, and the point $P$ on the axis.

Step 2The potential of one ring pair

The $dq$ ring sits at $x = l/2$ and the $-dq$ ring at $x = -l/2$ . For $x \gg l$ 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:

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 the ring pairs over the plate

The plates run from the axis to the rim, so (2) is integrated from $\rho = 0$ to $R$ :

\int_0^R\frac{\rho\,d\rho}{(\rho^2 + x^2)^{3/2}} = \left[-\frac{1}{\sqrt{\rho^2 + x^2}}\right]_0^R = \frac{1}{|x|} - \frac{1}{\sqrt{x^2 + R^2}}

Multiplying by the factor $\sigma l x/(2\varepsilon_0)$ , with $x/|x| = \pm 1$ on the two sides:

\varphi = \frac{\sigma l}{2\varepsilon_0}\left(\frac{x}{|x|} - \frac{x}{\sqrt{x^2 + R^2}}\right)

In terms of the distance $x > 0$ from the plates on either side:

\varphi = \pm\frac{\sigma l}{2\varepsilon_0}\left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right)

(3)

The upper sign holds on the side of the positive plate, the lower on the side of the negative plate.

Step 4The field on the axis

By symmetry the field on the axis lies along it, and its magnitude is the slope of (3):

\begin{aligned} \frac{d}{dx}\left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) &= -\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 = \frac{\sigma l R^2}{2\varepsilon_0(x^2 + R^2)^{3/2}}

(4)

On both sides the field points the same way, along the axis from the negative plate’s side toward the positive plate’s side: away from the capacitor beyond the positive plate, toward it beyond the negative plate.

Step 5Far from the capacitor

For $x \gg R$ expand the root, $x/\sqrt{x^2 + R^2} \approx 1 - R^2/(2x^2)$ , in (3), and drop $R^2$ beside $x^2$ in (4). With the dipole moment of the capacitor,

p = \sigma\pi R^2\cdot l

the charge of a plate times the gap,

\begin{gathered}\varphi \approx \pm\frac{\sigma l R^2}{4\varepsilon_0 x^2} = \pm\frac{p}{4\pi\varepsilon_0 x^2}\\ E \approx \frac{\sigma l R^2}{2\varepsilon_0 x^3} = \frac{p}{2\pi\varepsilon_0 x^3}\end{gathered}

(5)

These are the axial potential and field of a point dipole $p$ .

Check

Close to the plates, $l \ll x \ll R$ , (3) tends to $\pm\sigma l/(2\varepsilon_0)$ , so the two sides differ by $\sigma l/\varepsilon_0$ : the field $\sigma/\varepsilon_0$ inside the gap times its width $l$ , the voltage across the capacitor, as it must be. Far away, the capacitor must look like its dipole moment alone: on the axis a point dipole gives $p/(4\pi\varepsilon_0 x^2)$ and $2p/(4\pi\varepsilon_0 x^3)$ , which is (5) exactly.

Remark

The potential (3) is the surface dipole density $\sigma l$ times the solid angle the plate subtends at $P$ , divided by $4\pi\varepsilon_0$ : that solid angle is $2\pi(1 - x/\sqrt{x^2 + R^2})$ . The rule holds for a double layer of any shape, which is why the outside of an ideal capacitor is never quite field free: near the plates the field outside is weak, $\sigma l/(2\varepsilon_0 R)$ at most, but it decays only as a dipole field far away.