Irodov 3.33: From Infinite Plane to Point Charge – Amit & Deepali Sinha Foundation

Problem Irodov 3.33

A very thin flat disc of radius $R$ in vacuum carries charge spread uniformly over it with surface density $\sigma$ . Find the electric potential and the electric field on the disc’s axis as functions of the distance $l$ from its centre, and examine both results for $l \to 0$ and for $l \gg R$ .

The uniformly charged disc of radius R and a point P on its axis at distance l from the centre. — The uniformly charged disc of radius $R$ and a point $P$ on its axis at distance $l$ from the centre.

Given

radius $R$
uniform surface density $\sigma$
vacuum

Find

\varphi(l)

and

E(l)

on the axis; their forms for

l \to 0

and

l \gg R

Solution

Step 1The potential of one ring

Cut the disc into thin rings. The ring of radius $r$ and width $dr$ carries charge

dq = \sigma\cdot 2\pi r\,dr

and every part of it is the same distance $\sqrt{r^2 + l^2}$ from $P$ , so its potential at $P$ is

d\varphi = \frac{dq}{4\pi\varepsilon_0\sqrt{r^2 + l^2}} = \frac{\sigma}{2\varepsilon_0}\,\frac{r\,dr}{\sqrt{r^2 + l^2}} \tag{1}

The ring of radius r and width dr, and its field dE at P, which by the ring's symmetry lies along the axis. — The ring of radius $r$ and width $dr$ , and its field $d\vect{E}$ at $P$ , which by the ring’s symmetry lies along the axis.

Step 2Summing the rings

Integrating (1) over the disc, from $r = 0$ to $r = R$ , with $l \geq 0$ :

\varphi = \frac{\sigma}{2\varepsilon_0}\int_0^R\frac{r\,dr}{\sqrt{r^2 + l^2}} = \frac{\sigma}{2\varepsilon_0}\Big[\sqrt{r^2 + l^2}\,\Big]_0^R

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

(2)

Step 3The field from the potential

On the axis the field is axial by symmetry, and its component along $l$ is minus the derivative of (2):

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

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

(3)

For $\sigma > 0$ the field points along the axis away from the disc, on either side of it.

Step 4Close to the disc and far from it

Near the centre, $l \to 0$ , the square roots tend to $R$ and the fraction $l/\sqrt{l^2+R^2}$ to zero:

l \to 0:\quad \varphi = \frac{\sigma R}{2\varepsilon_0}, \quad E = \frac{\sigma}{2\varepsilon_0}

(4)

Far away, $l \gg R$ , expanding the square roots to first order in $R^2/l^2$ :

\begin{aligned} \sqrt{l^2 + R^2} - l &\approx \frac{R^2}{2l}\\ 1 - \frac{l}{\sqrt{l^2 + R^2}} &\approx \frac{R^2}{2l^2} \end{aligned}

With the disc’s charge $q = \sigma\pi R^2$ these give

l \gg R:\quad \varphi \approx \frac{q}{4\pi\varepsilon_0 l}, \quad E \approx \frac{q}{4\pi\varepsilon_0 l^2}

(5)

The field E (crimson, above) and the potential φ (teal, below) on the axis against l. Both start from their values at the centre, σ/2ε_0 (the plane's field) and σ R/2ε_0, and fall steadily as l grows past R. — The field $E$ (crimson, above) and the potential $\varphi$ (teal, below) on the axis against $l$ . Both start from their values at the centre, $\sigma/2\varepsilon_0$ (the plane’s field) and $\sigma R/2\varepsilon_0$ , and fall steadily as $l$ grows past $R$ .

Check

The two limits are the right ones. Close to the disc, $l \ll R$ , the disc looks like an infinite plane, whose field is $\sigma/2\varepsilon_0$ on each side, and that is what (3) gives. Far away the disc looks like a point charge $q$ , and both results reduce to Coulomb’s forms. As $R \to \infty$ , $E$ tends to $\sigma/2\varepsilon_0$ at every $l$ , the plane’s uniform field, while $\varphi$ grows without bound, as an infinite plane’s potential must when referred to infinity.

Remark

The field near the disc does not depend on $R$ at all, but the potential at its centre, $\sigma R/2\varepsilon_0$ , grows with the disc’s size. That is the plane’s paradox in miniature: the field is a local quantity set by the nearby charge, the potential a global one that adds up all of it. The bracket in (3) is $1 - \cos\alpha$ , where $\alpha$ is the half-angle the disc’s rim subtends at $P$ , so the field is $\sigma/4\pi\varepsilon_0$ times the solid angle $2\pi(1 - \cos\alpha)$ the disc fills as seen from $P$ .