Irodov 3.34: Polar Coordinates from the Rim – Amit & Deepali Sinha Foundation

Problem Irodov 3.34

A thin disc of radius $R$ carries charge spread uniformly over it with surface density $\sigma$ . Find the electric potential $\varphi$ at a point on the disc’s edge.

The uniformly charged disc of radius R seen face-on, and the point A on its edge where the potential is wanted. — The uniformly charged disc of radius $R$ seen face-on, and the point $A$ on its edge where the potential is wanted.

Given

radius $R$
uniform surface density $\sigma$

Find

the potential

\varphi

at a point

A

of the edge

Solution

Step 1Coordinates centred on the edge point

The potential at $A$ is the sum of $\sigma\,dS/(4\pi\varepsilon_0 r)$ over the disc, where $r$ is the distance from $A$ . The $1/r$ suggests polar coordinates centred on $A$ itself: the area element is then $dS = r\,dr\,d\theta$ , and the $r$ in it cancels the $1/r$ .

Polar coordinates about the edge point A, with θ measured from the diameter through A. The chord at angle θ has length 2Rθ, and the element dS lies on it at distance r from A. — Polar coordinates about the edge point $A$ , with $\theta$ measured from the diameter through $A$ . The chord at angle $\theta$ has length $2R\cos\theta$ , and the element $dS$ lies on it at distance $r$ from $A$ .

The chord from $A$ at angle $\theta$ to the diameter, together with the diameter, makes a triangle whose angle at the chord’s far end is a right angle (an angle in a semicircle), so its length is

\ell(\theta) = 2R\cos\theta, \qquad -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \tag{1}

Step 2The integral

Writing the sum with $dS = r\,dr\,d\theta$ , integrating $r$ along each chord from $0$ to (1), then $\theta$ across the disc:

\begin{aligned} \varphi &= \frac{\sigma}{4\pi\varepsilon_0}\int_{-\pi/2}^{\pi/2}\int_0^{2R\cos\theta}\frac{r\,dr}{r}\,d\theta\\ &= \frac{\sigma}{4\pi\varepsilon_0}\int_{-\pi/2}^{\pi/2}2R\cos\theta\,d\theta\\ &= \frac{\sigma}{4\pi\varepsilon_0}\cdot 2R\Big[\sin\theta\Big]_{-\pi/2}^{\pi/2} = \frac{\sigma}{4\pi\varepsilon_0}\cdot 4R \end{aligned}

\varphi = \frac{\sigma R}{\pi\varepsilon_0}

(2)

Check

Compare with the centre, where every ring of radius $r$ and width $dr$ is at distance $r$ :

\varphi_{\text{centre}} = \frac{\sigma}{4\pi\varepsilon_0}\int_0^R\frac{2\pi r\,dr}{r} = \frac{\sigma R}{2\varepsilon_0}

The ratio of edge to centre is $(1/\pi)/(1/2) = 2/\pi \approx 0.64$ : lower at the edge, as it must be, since the charge lies on average farther from a point of the edge than from the centre. Units: $\sigma R/\varepsilon_0$ is charge per length over $\varepsilon_0$ , a potential.

Remark

The potential of a uniformly charged disc is not the same over its surface, so a uniformly charged disc is not a conductor’s charge distribution: on a conducting disc the charge would flow outward until the potential evened out, piling up toward the rim (the conducting disc’s density grows as $1/\sqrt{R^2 - r^2}$ near the edge). For a field point in the plane of any uniformly charged flat region, the same trick of centring the coordinates on the point turns the potential into a sum of chord lengths.