Irodov 3.32: Half a Shell, a Quarter Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.32

A thin hemispherical shell of radius $R$ carries charge spread uniformly over its surface with density $\sigma$ . Find the electric potential and the magnitude of the electric field at the centre of the sphere the hemisphere belongs to.

The charged hemisphere in section through its symmetry axis: surface density σ, radius R, centre O. — The charged hemisphere in section through its symmetry axis: surface density $\sigma$ , radius $R$ , centre $O$ .

Given

radius $R$
uniform surface density $\sigma$

Find

\varphi

and

E

at the centre

O

Solution

Step 1The potential: every charge at the same distance

The potential is a scalar sum of $dq/(4\pi\varepsilon_0 r)$ , and every element of the hemisphere is at the same distance $R$ from $O$ , so only the total charge matters. The hemisphere’s area is $2\pi R^2$ , so its charge is $q = 2\pi R^2\sigma$ :

\varphi = \frac{q}{4\pi\varepsilon_0 R} = \frac{2\pi R^2\sigma}{4\pi\varepsilon_0 R}

\varphi = \frac{\sigma R}{2\varepsilon_0}

(1)

Step 2The field of one band

For the field the directions matter. Cut the hemisphere into narrow bands about its symmetry axis, the band at polar angle $\theta$ from the axis having width $R\,d\theta$ and circumference $2\pi R\sin\theta$ :

dq = \sigma\cdot 2\pi R\sin\theta\cdot R\,d\theta \tag{2}

The band at polar angle θ, seen in section as one element dq, a distance R from O; its field dE at O and the axial component dE_z. — The band at polar angle $\theta$ , seen in section as one element $dq$ , a distance $R$ from $O$ ; its field $d\vect{E}$ at $O$ and the axial component $dE_z$ .

Every element of the band is a distance $R$ from $O$ , and its field there makes the angle $\theta$ with the axis. Round the band the components across the axis cancel in pairs, and the axial components add:

dE_z = \frac{dq}{4\pi\varepsilon_0 R^2}\cos\theta \tag{3}

Step 3Summing the bands

Substituting (2) into (3) and integrating from the pole, $\theta = 0$ , to the rim, $\theta = \pi/2$ :

\begin{aligned} E &= \frac{2\pi R^2\sigma}{4\pi\varepsilon_0 R^2}\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta\\ &= \frac{\sigma}{2\varepsilon_0}\left[\frac{\sin^2\theta}{2}\right]_0^{\pi/2} \end{aligned}

E = \frac{\sigma}{4\varepsilon_0}

(4)

For $\sigma > 0$ the field at $O$ points along the axis away from the hemisphere, out through its open face.

Check

The field, by a second route that skips the band integral: an area element $dA$ contributes $\sigma\,dA\cos\theta/(4\pi\varepsilon_0R^2)$ along the axis, and $dA\cos\theta$ is its projection onto the plane of the rim. The projections of the whole hemisphere tile the base disc of area $\pi R^2$ once, so

E = \frac{\sigma}{4\pi\varepsilon_0R^2}\cdot\pi R^2 = \frac{\sigma}{4\varepsilon_0}

again. The potential checks against a whole sphere of charge $4\pi R^2\sigma$ , whose potential at the centre, $\sigma R/\varepsilon_0$ , the standard result, is twice the answer, as the two halves require. Units: $\sigma/\varepsilon_0$ is a field.

Remark

The projection argument of the check works for any piece of the hemisphere: its axial field at $O$ is $\sigma/(4\pi\varepsilon_0R^2)$ times the area of its shadow on the plane of the rim. A cap of half-angle $\alpha$ about the axis casts the disc $\pi R^2\sin^2\alpha$ and gives $(\sigma/4\varepsilon_0)\sin^2\alpha$ ; the field of the hemisphere is the cap with $\alpha = \pi/2$ , and a whole sphere gives zero because its lower half casts the same shadow with the opposite sign of $\cos\theta$ .