Irodov 3.16: A Cosine Shell Seen from Inside – Amit & Deepali Sinha Foundation

Problem Irodov 3.16

A spherical surface of radius $r$ carries surface charge with density $\sigma = \vect{a}\cdot\vect{r}$ , where $\vect{a}$ is a constant vector and $\vect{r}$ is the position vector of a point on the sphere measured from its center. Find the electric field vector at the center of the sphere.

The sphere of radius r in section through its center O along a, with σ = a· r: positive on the side a points to, negative opposite. — The sphere of radius $r$ in section through its center $O$ along $\vect a$ , with $\sigma = \vect a\cdot\vect r$ : positive on the side $\vect a$ points to, negative opposite.

Given

sphere of radius $r$
$\sigma = \vect a\cdot\vect r$ , $\vect a$ constant

Find

the field vector

\vect E

at the center

O

Solution

Step 1Cut the sphere into bands

Take the $x$ axis along $\vect a$ and let $\theta$ be the polar angle of a point of the sphere measured from it. On the sphere $\lvert\vect r\rvert = r$ , so the density depends on $\theta$ alone:

\sigma = a r\cos\theta \tag{1}

The band between $\theta$ and $\theta + d\theta$ has radius $r\sin\theta$ and width $r\,d\theta$ , so with (1) its charge is

dq = \sigma\cdot 2\pi r\sin\theta\cdot r\,d\theta = 2\pi a r^3\cos\theta\sin\theta\,d\theta \tag{2}

A band at polar angle θ from a, seen in section as the element dq, and its field d E at O with the component dE_x along a. — A band at polar angle $\theta$ from $\vect a$ , seen in section as the element $dq$ , and its field $d\vect E$ at $O$ with the component $dE_x$ along $\vect a$ .

Step 2The field of one band at the center

Every point of the band is the distance $r$ from $O$ , and the line from it to $O$ makes the angle $\theta$ with the $x$ axis. The components across the axis cancel round the band, and a positive band pushes the field at $O$ toward $-x$ , so the band contributes

dE_x = -\frac{1}{4\pi\varepsilon_0}\,\frac{dq}{r^2}\cos\theta

Substituting (2):

dE_x = -\frac{a r}{2\varepsilon_0}\cos^2\theta\sin\theta\,d\theta \tag{3}

Step 3Sum over the sphere

Integrating (3) from pole to pole:

\begin{aligned} E_x &= -\frac{a r}{2\varepsilon_0}\int_0^{\pi}\cos^2\theta\sin\theta\,d\theta = -\frac{a r}{2\varepsilon_0}\left[-\frac{\cos^3\theta}{3}\right]_0^{\pi}\\ &= -\frac{a r}{2\varepsilon_0}\cdot\frac{2}{3} = -\frac{a r}{3\varepsilon_0} \end{aligned}

The field lies along $\vect a$ , so the vector is

\vect E = -\frac{r\,\vect a}{3\varepsilon_0}

(4)

The field at the center has magnitude $ar/(3\varepsilon_0)$ and points against $\vect a$ , from the positive half of the sphere toward the negative half.

The total field E at the center, summed over the whole sphere: it points against a. — The total field $\vect E$ at the center, summed over the whole sphere: it points against $\vect a$ .

Check

Units: $a r$ is a surface density by (1), and a surface density over $\varepsilon_0$ is a field. Symmetry: reflecting through the equatorial plane turns $\theta$ into $\pi - \theta$ and flips the sign of $\sigma$ , so the negative hemisphere must push the field at $O$ exactly as hard, and the same way, as the positive one. Splitting the integral confirms it: $\int_0^{\pi/2}\cos^2\theta\sin\theta\,d\theta$ and $\int_{\pi/2}^{\pi}\cos^2\theta\sin\theta\,d\theta$ are both $1/3$ .

Remark

This $\cos\theta$ shell is the surface of a uniformly polarized ball. Its dipole moment is $\vect p = (4\pi r^4/3)\,\vect a$ , so the field at the center can be written $-\vect p/(4\pi\varepsilon_0 r^3)$ , and that value holds not just at the center but everywhere inside the sphere, where the field is uniform. Outside, the same shell has the field of the point dipole $\vect p$ .