Irodov 3.17: Two Shifted Balls Make a Shell – Amit & Deepali Sinha Foundation

Problem Irodov 3.17

A sphere of radius $R$ carries surface charge whose density depends on the polar angle $\theta$ as $\sigma = \sigma_0\cos\theta$ , with $\sigma_0$ a positive constant. Show that this distribution is what remains when two uniformly charged solid balls of radius $R$ , with charges equal in size and opposite in sign, are displaced slightly relative to each other. Using that picture, find the electric field vector inside the sphere.

The sphere of radius R in section through the z axis, with σ = σ_0θ: positive on the upper half, negative on the lower, and densest at the poles. — The sphere of radius $R$ in section through the $z$ axis, with $\sigma = \sigma_0\cos\theta$ : positive on the upper half, negative on the lower, and densest at the poles.

Given

sphere of radius $R$
$\sigma = \sigma_0\cos\theta$ , $\sigma_0 > 0$ , $\theta$ from the $z$ axis

Find

show the two-ball representation; the field vector

\vect E

inside

Solution

Step 1Two balls leave a cosine layer

Take a ball of radius $R$ with uniform density $-\rho$ centered at $O$ , and an identical ball of density $+\rho$ whose center is displaced from $O$ by a small vector $\vect l = l\,\vect k$ along $z$ . Where they overlap the charges cancel; only two thin crescents remain charged.

The ball of density -ρ and the ball of density +ρ, its center displaced by l. The overlap is neutral; the crescents, drawn with a large l for clarity, carry the charge. — The ball of density $-\rho$ and the ball of density $+\rho$ , its center displaced by $\vect l$ . The overlap is neutral; the crescents, drawn with a large $l$ for clarity, carry the charge.

Follow a direction $\vect n$ from $O$ at polar angle $\theta$ . The negative ball’s surface lies at distance $R$ , and the positive ball’s surface, whose center is shifted by $\vect l$ , lies farther by the projection of $\vect l$ on $\vect n$ , to first order in $l$ :

R + \vect l\cdot\vect n = R + l\cos\theta

So along $\vect n$ there is an uncompensated layer of thickness $l\cos\theta$ , positive where $\cos\theta > 0$ and negative where $\cos\theta < 0$ . A layer of density $\rho$ and thickness $l\cos\theta$ , seen from outside its thickness, is a surface charge of density

\sigma = \rho\, l\cos\theta \tag{1}

This is the given distribution when the shift and the density are chosen so that

\rho\, l = \sigma_0

(2)

Step 2The field inside a uniform ball

For one ball of density $\rho$ , by symmetry the field is radial and the same at every point a distance $r$ from its center; Gauss’s law on the concentric sphere through the point gives

E\cdot 4\pi r^2 = \frac{\rho\cdot\frac{4}{3}\pi r^3}{\varepsilon_0}

so inside the ball, with $\vect r$ drawn from its center to the point,

\vect E = \frac{\rho\,\vect r}{3\varepsilon_0} \tag{3}

Step 3Superpose the two balls inside

A point inside the sphere is inside both balls. Let $\vect r_-$ and $\vect r_+$ be its position vectors from the centers of the negative and the positive ball. The positive center is the negative one shifted by $\vect l$ , so

\vect r_- - \vect r_+ = \vect l \tag{4}

Adding the two fields (3), the second with density $-\rho$ , and using (4):

\vect E = \frac{\rho\,\vect r_+}{3\varepsilon_0} - \frac{\rho\,\vect r_-}{3\varepsilon_0} = -\frac{\rho\,(\vect r_- - \vect r_+)}{3\varepsilon_0} = -\frac{\rho\,\vect l}{3\varepsilon_0}

With $\vect l = l\,\vect k$ and (2):

\vect E = -\frac{\sigma_0}{3\varepsilon_0}\,\vect k

(5)

The field inside has magnitude $\sigma_0/(3\varepsilon_0)$ and points along $-z$ , from the positive pole toward the negative one. It does not depend on the point: the field is uniform throughout the interior.

The total field at three interior points, each summed over the whole sphere: equal and along -z everywhere inside. — The total field at three interior points, each summed over the whole sphere: equal and along $-z$ everywhere inside.

Check

At the center, directly. The band at $\theta$ carries $dq = \sigma_0\cos\theta\cdot 2\pi R^2\sin\theta\,d\theta$ , and every point of it is the distance $R$ from $O$ , so

E_z = -\frac{1}{4\pi\varepsilon_0R^2}\int_0^{\pi}2\pi R^2\sigma_0\cos^2\theta\sin\theta\,d\theta = -\frac{\sigma_0}{2\varepsilon_0}\cdot\frac{2}{3} = -\frac{\sigma_0}{3\varepsilon_0},

the same as (5). Units: $\sigma_0/\varepsilon_0$ is a field.

Remark

The two shifted balls are the model of a uniformly polarized ball: the polarization $\vect P = \rho\,\vect l$ leaves the bound surface charge $\sigma' = P\cos\theta$ and, by (5), the uniform depolarizing field $-\vect P/(3\varepsilon_0)$ inside. Outside, each ball acts as a point charge at its own center, so in the limit $l \to 0$ the sphere’s field outside is exactly that of a point dipole of moment $(4\pi R^3\sigma_0/3)\,\vect k$ at $O$ , at every outside point, not only far away.