Irodov 3.96: Two Shifted Balls Make One Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.96

A dielectric ball carries a uniform, permanent polarization $\vect P$ . Such a ball can be pictured as the result of shifting all of its positive charge by a small distance relative to all of its negative charge. Using that picture:

find the electric field $\vect E$ inside the ball;
show that outside the ball the field is that of a dipole at the ball’s center, with potential $\varphi = \vect p_0\cdot\vect r/4\pi\varepsilon_0 r^3$ , where $\vect p_0$ is the electric moment of the ball and $\vect r$ is the radius vector from its center.

The ball with its uniform, permanent polarization P. — The ball with its uniform, permanent polarization $\vect P$ .

Given

uniform permanent polarization $\vect P$ of the ball

Find

(a)

\vect E

inside · (b) the outside field is a dipole’s,

\varphi = \vect p_0\cdot\vect r/4\pi\varepsilon_0 r^3

Solution

Step 1Two charged balls, slightly apart

Let the ball have radius $R$ , a symbol the answers will not need. Its positive charge, of density $+\rho$ , fills one ball centered at $O_+$ , and its negative charge, $-\rho$ , an equal ball centered at $O_-$ , with $\vect a$ the small shift from $O_-$ to $O_+$ . A volume $V$ of the pair carries the dipole moment $\rho V\vect a$ , so the polarization, the moment per volume, is

\vect P = \rho\,\vect a \tag{1}

The polarized ball as two balls of charge density +ρ and -ρ whose centers are shifted by a (drawn far larger than it is). Where they overlap the charge cancels; only the thin crescents carry charge. — The polarized ball as two balls of charge density $+\rho$ and $-\rho$ whose centers are shifted by $\vect a$ (drawn far larger than it is). Where they overlap the charge cancels; only the thin crescents carry charge.

Step 2The field inside one uniformly charged ball

By symmetry the field of a ball of density $\rho$ is radial from its center and the same all over a concentric sphere, so Gauss’s theorem over the sphere through a point at $\vect s$ from the center gives

E\cdot 4\pi s^2 = \frac{1}{\varepsilon_0}\,\rho\,\frac{4}{3}\pi s^3

\vect E = \frac{\rho\,\vect s}{3\varepsilon_0} \tag{2}

Step 3(a) The two fields add to a uniform field

A point inside both balls lies at $\vect s_+$ from $O_+$ and at $\vect s_-$ from $O_-$ , and the two differ by the shift of the centers, $\vect s_- - \vect s_+ = \vect a$ . Applying (2) to each ball, the second with density $-\rho$ :

\vect E = \frac{\rho\,\vect s_+}{3\varepsilon_0} - \frac{\rho\,\vect s_-}{3\varepsilon_0} = -\frac{\rho\,(\vect s_- - \vect s_+)}{3\varepsilon_0} = -\frac{\rho\,\vect a}{3\varepsilon_0}

Substituting (1):

\vect E = -\frac{\vect P}{3\varepsilon_0}

(3)

The point dropped out, so the field is the same everywhere inside, and it points against $\vect P$ .

The uniform field inside: E points against P, with magnitude P/3ε_0. The bound charge sits in the crescents, positive on the cap P points out of, thinning to none at the equator. — The uniform field inside: $\vect E$ points against $\vect P$ , with magnitude $P/3\varepsilon_0$ . The bound charge sits in the crescents, positive on the cap $\vect P$ points out of, thinning to none at the equator.

Step 4(b) Outside, each ball is a point charge

Outside both balls, each of the two charged balls of the model figure in Step 1 acts as a point charge $\pm q$ at its center, with $q = 4\pi R^3\rho/3$ . With $\vect r$ from the midpoint of $O_-O_+$ , which for a vanishing shift is the ball’s center:

\varphi = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{|\vect r - \vect a/2|} - \frac{1}{|\vect r + \vect a/2|}\right)

To first order in $a$ :

\frac{1}{|\vect r \mp \vect a/2|} = \frac{1}{r} \pm \frac{\vect a\cdot\vect r}{2r^3}

Substituting, the $1/r$ terms cancel and the two corrections add:

\varphi = \frac{q\,\vect a\cdot\vect r}{4\pi\varepsilon_0 r^3}

The moment of the pair is $q\vect a$ ; by (1) it is the ball’s volume times $\vect P$ .

\vect p_0 = q\,\vect a = \frac{4}{3}\pi R^3\rho\,\vect a = \frac{4}{3}\pi R^3\,\vect P

\varphi = \frac{\vect p_0\cdot\vect r}{4\pi\varepsilon_0 r^3}

(4)

This is the potential of a point dipole $\vect p_0$ at the center, so outside the ball the field is exactly a dipole’s.

Check

Two independent tests. The potential must be continuous at the surface. Inside, (3) is minus the gradient of $\varphi_{\mathrm{in}} = \vect P\cdot\vect r/3\varepsilon_0$ , which is zero at the center as $\varphi$ is by symmetry; outside, at $r = R$ , (4) gives

\frac{\frac{4}{3}\pi R^3\,\vect P\cdot\vect r}{4\pi\varepsilon_0 R^3} = \frac{\vect P\cdot\vect r}{3\varepsilon_0},

the same. And the field at the center can be summed directly from the surface charge $\sigma' = P\cos\theta$ : rings at angle $\theta$ contribute along $-\vect P$

E = \int_0^\pi \frac{P\cos\theta\cdot 2\pi R^2\sin\theta\,d\theta}{4\pi\varepsilon_0 R^2}\cos\theta = \frac{P}{2\varepsilon_0}\cdot\frac{2}{3} = \frac{P}{3\varepsilon_0}

in agreement with (3).

Remark

Inside the ball $\vect D = \varepsilon_0\vect E + \vect P = 2\vect P/3$ , along $\vect P$ , while $\vect E$ runs against it: the field and the displacement point opposite ways, which no linear dielectric could do but a permanent polarization can. The uniform interior is a gift of the sphere’s shape, shared by every ellipsoid; a cube polarized uniformly has a non-uniform field inside.