Irodov 3.28: A Cavity Holds a Uniform Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.28

A ball carries charge spread uniformly through its volume with density $\rho$ , except for a spherical cavity cut out of it. The centre of the cavity is displaced from the centre of the ball by the vector $\vect{a}$ . Taking the permittivity to be unity everywhere, find the electric field $\vect{E}$ inside the cavity.

Section through both centres: the charged ball, centre O, and the empty cavity, centre O', displaced from O by a. — Section through both centres: the charged ball, centre $O$ , and the empty cavity, centre $O'$ , displaced from $O$ by $\vect{a}$ .

Given

$\rho$
$\vect{a}$
$\varepsilon = 1$

Find

\vect{E}

inside the cavity

Solution

Step 1Fill the cavity twice over

The charge has no spherical symmetry, so Gauss’s law cannot be applied to it directly. But a cavity is the same as a region holding charge $+\rho$ and charge $-\rho$ together. So the actual distribution is the sum of two complete, uniformly charged balls:

\text{ball with cavity} = \bigl(\text{full ball of } {+\rho}\text{, centre } O\bigr) + \bigl(\text{cavity-sized ball of } {-\rho}\text{, centre } O'\bigr)

Each of these has spherical symmetry, and their fields add.

Step 2The field inside a uniform ball

For a full ball of density $\rho$ , symmetry makes $\vect E$ radial and the same in magnitude all over a concentric sphere of radius $r$ inside it, which encloses the charge $4\pi r^3\rho/3$ :

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

With $\vect r$ the position vector from the ball’s centre, the field points along $\vect r$ for $\rho > 0$ , so in vector form

\vect E = \frac{\rho}{3\varepsilon_0}\,\vect r \tag{1}

This holds at every point inside that ball, and points inside the cavity are inside both balls.

Step 3Superpose at a point of the cavity

A point P inside the cavity, with its position vectors r from O and r' from O'. The triangle closes: r = a + r'. — A point $P$ inside the cavity, with its position vectors $\vect r$ from $O$ and $\vect r'$ from $O'$ . The triangle closes: $\vect r = \vect a + \vect r'$ .

Take a point $P$ in the cavity, at $\vect r$ from $O$ and $\vect r'$ from $O'$ . Applying (1) to each ball, with density $+\rho$ about $O$ and $-\rho$ about $O'$ , the total is

\vect E = \frac{\rho}{3\varepsilon_0}\,\vect r + \frac{-\rho}{3\varepsilon_0}\,\vect r' = \frac{\rho}{3\varepsilon_0}\,(\vect r - \vect r')

The figure shows that the two position vectors differ by the displacement of the centres:

\vect r - \vect r' = \vect a

so the position of $P$ drops out:

\vect E = \frac{\rho}{3\varepsilon_0}\,\vect a

(2)

The field is the same at every point of the cavity, with magnitude $\rho a/(3\varepsilon_0)$ , directed along $\vect a$ (from the ball’s centre toward the cavity’s centre) for $\rho > 0$ .

The field inside the cavity, computed from the two balls at two points: the arrows are equal and both point along a. — The field inside the cavity, computed from the two balls at two points: the arrows are equal and both point along $\vect a$ .

Check

Limit $\vect a \to 0$ : the cavity becomes concentric, and the answer gives $\vect E = 0$ , which Gauss’s law confirms directly, since a concentric sphere inside the cavity encloses no charge. Units: $\rho a/\varepsilon_0$ is a charge per area over $\varepsilon_0$ , the units of the field of a plane sheet $\sigma/(2\varepsilon_0)$ .

Remark

The result does not depend on the radius of either sphere, nor on whether the cavity reaches the ball’s surface, as long as the point stays inside both. The same superposition describes two overlapping balls of densities $+\rho$ and $-\rho$ : in their overlap the field is again uniform, pointing from the positive centre toward the negative one. Shrinking the shift while raising $\rho$ turns this into a uniformly polarised sphere, whose interior field is uniform for exactly this reason.