Irodov 3.88: Bound Charge Seen Through a Shadow – Amit & Deepali Sinha Foundation

Problem Irodov 3.88

Inside a ball of uniform isotropic dielectric with permittivity $\varepsilon = 5.00$ and radius $R = 3.0$ cm, the electric field is uniform, of strength $E = 100$ V/m. Find the largest surface density of the bound charge on the ball and the total bound charge of either sign.

The dielectric ball of radius R with the uniform field E inside it. — The dielectric ball of radius $R$ with the uniform field $\vect E$ inside it.

Given

$E = 100$ V/m inside
$\varepsilon = 5.00$
$R = 3.0$ cm

Find

the largest

\sigma'

, and the bound charge

q'

of one sign

Solution

Step 1The polarization

In a linear isotropic dielectric the polarization follows the field, $\vect P = \varepsilon_0(\varepsilon - 1)\vect E$ . The field inside is uniform, so the polarization is uniform too, along $\vect E$ , of magnitude

P = \varepsilon_0(\varepsilon - 1)E \tag{1}

Step 2The surface density and its maximum

The uniform polarization and the bound charge it leaves on the surface: positive on the hemisphere E points to, negative on the other, densest at the poles and thinning to zero at the equator. The angle θ is measured from the direction of E. — The uniform polarization and the bound charge it leaves on the surface: positive on the hemisphere $\vect E$ points to, negative on the other, densest at the poles and thinning to zero at the equator. The angle $\theta$ is measured from the direction of $\vect E$ .

The bound surface density is the normal component of $\vect P$ at the surface. Measuring the polar angle $\theta$ of a surface point from the direction of $\vect E$ , the outward normal makes the angle $\theta$ with $\vect P$ :

\sigma' = \vect P\cdot\vect n = P\cos\theta \tag{2}

It is largest at the pole $\theta = 0$ , where $\sigma' = P$ . With (1), numerically:

\sigma'_{\mathrm{max}} = 8.85\times10^{-12}\times(5.00 - 1)\times100\text{ C/m}^2

\sigma'_{\mathrm{max}} = \varepsilon_0(\varepsilon - 1)E = 3.5\text{ nC/m}^2

(3)

Step 3The bound charge on one hemisphere

The positive charge covers the hemisphere $0 \leq \theta \leq \pi/2$ . Integrating (2) over it, with the ring of area $dS = 2\pi R^2\sin\theta\,d\theta$ :

\begin{aligned} q' &= \int_0^{\pi/2}P\cos\theta\cdot 2\pi R^2\sin\theta\,d\theta = 2\pi R^2P\left[\frac{\sin^2\theta}{2}\right]_0^{\pi/2}\\ &= \pi R^2P \end{aligned}

Substituting (1) and the numbers:

q' = \pi\times(0.030)^2\times8.85\times10^{-12}\times4.00\times100\text{ C}

q' = \pi R^2\varepsilon_0(\varepsilon - 1)E = 10\text{ pC}

(4)

The negative hemisphere carries the same amount, so the ball stays neutral.

Check

Dimensions: $\varepsilon_0E$ is a surface charge density (F/m times V/m gives C/m $^2$ ), and $\pi R^2$ times it a charge. As $\varepsilon \to 1$ both results vanish, as they must with no medium. Independently, the surface charge must carry the dipole moment of the polarized ball, $P$ times its volume. The moment of $\sigma' = P\cos\theta$ about the center, along $\vect E$ , is

\int R\cos\theta\,\sigma'\,dS = PR^3\int_0^{\pi}\cos^2\theta\,2\pi\sin\theta\,d\theta = \frac{4}{3}\pi R^3P,

which is $P$ times the volume, confirming (2).

Remark

A uniform field inside a dielectric ball is exactly what a uniform external field produces: the surface charge $P\cos\theta$ of a uniformly polarized ball makes the uniform inside field $-\vect P/(3\varepsilon_0)$ , which partly cancels the applied field $\vect E_0$ , and the two combine to $\vect E = 3\vect E_0/(\varepsilon + 2)$ . The field inside here therefore corresponds to an applied field $E_0 = (\varepsilon + 2)E/3 \approx 230$ V/m far from the ball.