Irodov 3.98: A Dielectric Ball Weakens Its Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.98

A ball of uniform dielectric with permittivity $\varepsilon$ is placed in a uniform electric field $\vect E_0$ , and it becomes uniformly polarized. Using the field of a uniformly polarized ball (Problem 3.96), find the field $\vect E$ inside the ball and the polarization $\vect P$ of the dielectric.

The dielectric ball of permittivity ε placed in the uniform field E_0. — The dielectric ball of permittivity $\varepsilon$ placed in the uniform field $\vect E_0$ .

Given

uniform field $\vect E_0$
dielectric ball, permittivity $\varepsilon$
the ball polarizes uniformly

Find

\vect E

and

\vect P

inside the ball

Solution

Step 1The field inside as a superposition

The field inside is the applied $\vect E_0$ plus the field of the ball’s own bound charge. The polarization is uniform, so by Problem 3.96 the bound charge adds the uniform field $-\vect P/3\varepsilon_0$ inside.

\vect E = \vect E_0 - \frac{\vect P}{3\varepsilon_0} \tag{1}

Step 2The dielectric closes the loop

The polarization is itself set by the field inside, through the linear relation of an isotropic dielectric.

\vect P = \varepsilon_0(\varepsilon - 1)\vect E \tag{2}

Substituting (2) into (1) and collecting the terms in $\vect E$ :

\begin{aligned} \vect E &= \vect E_0 - \frac{\varepsilon - 1}{3}\,\vect E\\ \frac{\varepsilon + 2}{3}\,\vect E &= \vect E_0 \end{aligned}

\vect E = \frac{3\vect E_0}{\varepsilon + 2}

(3)

The field inside is uniform, parallel to $\vect E_0$ , and weaker by the factor $3/(\varepsilon+2)$ .

The field lines for ε = 3. Inside the ball the field is uniform and weaker, so only some of the lines that reach the ball go through; the others end on negative bound charge on the near side, and as many start again from positive bound charge on the far side. — The field lines for $\varepsilon = 3$ . Inside the ball the field is uniform and weaker, so only some of the lines that reach the ball go through; the others end on negative bound charge on the near side, and as many start again from positive bound charge on the far side.

Step 3The polarization

Substituting (3) into (2):

\vect P = \frac{3\varepsilon_0(\varepsilon - 1)}{\varepsilon + 2}\,\vect E_0

(4)

Check

With $\varepsilon = 1$ there is no dielectric: $\vect E = \vect E_0$ and $\vect P = 0$ . As $\varepsilon \to \infty$ the ball behaves as a conductor: $\vect E \to 0$ and $\vect P \to 3\varepsilon_0\vect E_0$ , so the moment of a ball of radius $R$ , its volume times $\vect P$ , tends to

\frac{4}{3}\pi R^3\cdot 3\varepsilon_0\vect E_0 = 4\pi\varepsilon_0 R^3\,\vect E_0,

the moment induced on a conducting ball of the same radius in a uniform field.

Remark

Outside, the ball adds the field of a point dipole $4\pi R^3\vect P/3$ at its center, which is what bends the lines toward it in the figure. Inside, $\vect D = \varepsilon_0\varepsilon\vect E = 3\varepsilon\varepsilon_0\vect E_0/(\varepsilon + 2)$ is larger than the applied $\varepsilon_0\vect E_0$ : the ball gathers displacement flux while thinning the field, the same contrast as at a flat boundary, where $E_n$ drops by $\varepsilon$ and $D_n$ is continuous.