Irodov 3.97: Removing a Ball Adds a Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.97

An infinite uniform dielectric carries a permanent polarization $\vect P$ , and far from a spherical cavity inside it the field is $\vect E$ . Using the field of a uniformly polarized ball (Problem 3.96), find the field $\vect E_0$ inside the cavity.

The permanently polarized dielectric with its spherical cavity. Far from the cavity the field is E, drawn at an arbitrary angle to P. — The permanently polarized dielectric with its spherical cavity. Far from the cavity the field is $\vect E$ , drawn at an arbitrary angle to $\vect P$ .

Given

uniform permanent polarization $\vect P$
field $\vect E$ far from the cavity
spherical cavity

Find

the field

\vect E_0

inside the cavity

Solution

Step 1The cavity as a superposition

The polarization is permanent, so taking matter out leaves the rest unchanged and the bound charges simply add. The dielectric with the cavity is therefore the intact dielectric plus a ball, filling the cavity, polarized with $-\vect P$ : the two polarizations cancel inside the cavity. The intact dielectric has the uniform field $\vect E$ everywhere, the cavity’s place included.

\vect E_0 = \vect E + \vect E_{\mathrm{ball}} \tag{1}

Step 2The field of the reversed ball

Problem 3.96 gives the field inside a uniformly polarized ball as minus its polarization over $3\varepsilon_0$ ; for a polarization of $-\vect P$ :

\vect E_{\mathrm{ball}} = -\frac{-\vect P}{3\varepsilon_0} = \frac{\vect P}{3\varepsilon_0}

Substituting into (1):

\vect E_0 = \vect E + \frac{\vect P}{3\varepsilon_0}

(2)

The field in the cavity is the far field E plus P/3ε_0 (both dashed), which comes from the bound charge on the cavity wall: positive where P leaves the material into the cavity, negative where it enters it again. — The field in the cavity is the far field $\vect E$ plus $\vect P/3\varepsilon_0$ (both dashed), which comes from the bound charge on the cavity wall: positive where $\vect P$ leaves the material into the cavity, negative where it enters it again.

Check

The sign can be read off the wall. The outward normal of the dielectric at the wall points into the cavity, toward its center, so the bound charge is $\sigma' = \vect P\cdot\vect n = -P\cos\theta$ , with $\theta$ measured from $\vect P$ at the center. The wall is negative on the side $\vect P$ points to and positive on the other, and the field of that charge inside the cavity runs from the positive side to the negative one: along $\vect P$ , as the boxed $+\vect P/3\varepsilon_0$ says. A form with $-\vect P/3\varepsilon_0$ would point from the negative wall toward the positive one, which no field of these charges does.

Remark

The extra term depends on the cavity’s shape. A long needle-shaped cavity along $\vect P$ has no bound charge on its sides and feels just $\vect E$ ; a thin disc-shaped cavity across $\vect P$ sits between two sheets $\pm P$ and feels $\vect E + \vect P/\varepsilon_0$ . The sphere falls in between with a third. The same $\vect E + \vect P/3\varepsilon_0$ is the field that acts on a single molecule in a polarized medium, and it is the step that leads to the Clausius and Mossotti relation between permittivity and molecular polarizability.