Irodov 3.86: Half Filled, Still Radial – Amit & Deepali Sinha Foundation

Problem Irodov 3.86

A spherical capacitor has two concentric electrodes. A plane through their common center divides the space between them into two halves, and one half is filled with a uniform isotropic dielectric of permittivity $\varepsilon$ while the other is empty. The capacitor carries the charge $q$ . Find the magnitude of the electric field between the electrodes as a function of the distance $r$ from the common center.

The capacitor in section. A plane through the center splits the gap between the electrodes; the lower half is filled with the dielectric ε, the upper half is empty. — The capacitor in section. A plane through the center splits the gap between the electrodes; the lower half is filled with the dielectric $\varepsilon$ , the upper half is empty.

Given

charge $q$
half the gap filled with permittivity $\varepsilon$ , bounded by a plane through the center

Find

E(r)

between the electrodes

Solution

Step 1A radial field, the same in both halves

Try a field that is radial and depends on $r$ alone, the same in the dielectric half and in the empty half. Each electrode is then an equipotential, as a conductor must be. At the plane dividing the halves such a field is purely tangential, and tangential $\vect E$ must be continuous across a boundary, which a field equal on both sides satisfies. The interface carries no bound charge either, because $\vect P$ in the dielectric is radial and so parallel to the interface.

So at distance $r$ the field has one magnitude $E(r)$ everywhere between the electrodes, while the displacement differs between the halves:

D_{\mathrm{empty}} = \varepsilon_0E, \qquad D_{\mathrm{diel}} = \varepsilon_0\varepsilon E \tag{1}

Step 2Gauss’s theorem for D over a sphere

The field between the electrodes: radial, with one line density all round, in the empty half and in the dielectric alike. The dashed sphere of radius r is the Gaussian surface. — The field between the electrodes: radial, with one line density all round, in the empty half and in the dielectric alike. The dashed sphere of radius $r$ is the Gaussian surface.

Take a sphere of radius $r$ between the electrodes. Half of its area, $2\pi r^2$ , lies in the empty half and half in the dielectric, and it encloses the free charge $q$ of the inner electrode. With (1):

\begin{aligned} \oint\vect D\cdot d\vect S &= \varepsilon_0E\cdot 2\pi r^2 + \varepsilon_0\varepsilon E\cdot 2\pi r^2 = q\\ 2\pi\varepsilon_0(1 + \varepsilon)\,r^2E &= q \end{aligned}

Solving for $E$ :

E = \frac{q}{2\pi\varepsilon_0(\varepsilon + 1)\,r^2}

(2)

Check

With $\varepsilon = 1$ the gap is empty and the result becomes $q/(4\pi\varepsilon_0r^2)$ , the field of a point charge $q$ at the center, as it must be. With the whole gap filled the field would be $q/(4\pi\varepsilon_0\varepsilon r^2)$ ; the half-filled value lies between the two, the field set by the average permittivity $(\varepsilon + 1)/2$ .

Remark

The free charge on the inner electrode is not uniform. By (1) the hemisphere against the dielectric carries the fraction $\varepsilon/(\varepsilon + 1)$ of $q$ and the other only $1/(\varepsilon + 1)$ ; that uneven free charge, together with the bound charge on the dielectric’s faces at the electrodes, is exactly what keeps the total field the same in both halves. The two halves act as two hemispherical capacitors in parallel.