Irodov 3.105: A Dielectric That Flattens the Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.105

A spherical capacitor has concentric electrodes of radii $R_1$ and $R_2$ , with $R_2 > R_1$ . The space between them is filled with an isotropic dielectric whose permittivity depends on the distance $r$ from the common center as $\varepsilon = a/r$ , with $a$ a constant. What is the capacitance of this capacitor?

The capacitor in section: inner electrode of radius R_1, outer electrode of inner radius R_2, and between them the dielectric whose permittivity falls off as a/r from the center O. — The capacitor in section: inner electrode of radius $R_1$ , outer electrode of inner radius $R_2$ , and between them the dielectric whose permittivity falls off as $a/r$ from the center $O$ .

Given

$R_1$
$R_2 > R_1$
$\varepsilon = a/r$

Find

the capacitance

C

Solution

Step 1The displacement from Gauss’s theorem

Put a charge $q$ on the inner electrode and $-q$ on the outer. The permittivity depends on $r$ alone, so the arrangement keeps full spherical symmetry and $\vect D$ is radial with a magnitude that depends only on $r$ . Gauss’s theorem for $\vect D$ over a sphere of radius $r$ between the electrodes sees only the free charge $q$ :

\begin{aligned} \oint \vect D \cdot d\vect S &= 4\pi r^2 D = q\\ D &= \frac{q}{4\pi r^2} \end{aligned}

Step 2The field inside the graded dielectric

In an isotropic dielectric $\vect D = \varepsilon_0\varepsilon\vect E$ . Dividing by $\varepsilon_0\varepsilon$ with $\varepsilon = a/r$ gives

E = \frac{D}{\varepsilon_0\varepsilon} = \frac{q}{4\pi r^2}\cdot\frac{r}{\varepsilon_0 a} = \frac{q}{4\pi\varepsilon_0 a\,r} \tag{1}

The field falls only as $1/r$ , not as the $1/r^2$ of an empty spherical capacitor: the permittivity drops outward just fast enough to undo one power of $r$ .

Step 3The voltage across the gap

The potential difference between the electrodes is the line integral of (1) along a radius from $R_1$ to $R_2$ :

\begin{aligned} V &= \int_{R_1}^{R_2} E\,dr = \frac{q}{4\pi\varepsilon_0 a}\int_{R_1}^{R_2}\frac{dr}{r}\\ &= \frac{q}{4\pi\varepsilon_0 a}\,\ln\frac{R_2}{R_1} \end{aligned}

Step 4The capacitance

The capacitance is the charge per unit voltage, $C = q/V$ , and $q$ cancels:

C = \dfrac{4\pi\varepsilon_0 a}{\ln(R_2/R_1)}

(2)

Step 5Where the extra field lines begin

The field in section. The number of E lines grows outward as the permittivity falls, so new lines begin inside the dielectric, each on a positive bound charge (for q > 0). — The field in section. The number of $\vect E$ lines grows outward as the permittivity falls, so new lines begin inside the dielectric, each on a positive bound charge (for $q > 0$ ).

The flux of $\vect E$ through a sphere of radius $r$ is $4\pi r^2 E$ , which by (1) grows in proportion to $r$ , so lines of $\vect E$ must begin inside the dielectric. Their sources are the bound charges, of density $\rho' = -\nabla\cdot\vect P$ , with $\vect P = \vect D - \varepsilon_0\vect E$ radial:

\begin{aligned} P &= \frac{q}{4\pi r^2} - \frac{q}{4\pi a r} = \frac{q}{4\pi r^2}\left(1 - \frac{r}{a}\right)\\ \rho' &= -\frac{1}{r^2}\frac{d}{dr}\left(r^2 P\right) = -\frac{1}{r^2}\frac{d}{dr}\left[\frac{q}{4\pi}\left(1 - \frac{r}{a}\right)\right] = \frac{q}{4\pi a r^2} \end{aligned}

The bound volume charge has the sign of $q$ , and it is what feeds the growing flux of $\vect E$ .

Check

Dimensions: $\varepsilon = a/r$ is a pure number, so $a$ is a length and $4\pi\varepsilon_0 a$ is a capacitance (F/m times m); the logarithm is a pure number. Thin gap: with $R_2 = R_1 + h$ and $h \ll R_1$ , $\ln(R_2/R_1) \approx h/R_1$ , and

C \approx \frac{4\pi\varepsilon_0 a R_1}{h} = \frac{\varepsilon_0 (a/R_1)(4\pi R_1^2)}{h},

which is the parallel-plate result $\varepsilon_0\varepsilon S/h$ for the area $S = 4\pi R_1^2$ and the local permittivity $a/R_1$ .

Remark

The bound charge balances, as it must for a neutral dielectric. The volume charge $\rho'$ integrated over the shell gives $q(R_2 - R_1)/a$ , and the surface densities $-P(R_1)$ and $+P(R_2)$ , times the areas $4\pi R_1^2$ and $4\pi R_2^2$ , give $-q(1 - R_1/a)$ and $+q(1 - R_2/a)$ ; the three sum to zero. Since $\varepsilon \geq 1$ requires $a \geq R_2$ , the inner surface carries negative bound charge and the outer surface positive, as in any dielectric polarized outward.