Irodov 3.101: A Coat That Raises Capacitance – Amit & Deepali Sinha Foundation

Problem Irodov 3.101

An isolated conducting ball of radius $R_1$ is coated with a concentric layer of dielectric of permittivity $\varepsilon$ that touches the ball and extends out to the radius $R_2$ . Outside the layer there is vacuum. Find the capacitance of the ball.

The conducting ball of radius R_1 and the dielectric layer around it, out to the radius R_2; beyond it, vacuum. — The conducting ball of radius $R_1$ and the dielectric layer around it, out to the radius $R_2$ ; beyond it, vacuum.

Given

ball radius $R_1$
layer: permittivity $\varepsilon$ , outer radius $R_2$
vacuum outside

Find

the capacitance

C

Solution

Step 1The displacement and the field

Give the ball a charge $q$ . The arrangement is spherically symmetric, so $\vect D$ is radial and depends only on $r$ , and Gauss’s theorem for $\vect D$ on a concentric sphere of radius $r > R_1$ sees only the free charge $q$ , whether the sphere lies in the layer or outside it:

D\cdot 4\pi r^2 = q

Dividing by $\varepsilon_0\varepsilon$ in the layer and by $\varepsilon_0$ in the vacuum gives the field:

\begin{gathered} E(r) = \frac{q}{4\pi\varepsilon_0\varepsilon r^2}\quad (R_1 < r < R_2) \\ E(r) = \frac{q}{4\pi\varepsilon_0 r^2}\quad (r > R_2) \end{gathered} \tag{1}

Step 2The potential of the ball

An isolated conductor’s potential is measured from infinity, where it is zero. Integrate (1) inward from infinity to the ball’s surface, through the vacuum first and then through the layer:

\begin{aligned} \varphi &= \int_{R_2}^{\infty}\frac{q\,dr}{4\pi\varepsilon_0 r^2} + \int_{R_1}^{R_2}\frac{q\,dr}{4\pi\varepsilon_0\varepsilon r^2}\\ &= \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{R_2} + \frac{1}{\varepsilon}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\right] \end{aligned}

The field (top) and the potential (bottom) against r, drawn for ε = 2 and R_2 = 2R_1. The field is weakened inside the layer and jumps up at its outer surface; the potential is flat inside the metal. — The field (top) and the potential (bottom) against $r$ , drawn for $\varepsilon = 2$ and $R_2 = 2R_1$ . The field is weakened inside the layer and jumps up at its outer surface; the potential is flat inside the metal.

Put the bracket over the common denominator $\varepsilon R_1 R_2$ :

\frac{1}{R_2} + \frac{1}{\varepsilon}\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{\varepsilon R_1 + R_2 - R_1}{\varepsilon R_1 R_2} = \frac{R_2 + (\varepsilon - 1)R_1}{\varepsilon R_1 R_2}

Step 3The capacitance

The capacitance of an isolated conductor is $C = q/\varphi$ , and $q$ cancels. Dividing the numerator and the denominator by $R_2$ :

C = \dfrac{4\pi\varepsilon_0\varepsilon R_1}{1 + (\varepsilon - 1)R_1/R_2}

(2)

Check

Three limits. No dielectric, $\varepsilon = 1$ : $C = 4\pi\varepsilon_0 R_1$ , the bare ball. An infinitely thick layer, $R_2 \to \infty$ : $C = 4\pi\varepsilon_0\varepsilon R_1$ , the ball immersed in an unbounded dielectric. A vanishingly thin layer, $R_2 \to R_1$ : the denominator tends to $\varepsilon$ and $C \to 4\pi\varepsilon_0 R_1$ again, as it must when the coat has no thickness.

Remark

The coated ball is two capacitors in series: a spherical capacitor $4\pi\varepsilon_0\varepsilon R_1R_2/(R_2 - R_1)$ filled with the dielectric, and the bare sphere $4\pi\varepsilon_0 R_2$ facing infinity; adding their reciprocals gives (2). Because the outer term never goes away, even a thick coat of high permittivity cannot raise the capacitance beyond $4\pi\varepsilon_0 R_2$ , the capacitance of a metal ball the size of the coat.