Irodov 3.110: Two Distant Balls in Series – Amit & Deepali Sinha Foundation

Problem Irodov 3.110

Two identical metal balls of radius $a$ have their centers a distance $b$ apart, with $b \gg a$ . Both are immersed in a uniform dielectric of permittivity $\varepsilon$ that fills all the surrounding space. Find the capacitance of this two-ball system.

The two metal balls of radius a, centers a distance b apart, in the dielectric ε (drawn schematically: the problem has b a). — The two metal balls of radius $a$ , centers a distance $b$ apart, in the dielectric $\varepsilon$ (drawn schematically: the problem has $b \gg a$ ).

Given

ball radius $a$
center distance $b \gg a$
surrounding dielectric $\varepsilon$

Find

the capacitance

C

of the pair

Solution

Step 1The potential of one charged ball

Give the balls the charges $q$ and $-q$ . With $b \gg a$ , each ball sits in the weak, nearly uniform field of the other, which shifts its charge only slightly, so the charge spreads practically uniformly over each surface. For a ball carrying $q$ in an unbounded uniform dielectric, by symmetry $\vect D$ is radial and the same all round, and Gauss’s theorem on a concentric sphere of radius $r$ gives

D\cdot 4\pi r^2 = q

so its field $D/\varepsilon_0\varepsilon$ and its potential are those in vacuum divided by $\varepsilon$ :

\varphi(r) = \frac{q}{4\pi\varepsilon_0\varepsilon r}\quad (r \geq a) \tag{1}

Step 2The voltage between the balls

The balls charged q and -q, practically uniformly since b a; each one's potential has its own term and the term of the other ball at the distance b. — The balls charged $q$ and $-q$ , practically uniformly since $b \gg a$ ; each one’s potential has its own term and the term of the other ball at the distance $b$ .

Each ball’s potential is its own term (1) at $r = a$ plus the other ball’s at $r = b$ . Averaged over a ball’s surface the partner’s potential takes exactly its value at the ball’s center, so

\begin{gathered} \varphi_1 = \frac{q}{4\pi\varepsilon_0\varepsilon}\left(\frac{1}{a} - \frac{1}{b}\right) \\ \varphi_2 = -\frac{q}{4\pi\varepsilon_0\varepsilon}\left(\frac{1}{a} - \frac{1}{b}\right) \end{gathered}

The voltage is the difference:

V = \varphi_1 - \varphi_2 = \frac{2q}{4\pi\varepsilon_0\varepsilon}\left(\frac{1}{a} - \frac{1}{b}\right) = \frac{q}{2\pi\varepsilon_0\varepsilon a}\left(1 - \frac{a}{b}\right)

Step 3The capacitance

The capacitance is $C = q/V$ , and $q$ cancels:

C = \frac{2\pi\varepsilon_0\varepsilon a}{1 - a/b}

For $b \gg a$ the correction $a/b$ is negligible:

C \approx 2\pi\varepsilon_0\varepsilon a

(2)

Check

The result is two capacitors in series, each the capacitance $4\pi\varepsilon_0\varepsilon a$ of an isolated ball in the dielectric: in series they give $(4\pi\varepsilon_0\varepsilon a)/2 = 2\pi\varepsilon_0\varepsilon a$ . This is right physically, since with the balls far apart the field lines of $q$ run out to distances of order $b$ before turning toward $-q$ , so each ball behaves as if alone with a partner at infinity. Without the dielectric, $\varepsilon = 1$ , it reduces to $2\pi\varepsilon_0 a$ .

Remark

To this order the distance $b$ does not enter at all: two balls a meter apart and two balls a kilometer apart have the same capacitance, set by the radii alone. The first correction raises it by the factor $1/(1 - a/b)$ , about $1 + a/b$ , because each ball’s potential is lowered by its oppositely charged partner.