Irodov 3.113: The Wiring Decides Which Gaps Pair – Amit & Deepali Sinha Foundation

Problem Irodov 3.113

Four identical parallel metal plates, each of area $S$ , are stacked in air with the same spacing $d$ between neighbors. Number them 1 to 4 from the top. Find the capacitance between the terminals $A$ and $B$ for two ways of wiring them:

plates 1 and 3 are joined by a wire, $A$ is connected to plate 2 and $B$ to plate 4;
plates 1 and 4 are joined by a wire, $A$ is connected to plate 2 and $B$ to plate 3.

(a) Plates 1 and 3 joined on the left; A on plate 2, B on plate 4. — (a) Plates 1 and 3 joined on the left; $A$ on plate 2, $B$ on plate 4.

(b) The outer plates 1 and 4 joined on the left; A on plate 2, B on plate 3. — (b) The outer plates 1 and 4 joined on the left; $A$ on plate 2, $B$ on plate 3.

Given

$4$ plates
area $S$
spacing $d$
air

Find

C_{AB}

for wirings (a) and (b)

Solution

Step 1Every gap is one elementary capacitor

Neglecting edge effects, the field between two neighboring plates is uniform and lives only in that gap, so each of the three gaps behaves as a parallel-plate capacitor whose two plates are the facing surfaces.

C_0 = \frac{\varepsilon_0 S}{d} \tag{1}

Step 2Wiring (a): two gaps in parallel, then one in series

Plates 1 and 3 form one conductor, call it $F$ , which is connected to neither terminal. Gap 1-2 and gap 2-3 both lie between plate 2 and $F$ , that is between $A$ and $F$ ; gap 3-4 lies between $F$ and $B$ . The first two add in parallel:

C_{AF} = C_0 + C_0 = 2C_0, \qquad C_{FB} = C_0

The pair is in series with the third gap, because the isolated conductor $F$ carries no net charge, so the charge that leaves $A$ is the charge that arrives at $B$ :

C_a = \frac{C_{AF}\,C_{FB}}{C_{AF} + C_{FB}} = \frac{2C_0 \cdot C_0}{2C_0 + C_0} = \frac{2}{3}\,C_0

Substituting (1):

C_a = \dfrac{2\varepsilon_0 S}{3d}

(2)

Step 3The field and the charges in wiring (a)

Wiring (a) with A at potential V and B at 0: F floats at 2V/3. Lines and face charges are drawn on one scale, so gap 3-4 carries twice the field of the upper two, and F is neutral. — Wiring (a) with $A$ at potential $V$ and $B$ at $0$ : $F$ floats at $2V/3$ . Lines and face charges are drawn on one scale, so gap 3-4 carries twice the field of the upper two, and $F$ is neutral.

With $A$ at potential $V$ and $B$ at $0$ , the neutrality of $F$ fixes its potential $\varphi_F$ : the charge $F$ gains on its faces in gaps 1-2 and 2-3 must cancel the charge on its face in gap 3-4.

\begin{aligned} 2C_0(V - \varphi_F) &= C_0(\varphi_F - 0)\\ \varphi_F &= \frac{2}{3}\,V \end{aligned}

So each upper gap holds the voltage $V/3$ and the lower gap $2V/3$ , which is why the lower gap carries twice the field and twice the charge per face.

Step 4Wiring (b): one gap bridged by two in series

Now plates 1 and 4 form the isolated conductor $F$ . Gap 2-3 lies directly between $A$ and $B$ . Gap 1-2 lies between $A$ and $F$ and gap 3-4 between $F$ and $B$ , so those two are in series with each other, and that chain is in parallel with gap 2-3:

C_b = C_0 + \frac{C_0 \cdot C_0}{C_0 + C_0} = C_0 + \frac{C_0}{2} = \frac{3}{2}\,C_0

Substituting (1):

C_b = \dfrac{3\varepsilon_0 S}{2d}

(3)

Wiring (b) with A at V and B at 0: the joined outer plates float at V/2. The middle gap holds the full V and so carries twice the field of either outer gap. — Wiring (b) with $A$ at $V$ and $B$ at $0$ : the joined outer plates float at $V/2$ . The middle gap holds the full $V$ and so carries twice the field of either outer gap.

Check

Count the charge on plate 2 directly. In (a), with $\varphi_F = 2V/3$ , both faces of plate 2 face $F$ across a voltage $V/3$ , so plate 2 holds $2\cdot C_0V/3 = (2/3)C_0V$ , giving $C_a = (2/3)C_0$ as in (2). In (b), $F$ floats at $V/2$ by symmetry; plate 2 faces $B$ across $V$ and $F$ across $V/2$ , so it holds $C_0V + C_0V/2 = (3/2)C_0V$ , giving (3). In both cases the isolated conductor’s face charges sum to zero: in (a) $-C_0V/3 - C_0V/3 + 2C_0V/3 = 0$ , in (b) $-C_0V/2 + C_0V/2 = 0$ .

Remark

The same four plates give $(2/3)C_0$ in one wiring and $(3/2)C_0$ in the other, and the wires alone make the difference. Joining plates 1 and 3 to $A$ and plates 2 and 4 to $B$ puts all three gaps in parallel, $3C_0$ , the most the stack can give: this interleaved comb is how a variable tuning capacitor packs a large capacitance into a small volume, every plate working on both faces.