Irodov 3.124: Three Plates Sharing One Island – Amit & Deepali Sinha Foundation

Problem Irodov 3.124

A rectangular circuit has a top wire holding, from left to right, a capacitor $C_1$ , a junction $A$ and a capacitor $C_2$ , and a bottom wire holding a cell of emf $\mathcal{E}_1$ , a junction $B$ and a cell of emf $\mathcal{E}_2$ . Both cells have their positive terminal on the right, so $\mathcal{E}_1$ has its positive terminal at $B$ and $\mathcal{E}_2$ its negative terminal at $B$ . A capacitor $C_3$ joins $A$ to $B$ , and plain wires join the left ends and the right ends of the top and bottom wires. Find the potential difference $\varphi_A - \varphi_B$ .

The circuit. Both cells have the long (+) plate on the right; C_3 joins the top junction A to the bottom junction B. — The circuit. Both cells have the long ( $+$ ) plate on the right; $C_3$ joins the top junction $A$ to the bottom junction $B$ .

Given

$\mathcal{E}_1$ , $\mathcal{E}_2$
$C_1$ , $C_2$ , $C_3$
capacitors uncharged before the circuit was assembled (assumed)

Find

\varphi_A - \varphi_B

Solution

Step 1The potentials of the two side wires

Take $\varphi_B = 0$ . In equilibrium no current flows, so the potential changes only across the cells. Crossing $\mathcal{E}_1$ from its negative plate on the left wire to its positive plate at $B$ rises by $\mathcal{E}_1$ , and crossing $\mathcal{E}_2$ from its negative plate at $B$ to its positive plate on the right wire rises by $\mathcal{E}_2$ :

\varphi_L = \varphi_B - \mathcal{E}_1 = -\mathcal{E}_1, \qquad \varphi_R = \varphi_B + \mathcal{E}_2 = \mathcal{E}_2 \tag{1}

where $\varphi_L$ and $\varphi_R$ are the potentials of the left and right side wires.

Step 2The island at

A

carries no net charge

Each plate on $A$ carries its capacitance times the excess of $\varphi_A$ over the potential of the facing plate, and the three charges add to zero:

C_1(\varphi_A - \varphi_L) + C_2(\varphi_A - \varphi_R) + C_3(\varphi_A - \varphi_B) = 0

Substituting (1) and collecting $\varphi_A$ :

\begin{aligned} C_1(\varphi_A + \mathcal{E}_1) + C_2(\varphi_A - \mathcal{E}_2) + C_3\varphi_A &= 0\\[2pt] (C_1 + C_2 + C_3)\,\varphi_A &= C_2\mathcal{E}_2 - C_1\mathcal{E}_1 \end{aligned}

Since $\varphi_B = 0$ , this is the asked difference:

\varphi_A - \varphi_B = \frac{C_2\mathcal{E}_2 - C_1\mathcal{E}_1}{C_1 + C_2 + C_3}

(2)

Check

Limits. With $C_3 \to \infty$ the big capacitor pins $A$ to $B$ and (2) goes to zero, as it must. With $C_1 = C_2 = C$ and $\mathcal{E}_1 = \mathcal{E}_2$ , the two side wires sit at $-\mathcal{E}$ and $+\mathcal{E}$ , symmetric about $B$ , and $A$ halfway between them by symmetry is at $0$ : the formula gives $C(\mathcal{E} - \mathcal{E})/(2C + C_3) = 0$ . With $C_2 = 0$ (the right capacitor removed), $A$ is fed only through $C_1$ from the wire at $-\mathcal{E}_1$ , and $C_1$ , $C_3$ in series divide that $-\mathcal{E}_1$ so that $C_3$ takes the fraction $C_1/(C_1 + C_3)$ , which is the formula’s value.

Remark

The answer is a capacitance-weighted average: $\varphi_A$ is the mean of the three potentials $\varphi_L = -\mathcal{E}_1$ , $\varphi_R = \mathcal{E}_2$ and $\varphi_B = 0$ seen across the gaps, each weighted by its capacitance, exactly as the centre of mass weights positions by masses. Any isolated node joined to fixed potentials only through capacitors sits at such a weighted mean, which is the capacitor analogue of Millman’s theorem for resistors.