Irodov 3.119: One Charge for Both Capacitors – Amit & Deepali Sinha Foundation

Problem Irodov 3.119

A single rectangular loop holds a capacitor $C_1$ in its top side, a capacitor $C_2$ in its bottom side, a source of emf $\mathcal{E}_1$ in its left side and a source of emf $\mathcal{E}_2$ in its right side. Both sources have their positive terminal toward the bottom side and their negative terminal toward the top side. Find the charge of each capacitor.

The loop: C_1 on top, C_2 below, E_1 on the left and E_2 on the right, each source with its long (+) plate toward the bottom wire. — The loop: $C_1$ on top, $C_2$ below, $\mathcal{E}_1$ on the left and $\mathcal{E}_2$ on the right, each source with its long ( $+$ ) plate toward the bottom wire.

Given

$\mathcal{E}_1, \mathcal{E}_2$
$C_1, C_2$
capacitors uncharged before connection (assumed)

Find

the charges

q_1

q_2

of the two capacitors

Solution

Step 1The two capacitors carry equal charges

The left plates of both capacitors, the source $\mathcal{E}_1$ and the wires joining them form one conductor, cut off from the rest of the loop by the two capacitor gaps. It was neutral before the circuit was closed and stays neutral, so whatever charge sits on the left plate of $C_1$ is matched by the opposite charge on the left plate of $C_2$ . Each capacitor’s two plates carry opposite charges, so both capacitors hold the same magnitude $q$ :

q_1 = q_2 = q

Step 2Walk round the loop

Let $q$ be the charge on the left plate of $C_1$ ; by Step 1 the left plate of $C_2$ carries $-q$ . Call the potentials of the top-left, top-right, bottom-right and bottom-left corners $\varphi_a$ , $\varphi_b$ , $\varphi_c$ , $\varphi_d$ . Each capacitor’s voltage, left plate minus right plate, is its left-plate charge over its capacitance:

\varphi_a - \varphi_b = \frac{q}{C_1}, \qquad \varphi_d - \varphi_c = -\frac{q}{C_2} \tag{1}

Each source raises the potential by its emf from its negative terminal at the top to its positive terminal at the bottom:

\varphi_d = \varphi_a + \mathcal{E}_1, \qquad \varphi_c = \varphi_b + \mathcal{E}_2 \tag{2}

Subtracting the two relations (2) and substituting (1) on both sides:

\begin{aligned} \varphi_d - \varphi_c &= (\varphi_a - \varphi_b) + \mathcal{E}_1 - \mathcal{E}_2\\ -\frac{q}{C_2} &= \frac{q}{C_1} + \mathcal{E}_1 - \mathcal{E}_2 \end{aligned}

Step 3The charge

Solving for $q$ over the common denominator $C_1C_2$ :

q\,\frac{C_1 + C_2}{C_1C_2} = \mathcal{E}_2 - \mathcal{E}_1

The sign of $q$ only says which plate is positive; the charge of each capacitor is its magnitude:

q_1 = q_2 = \frac{|\mathcal{E}_1 - \mathcal{E}_2|\,C_1C_2}{C_1 + C_2}

(3)

The plate charges for E_2 > E_1: the left plate of C_1 and the right plate of C_2 are positive, and each half of the loop stays neutral. — The plate charges for $\mathcal{E}_2 > \mathcal{E}_1$ : the left plate of $C_1$ and the right plate of $C_2$ are positive, and each half of the loop stays neutral.

For $\mathcal{E}_2 > \mathcal{E}_1$ , $q > 0$ : the left plate of $C_1$ and the right plate of $C_2$ are the positive ones, as drawn. For $\mathcal{E}_1 > \mathcal{E}_2$ every sign reverses.

Check

Units: emf times capacitance is charge. With equal emfs $q = 0$ , the sources balancing round the loop; with $C_2 \to \infty$ , $q \to C_1|\mathcal{E}_2 - \mathcal{E}_1|$ , the full net emf across $C_1$ alone, as if $C_2$ were a wire.

Remark

The loop is two capacitors in series, of capacitance $C_1C_2/(C_1 + C_2)$ , on a single effective source $\mathcal{E}_2 - \mathcal{E}_1$ : the charge is that capacitance times that emf. Series capacitors always share one charge, whatever their sizes, because the conductor between them can only shift its charge, never gain any.