Irodov 3.118: Two Sources Pulling Against Each Other

Problem Irodov 3.118

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. For each capacitor find the potential difference between its left plate and its right plate.

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

V_1 = \varphi_{\mathrm{left}} - \varphi_{\mathrm{right}}

for

C_1

and

V_2

likewise for

C_2

Solution

Step 1Name the four corners

Call the potentials of the top-left, top-right, bottom-right and bottom-left corners $\varphi_a$ , $\varphi_b$ , $\varphi_c$ , $\varphi_d$ . The left plate of $C_1$ is at $\varphi_a$ and its right plate at $\varphi_b$ ; the left plate of $C_2$ is at $\varphi_d$ and its right plate at $\varphi_c$ . So the two quantities asked for are

V_1 = \varphi_a - \varphi_b, \qquad V_2 = \varphi_d - \varphi_c \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}

Step 2The two voltages are tied by the sources

Subtracting the two relations (2) and using (1):

\begin{align*} \varphi_d - \varphi_c &= (\varphi_a - \varphi_b) + \mathcal{E}_1 - \mathcal{E}_2 \\ V_2 &= V_1 - (\mathcal{E}_2 - \mathcal{E}_1) \tag{3}\end{align*}

Step 3The left half of the loop carries no net charge

The left plates of both capacitors, the source $\mathcal{E}_1$ and the wires between them form one conductor, isolated from the rest of the loop by the two capacitor gaps. Uncharged before the circuit was closed, it stays neutral, so the charges on the two left plates cancel:

C_1V_1 + C_2V_2 = 0 \tag{4}

Substituting (3):

\begin{aligned} C_1V_1 + C_2V_1 - C_2(\mathcal{E}_2 - \mathcal{E}_1) &= 0\\ V_1 &= \frac{C_2(\mathcal{E}_2 - \mathcal{E}_1)}{C_1 + C_2} \end{aligned}

Dividing numerator and denominator by $C_2$ :

V_1 = \frac{\mathcal{E}_2 - \mathcal{E}_1}{1 + C_1/C_2}

(5)

Step 4The voltage on the lower capacitor

From (4), $V_2$ is fixed by $V_1$ :

V_2 = -\frac{C_1}{C_2}\,V_1 = -\frac{C_1(\mathcal{E}_2 - \mathcal{E}_1)}{C_1 + C_2}

Dividing numerator and denominator by $C_1$ :

V_2 = \frac{\mathcal{E}_1 - \mathcal{E}_2}{1 + C_2/C_1}

(6)

For $\mathcal{E}_2 > \mathcal{E}_1$ the left plate of $C_1$ is the higher one and the right plate of $C_2$ is the higher one.

Check

The tie (3) must hold: $V_1 - V_2 = (\mathcal{E}_2 - \mathcal{E}_1)\,[C_2 + C_1]/(C_1 + C_2) = \mathcal{E}_2 - \mathcal{E}_1$ , as required. With equal emfs both voltages vanish: the two sources balance round the loop and nothing charges. With $C_1 \gg C_2$ , $V_1 \to 0$ and $V_2 \to \mathcal{E}_1 - \mathcal{E}_2$ : the large capacitor behaves like a wire and the small one takes the whole net emf, as two capacitors in series always share a voltage inversely to their capacitances.

Remark

The loop is just two capacitors in series driven by one effective source of emf $\mathcal{E}_2 - \mathcal{E}_1$ , the two real sources being in opposition. Turning either source round would replace the difference by the sum $\mathcal{E}_1 + \mathcal{E}_2$ , which is why the polarity drawn in the figure decides the answer.