Irodov 3.123: The Middle Wire Splits the Series Pair

Problem Irodov 3.123

Two batteries, each of emf $\mathcal{E} = 60$ V, are connected in series in a left vertical branch, both with the positive terminal upward. A right vertical branch holds two capacitors in series, $C_1 = 2.0$ $\mu$ F above and $C_2 = 3.0$ $\mu$ F below. The top wire (section $1$ ) joins the tops of the two branches and the bottom wire (section $2$ ) joins their bottoms. A middle wire runs from the junction between the batteries, through an open switch $Sw$ and then along section $3$ , to the junction between the capacitors. The switch is now closed. Find the charges that flow through sections $1$ , $2$ and $3$ , each counted positive when it flows from the battery side toward the capacitor side.

The circuit with the switch open. Both batteries have the long (+) plate up; the arrows give the positive senses for the charges through sections 1, 2 and 3. — The circuit with the switch open. Both batteries have the long ( $+$ ) plate up; the arrows give the positive senses for the charges through sections $1$ , $2$ and $3$ .

Given

$\mathcal{E} = 60$ V (each battery)
$C_1 = 2.0~\mu$ F
$C_2 = 3.0~\mu$ F
$Sw$ open, then closed
capacitors uncharged before the circuit was assembled (assumed)

Find

q_1

q_2

q_3

, in the senses of the arrows

Solution

Step 1The charges with the switch open

With the switch open the middle wire carries nothing, so $C_1$ and $C_2$ are in series across the two batteries, $2\mathcal{E}$ in all. The junction between them, with the two plates on it, is isolated and neutral (the capacitors are assumed uncharged to begin with), so both capacitors carry the same charge:

q_0 = \frac{C_1C_2}{C_1 + C_2}\cdot 2\mathcal{E} \tag{1}

The upper plate of $C_1$ , on the positive top wire, carries $+q_0$ ; the lower plate of $C_2$ carries $-q_0$ ; the two plates on the middle junction carry $-q_0$ and $+q_0$ .

Step 2The charges with the switch closed

Closing the switch ties the junction between the capacitors to the junction between the batteries. Each capacitor then sits across one battery:

q_{C_1}' = C_1\mathcal{E}, \qquad q_{C_2}' = C_2\mathcal{E} \tag{2}

The upper plate of $C_1$ carries $+C_1\mathcal{E}$ , the lower plate of $C_2$ carries $-C_2\mathcal{E}$ , and the middle junction holds $-C_1\mathcal{E} + C_2\mathcal{E}$ on its two plates.

Step 3Charge through each section

Subtracting the plate charges $\pm q_0$ of (1) from those of (2):

\begin{aligned} q_1 &= C_1\mathcal{E} - \frac{2\mathcal{E}C_1C_2}{C_1 + C_2} = \frac{\mathcal{E}C_1(C_1 + C_2 - 2C_2)}{C_1 + C_2}\\[2pt] q_2 &= -C_2\mathcal{E} - \left(-\frac{2\mathcal{E}C_1C_2}{C_1 + C_2}\right) = \frac{\mathcal{E}C_2(2C_1 - C_1 - C_2)}{C_1 + C_2}\\[2pt] q_3 &= (C_2 - C_1)\mathcal{E} - (q_0 - q_0) \end{aligned}

With $C_1 + C_2 = 5.0~\mu$ F and $C_1 - C_2 = -1.0~\mu$ F, numerically:

\begin{aligned} q_1 &= \frac{60 \times 2.0 \times (-1.0)}{5.0}~\mu\text{C}\\[2pt] q_2 &= \frac{60 \times 3.0 \times (-1.0)}{5.0}~\mu\text{C}\\[2pt] q_3 &= 60 \times (3.0 - 2.0)~\mu\text{C} \end{aligned}

\begin{gathered} q_1 = \frac{\mathcal{E}C_1(C_1 - C_2)}{C_1 + C_2} = -24~\mu\text{C}\\[2pt] q_2 = \frac{\mathcal{E}C_2(C_1 - C_2)}{C_1 + C_2} = -36~\mu\text{C}\\[2pt] q_3 = \mathcal{E}(C_2 - C_1) = +60~\mu\text{C} \end{gathered}

(3)

So $24~\mu$ C flows back along section $1$ and $36~\mu$ C back along section $2$ , while $60~\mu$ C flows along the arrow of section $3$ .

Check

The three sections together carry everything that reaches the capacitor plates, and the total charge on all four plates is zero before and after, so the three charges must add up to zero. Indeed

q_1 + q_2 + q_3 = \frac{\mathcal{E}(C_1 - C_2)(C_1 + C_2)}{C_1 + C_2} + \mathcal{E}(C_2 - C_1) = 0

and in numbers $-24 - 36 + 60 = 0$ . For equal capacitors the series pair already splits the $2\mathcal{E}$ evenly, so closing the switch changes nothing, and all three charges vanish.

Remark

The middle wire lets each capacitor take the voltage of its own battery instead of sharing $2\mathcal{E}$ in proportion to $1/C$ . With the switch open the smaller capacitor $C_1$ held 72 V, more than its battery’s 60 V, so it gives charge back through section $1$ ; the larger $C_2$ held only 48 V and charges further, fed through section $3$ at its upper plate while positive charge drains back along section $2$ from its lower one.