Irodov 3.122: Closing the Switch Empties One Capacitor

Problem Irodov 3.122

A circuit has two meshes that share a middle branch. The left branch is a cell of emf $\mathcal{E}$ with its positive terminal at the top. From that terminal the top wire runs through an open switch $Sw$ to a junction, then through a capacitor $C_1$ to the top right corner. The middle branch joins that junction, through a capacitor $C_2$ , to a junction on the bottom wire. The right branch is a second cell of emf $\mathcal{E}$ , also with its positive terminal at the top, on the side of $C_1$ . Section $1$ of the bottom wire runs from the middle junction to the left cell, and section $2$ from the middle junction to the right cell. The switch is now closed. Find the charges that flow through sections $1$ and $2$ , each counted positive when it flows away from the middle junction.

The circuit with the switch open. Both cells have the long (+) plate on top; the arrows on the bottom wire give the positive senses for the charges through sections 1 and 2. — The circuit with the switch open. Both cells have the long ( $+$ ) plate on top; the arrows on the bottom wire give the positive senses for the charges through sections $1$ and $2$ .

Given

$\mathcal{E}$ (each cell)
$C_1$ , $C_2$
$Sw$ open, then closed
capacitors uncharged before the circuit was assembled (assumed)

Find

q_1

and

q_2

, in the senses of the arrows

Solution

Step 1The charges with the switch open

Call $J$ the upper junction, where the switch, $C_1$ and $C_2$ meet, and $M$ the lower junction on the bottom wire; take $\varphi_M = 0$ . With the switch open the left cell is cut off, so only the right cell acts, and it holds the top right corner at the potential $\mathcal{E}$ . The junction $J$ with the two plates on it is an isolated conductor which, assuming the capacitors started uncharged, carries no net charge:

\begin{align*} C_1(\varphi_J - \mathcal{E}) + C_2(\varphi_J - 0) &= 0 \\ \varphi_J &= \frac{C_1\mathcal{E}}{C_1 + C_2} \tag{1}\end{align*}

The right plate of $C_1$ , on the corner at $\mathcal{E}$ , therefore carries, by (1),

C_1(\mathcal{E} - \varphi_J) = C_1\mathcal{E}\left(1 - \frac{C_1}{C_1 + C_2}\right)

q_0 = \frac{C_1C_2\,\mathcal{E}}{C_1 + C_2}

(2)

and the two plates on $J$ carry $-q_0$ (on $C_1$ ) and $+q_0$ (on $C_2$ ).

Step 2The charges with the switch closed

Closing the switch joins $J$ to the positive terminal of the left cell, whose negative terminal is on the bottom wire at $\varphi_M = 0$ :

\varphi_J' = \mathcal{E}

Both plates of $C_1$ now sit at $\mathcal{E}$ , and $C_2$ holds the full emf:

q_{C_1}' = C_1(\mathcal{E} - \mathcal{E}) = 0, \qquad q_{C_2}' = C_2\mathcal{E} \tag{3}

Step 3Section 2 feeds only the right plate of

C_1

The right plate of $C_1$ is joined to the rest of the circuit only through the right cell and section $2$ . Charge moving along the arrow of section $2$ (away from $M$ , up through the right cell) lands on that plate, so it equals the change of that plate’s charge, from $q_0$ of (2) to zero by (3):

q_2 = 0 - q_0

q_2 = -\frac{C_1C_2\,\mathcal{E}}{C_1 + C_2}

(4)

The sign says the charge actually flows against the arrow, toward $M$ .

Step 4Section 1 feeds the junction

J

Charge moving along the arrow of section $1$ (away from $M$ , up through the left cell and the closed switch) arrives at $J$ . It equals the change of the total charge on the two plates attached to $J$ : before closing they carry $-q_0 + q_0$ , after closing $0 + C_2\mathcal{E}$ , by (3):

q_1 = (0 + C_2\mathcal{E}) - (-q_0 + q_0)

q_1 = C_2\,\mathcal{E}

(5)

Check

Charge is conserved at $M$ . The lower plate of $C_2$ goes from $-q_0$ to $-C_2\mathcal{E}$ , so the charge that leaves $M$ along the bottom wire must be $C_2\mathcal{E} - q_0$ . The two answers give

q_1 + q_2 = C_2\mathcal{E} - \frac{C_1C_2\,\mathcal{E}}{C_1 + C_2} = C_2\mathcal{E} - q_0

as required. A limit: as $C_1 \to 0$ the right branch is cut off, $q_2 \to 0$ , and only the left cell charges $C_2$ , sending $C_2\mathcal{E}$ through section $1$ , as (5) says for every $C_1$ .

Remark

Before the switch closes, the isolated junction $J$ forces $C_1$ and $C_2$ to share the right cell’s emf in series; closing the switch removes that constraint, and the junction takes on the net charge $C_2\mathcal{E}$ from the left cell. That is why $q_1$ does not depend on $C_1$ at all: the junction starts neutral and ends holding exactly the charge of $C_2$ at the full emf, whatever $C_1$ held before.