Irodov 3.132: A Mirror Move That Still Heats – Amit & Deepali Sinha Foundation

Problem Irodov 3.132

Three capacitors stand side by side, of capacitances $C$ , $C_0$ and $C$ , with their upper plates joined by a common wire. The lower plate of the left $C$ is joined to contact $1$ and the lower plate of the right $C$ to contact $2$ , and a cell of emf $\mathcal{E}$ joins these two lower plates, its positive terminal on the side of the right $C$ . The lower plate of $C_0$ is connected to the arm of a switch $Sw$ , which stands at contact $1$ . How much heat is generated after the switch is moved from contact $1$ to contact $2$ ?

The circuit. The solid arrow shows the switch on contact 1, the dashed arrow the contact 2 it is moved to; the cell's long (+) plate faces the right-hand C. — The circuit. The solid arrow shows the switch on contact $1$ , the dashed arrow the contact $2$ it is moved to; the cell’s long ( $+$ ) plate faces the right-hand $C$ .

Given

$C$ , $C_0$ , $\mathcal{E}$
steady state before the switch moves (assumed)
capacitors uncharged before the circuit was assembled (assumed)

Find

the heat

Q

generated

Solution

Step 1The potentials before the switch moves

Take the lower node of the left $C$ (wired to contact $1$ ) at $\varphi_1 = 0$ ; the cell puts the lower node of the right $C$ (wired to contact $2$ ) at $\varphi_2 = \mathcal{E}$ . The common top wire $T$ and the three upper plates form an isolated conductor, neutral since the capacitors started uncharged. With $C_0$ on contact $1$ , the left $C$ and $C_0$ both face $\varphi_1$ :

\begin{align*} (C + C_0)(\varphi_T - 0) + C(\varphi_T - \mathcal{E}) &= 0 \\ \varphi_T &= \frac{C}{2C + C_0}\,\mathcal{E} \tag{1}\end{align*}

Step 2The potentials after the switch moves

Now $C_0$ faces $\varphi_2$ together with the right $C$ , and the neutrality of $T$ reads

\begin{align*} C(\varphi_T' - 0) + (C + C_0)(\varphi_T' - \mathcal{E}) &= 0 \\ \varphi_T' &= \frac{C + C_0}{2C + C_0}\,\mathcal{E} \tag{2}\end{align*}

Step 3The stored energy does not change

Before the move, the group $C + C_0$ holds the voltage $\varphi_T$ and the right $C$ the voltage $\mathcal{E} - \varphi_T$ . With $D = 2C + C_0$ and (1):

\begin{aligned} W_i &= \frac{1}{2}(C + C_0)\left(\frac{C\mathcal{E}}{D}\right)^2 + \frac{1}{2}\,C\left(\frac{(C + C_0)\mathcal{E}}{D}\right)^2\\[2pt] &= \frac{C(C + C_0)(C + C + C_0)\,\mathcal{E}^2}{2D^2} = \frac{C(C + C_0)\,\mathcal{E}^2}{2(2C + C_0)} \end{aligned}

After the move, by (2), the left $C$ holds the voltage $(C + C_0)\mathcal{E}/D$ and the group of $C_0$ with the right $C$ holds $C\mathcal{E}/D$ . A capacitance $C$ again carries $(C + C_0)\mathcal{E}/D$ and a capacitance $C + C_0$ again carries $C\mathcal{E}/D$ , so $W_f$ is the same two terms and

\Delta W = W_f - W_i = 0 \tag{3}

Step 4The charge driven through the cell

Node $1$ is the negative side of the cell, and after the arm has left it holds only the lower plate of the left $C$ , whose charge is $C(\varphi_1 - \varphi_T)$ . Its change, from (1) to (2):

\Delta q_1 = C(0 - \varphi_T') - C(0 - \varphi_T) = -\frac{C(C + C_0) - C^2}{2C + C_0}\,\mathcal{E} = -\frac{CC_0}{2C + C_0}\,\mathcal{E}

So the charge $CC_0\mathcal{E}/(2C + C_0)$ leaves node $1$ and passes through the cell from its negative to its positive terminal. The cell does the work

A = \mathcal{E}\cdot\frac{CC_0\,\mathcal{E}}{2C + C_0} \tag{4}

Step 5The heat

Energy conservation: the work of the cell goes into the field energy and into heat. With $\Delta W = 0$ from (3) and $A$ from (4):

Q = A - \Delta W

Q = \frac{CC_0\,\mathcal{E}^2}{2C + C_0}

(5)

Check

The charge bookkeeping at node $2$ must agree. Before the move it holds the lower plate of the right $C$ , the arriving arm brings the lower plate of $C_0$ , and at the end the node holds both:

\begin{aligned} q_{\mathrm{before}} &= C(\mathcal{E} - \varphi_T)\\[2pt] q_{\mathrm{arm}} &= C_0(0 - \varphi_T)\\[2pt] q_{\mathrm{after}} &= (C + C_0)(\mathcal{E} - \varphi_T') \end{aligned}

Using (1) and (2), the gain is

\frac{(C + C_0)C - C(C + C_0) + C_0C}{2C + C_0}\,\mathcal{E} = \frac{CC_0}{2C + C_0}\,\mathcal{E}

exactly what left node $1$ . Limits: with $C_0 = 0$ nothing moves and no heat is generated; with $C_0 \to \infty$ the heat tends to $C\mathcal{E}^2$ .

Remark

The energy stored in the field is the same before and after, yet heat $CC_0\mathcal{E}^2/(2C + C_0)$ appears: all of it is work done by the cell. A switching process can leave the capacitors in a state of the same energy and still be irreversible; the charge had to be pushed round through the cell, and the resistance of the wires turns that work into heat, whatever its value.