Irodov 3.125: The Floating Wire Keeps Zero Charge – Amit & Deepali Sinha Foundation

Problem Irodov 3.125

Three parallel branches join a wire on the left to a wire on the right. Branch $k$ ( $k = 1, 2, 3$ ) holds, from left to right, a cell of emf $\mathcal{E}_k$ whose positive terminal faces the next element, a marked point $k$ , and a capacitor of capacitance $C_k$ . The middle junction of the right-hand wire is the point $O$ , and its potential is taken as zero. Find the potential at point $1$ . Then use the symmetry of that formula to write down the potentials at points $2$ and $3$ .

The three branches. Each cell's long (+) plate faces its numbered point; the left-hand wire is joined to O only through the capacitors, and O is the zero of potential. — The three branches. Each cell’s long ( $+$ ) plate faces its numbered point; the left-hand wire is joined to $O$ only through the capacitors, and $O$ is the zero of potential.

Given

$\mathcal{E}_1, \mathcal{E}_2, \mathcal{E}_3$
$C_1, C_2, C_3$
$\varphi_O = 0$
capacitors uncharged at the start (assumed)

Find

\varphi_1

, then

\varphi_2

and

\varphi_3

Solution

Step 1Each cell fixes a jump of potential

Call $\varphi_L$ the potential of the left-hand wire. Crossing cell $k$ from its short ( $-$ ) plate on that wire to its long ( $+$ ) plate at point $k$ raises the potential by $\mathcal{E}_k$ . No current flows once the capacitors are charged, so nothing else changes the potential along the way.

\varphi_k = \varphi_L + \mathcal{E}_k, \qquad k = 1, 2, 3 \tag{1}

Step 2The charges on the capacitors

Capacitor $C_k$ sits between point $k$ and the right-hand wire, which is at $\varphi_O = 0$ . Its plate on the side of point $k$ therefore carries

q_k = C_k(\varphi_k - \varphi_O) = C_k\varphi_k \tag{2}

Step 3The isolated conductor carries no net charge

We assume, as the problem intends, that the capacitors were uncharged before the circuit was assembled; the charges on the three left plates then add up to zero. Using (2):

C_1\varphi_1 + C_2\varphi_2 + C_3\varphi_3 = 0

(3)

Substituting (1) for each $\varphi_k$ and solving for the one unknown $\varphi_L$ :

\begin{aligned} C_1(\varphi_L + \mathcal{E}_1) + C_2(\varphi_L + \mathcal{E}_2) + C_3(\varphi_L + \mathcal{E}_3) &= 0\\[2pt] (C_1 + C_2 + C_3)\,\varphi_L &= -(C_1\mathcal{E}_1 + C_2\mathcal{E}_2 + C_3\mathcal{E}_3) \end{aligned}

\varphi_L = -\frac{C_1\mathcal{E}_1 + C_2\mathcal{E}_2 + C_3\mathcal{E}_3}{C_1 + C_2 + C_3}

(4)

Step 4The potential at point 1

Point $1$ sits $\mathcal{E}_1$ above the left-hand wire, by (1). Adding $\mathcal{E}_1$ to (4) over the common denominator:

\varphi_1 = \frac{\mathcal{E}_1(C_1 + C_2 + C_3) - C_1\mathcal{E}_1 - C_2\mathcal{E}_2 - C_3\mathcal{E}_3}{C_1 + C_2 + C_3}

The terms $C_1\mathcal{E}_1$ cancel, leaving

\varphi_1 = \frac{\mathcal{E}_1(C_2 + C_3) - \mathcal{E}_2C_2 - \mathcal{E}_3C_3}{C_1 + C_2 + C_3}

(5)

Step 5Points 2 and 3 by symmetry

The three branches enter the argument on an equal footing: nothing above depended on which branch was called $1$ . Exchanging the labels $1$ and $2$ in (5) gives the potential at point $2$ , and exchanging $1$ and $3$ gives the potential at point $3$ .

\varphi_2 = \frac{\mathcal{E}_2(C_1 + C_3) - \mathcal{E}_1C_1 - \mathcal{E}_3C_3}{C_1 + C_2 + C_3}

(6)

\varphi_3 = \frac{\mathcal{E}_3(C_1 + C_2) - \mathcal{E}_1C_1 - \mathcal{E}_2C_2}{C_1 + C_2 + C_3}

(7)

Check

The three answers must satisfy the neutrality condition (3). Collecting the coefficient of $\mathcal{E}_1$ in $C_1\varphi_1 + C_2\varphi_2 + C_3\varphi_3$ (times the common denominator) gives $C_1(C_2 + C_3) - C_2C_1 - C_3C_1 = 0$ , and likewise for $\mathcal{E}_2$ and $\mathcal{E}_3$ , so the sum vanishes identically. A special case: with three equal emfs $\mathcal{E}$ , every numerator is $\mathcal{E}(C_2 + C_3) - \mathcal{E}C_2 - \mathcal{E}C_3 = 0$ , so all three points sit at the potential of $O$ and no capacitor is charged, which is right, since the cells then simply hold the left-hand wire at $-\mathcal{E}$ . A limit: as $C_1 \to \infty$ , $\varphi_1 \to 0$ , because a huge capacitor pins point $1$ to $O$ .

Remark

Written through (4), every answer has the form $\varphi_k = \mathcal{E}_k - \bar{\mathcal{E}}$ , with $\bar{\mathcal{E}} = \sum C_i\mathcal{E}_i/\sum C_i$ the capacitance-weighted mean emf: each point stands above or below $O$ by how far its own cell exceeds or falls short of that mean. If all three cells are turned around, with every $+$ plate on the left-hand wire, each $\mathcal{E}_k$ changes sign and so does every potential:

\varphi_1 = \frac{\mathcal{E}_2C_2 + \mathcal{E}_3C_3 - \mathcal{E}_1(C_2 + C_3)}{C_1 + C_2 + C_3}

and likewise for points $2$ and $3$ .