Irodov 3.120: A Bridge Balanced by Capacitance Ratios

Problem Irodov 3.120

Three branches join a left-hand wire to a right-hand wire. The top branch holds, from left to right, a capacitor $C_1$ , a point $A$ and a capacitor $C_2$ . The middle branch holds a capacitor $C_3$ , a point $B$ and a capacitor $C_4$ . The bottom branch holds a source of emf $\mathcal{E}$ whose positive terminal faces the right-hand wire. Find the potential difference $\varphi_A - \varphi_B$ . Under what condition is it zero?

Two capacitor branches and the source between the side wires: A between C_1 and C_2, B between C_3 and C_4; the source's long (+) plate faces the right-hand wire. — Two capacitor branches and the source between the side wires: $A$ between $C_1$ and $C_2$ , $B$ between $C_3$ and $C_4$ ; the source’s long ( $+$ ) plate faces the right-hand wire.

Given

$C_1, C_2, C_3, C_4$
$\mathcal{E}$
capacitors uncharged before connection (assumed)

Find

\varphi_A - \varphi_B

, and when it vanishes

Solution

Step 1The source fixes the side wires

Call $\varphi_L$ and $\varphi_R$ the potentials of the left-hand and right-hand wires. No current flows once the capacitors are charged, so the source simply holds its two terminals apart by its emf, the positive one on the right:

\varphi_R - \varphi_L = \mathcal{E} \tag{1}

Step 2Point

A

sits on a neutral conductor

Point $A$ , the wire through it, and the plates of $C_1$ and $C_2$ that touch it form an isolated conductor, which was neutral before the circuit was assembled and stays so. Its charge on $C_1$ is $C_1(\varphi_A - \varphi_L)$ and on $C_2$ is $C_2(\varphi_A - \varphi_R)$ :

\begin{aligned} C_1(\varphi_A - \varphi_L) + C_2(\varphi_A - \varphi_R) &= 0\\ \varphi_A &= \frac{C_1\varphi_L + C_2\varphi_R}{C_1 + C_2} \end{aligned}

Point $B$ is the same construction with $C_3$ and $C_4$ :

\varphi_B = \frac{C_3\varphi_L + C_4\varphi_R}{C_3 + C_4}

Step 3The difference depends only on the emf

Write each mean as the left-hand potential plus a fraction of the step to the right-hand wire, which is (1):

\begin{aligned} \varphi_A &= \varphi_L + \frac{C_2}{C_1 + C_2}\,(\varphi_R - \varphi_L) = \varphi_L + \frac{C_2}{C_1 + C_2}\,\mathcal{E}\\ \varphi_B &= \varphi_L + \frac{C_4}{C_3 + C_4}\,\mathcal{E} \end{aligned}

Subtracting, $\varphi_L$ cancels; over the common denominator:

\varphi_A - \varphi_B = \mathcal{E}\,\frac{C_2(C_3 + C_4) - C_4(C_1 + C_2)}{(C_1 + C_2)(C_3 + C_4)}

The products $C_2C_4$ cancel in the numerator:

\varphi_A - \varphi_B = \mathcal{E}\,\frac{C_2C_3 - C_1C_4}{(C_1 + C_2)(C_3 + C_4)}

(2)

Step 4When the two points are at one potential

The difference vanishes when its numerator does:

C_2C_3 - C_1C_4 = 0

\frac{C_1}{C_2} = \frac{C_3}{C_4}

(3)

Check

Symmetric case: with $C_1 = C_3$ and $C_2 = C_4$ the two branches are identical, so $A$ and $B$ must be at one potential, and indeed the numerator $C_2C_1 - C_1C_2$ vanishes. A limit: as $C_2 \to \infty$ , $A$ is tied to the right-hand wire and $\varphi_A - \varphi_B \to \mathcal{E}\,C_3/(C_3 + C_4)$ , which is $\varphi_R - \varphi_B$ from the expression for $\varphi_B$ . The sign: if only $C_2$ is increased, $A$ is pulled toward the positive right-hand wire and $\varphi_A - \varphi_B$ grows, as the $C_2C_3$ term says.

Remark

This is the capacitor version of the Wheatstone bridge: $A$ and $B$ are the two mid-points, and connecting any element between them would draw no charge when $C_1/C_2 = C_3/C_4$ . Resistors balance when $R_1/R_2 = R_3/R_4$ ; a capacitor’s opposition to charging goes as $1/C$ , so the ratios of $1/C$ balance in the same way, which is the same condition written for capacitances. A capacitive bridge driven by an alternating source is how an unknown capacitance is measured against a standard.