Irodov 3.121: Sharing Charge with a Series Pair – Amit & Deepali Sinha Foundation

Problem Irodov 3.121

A capacitor $C_1 = 1.0$ $\mu$ F has been charged to the voltage $V = 110$ V. It is then connected in parallel across the ends of a chain of two uncharged capacitors in series, of capacitances $C_2 = 2.0$ $\mu$ F and $C_3 = 3.0$ $\mu$ F. How much charge flows through the connecting wires?

The charged capacitor C_1 and the uncharged series pair C_2, C_3, before their terminals are joined in parallel. — The charged capacitor $C_1$ and the uncharged series pair $C_2$ , $C_3$ , before their terminals are joined in parallel.

Given

$C_1 = 1.0~\mu\text{F}$ charged to $V = 110$ V
$C_2 = 2.0~\mu\text{F}$
$C_3 = 3.0~\mu\text{F}$
$C_2$ , $C_3$ uncharged

Find

the charge

\Delta q

through the connecting wires

Solution

Step 1The series pair acts as one capacitor

The inner plates of $C_2$ and $C_3$ and the wire between them form an isolated conductor that starts neutral, so both capacitors always carry one charge and the pair behaves as a single capacitor:

C_s = \frac{C_2C_3}{C_2 + C_3} \tag{1}

Numerically:

C_s = \frac{2.0 \times 3.0}{2.0 + 3.0}~\mu\text{F} = 1.2~\mu\text{F}

Step 2Charge is shared at one common voltage

Before the connection $C_1$ holds $q_0 = C_1V$ . Afterwards the upper plate of $C_1$ and the upper plate of the pair are one conductor, with no path for charge to leave it, so the charge $q_0$ is now spread over $C_1$ and $C_s$ at a common voltage $U$ :

\begin{aligned} C_1V &= C_1U + C_sU\\ U &= \frac{C_1V}{C_1 + C_s} \end{aligned}

Step 3The charge that moved

Everything now on the pair arrived through the connecting wires, since the pair started uncharged:

\Delta q = C_sU = \frac{C_1C_s}{C_1 + C_s}\,V

Writing it with reciprocals and substituting (1), $1/C_s = 1/C_2 + 1/C_3$ :

\Delta q = \frac{V}{1/C_1 + 1/C_s}

Numerically, the reciprocals add to $(1.0 + 0.50 + 0.333)~\mu\text{F}^{-1} = 1.83~\mu\text{F}^{-1}$ :

\Delta q = \frac{110~\text{V}}{1.83~\mu\text{F}^{-1}}

\Delta q = \frac{V}{1/C_1 + 1/C_2 + 1/C_3} = 60~\mu\text{C}

(2)

After the connection, drawn for the upper plate of C_1 positive: C_1 keeps 50 , and C_2 and C_3 each carry the 60 that flowed. — After the connection, drawn for the upper plate of $C_1$ positive: $C_1$ keeps $50~\mu\text{C}$ , and $C_2$ and $C_3$ each carry the $60~\mu\text{C}$ that flowed.

Check

The common voltage is $U = \Delta q/C_s = 60~\mu\text{C}/1.2~\mu\text{F} = 50$ V, so $C_1$ keeps $C_1U = 50~\mu\text{C}$ , and $50 + 60 = 110~\mu\text{C} = C_1V$ : no charge is lost. Across the pair, $C_2$ holds $60/2.0 = 30$ V and $C_3$ holds $60/3.0 = 20$ V, which add to $U = 50$ V. A limit: if the pair were huge, $\Delta q \to C_1V$ , all of $C_1$ ’s charge would leave it; if it were tiny, almost none would.

Remark

The energy does not survive the sharing. Before, $C_1V^2/2 = 6.05$ mJ; after, $(C_1 + C_s)U^2/2 = 2.75$ mJ. The missing $3.3$ mJ, which is $\Delta q\,V/2$ , goes into heat in the wires and radiation, however small their resistance: it is the same loss as in an inelastic collision, fixed by the conservation law alone.