Question

Consider the circuit shown in the figure below, where C1 = 4.00 mu F, C2 = 9.00 mu F, and triangleV = 22.0 V. Capacitor C1 is first charged by closing switch S1. Switch S1 is then opened, and the charged capacitor is connected to the uncharged capacitor by closing S2.

Answer 1

When switch $S_1$ is closed, capacitor $C_1$ is charged until potential difference across it is equal to $\Delta{V}$. Hence the charge on $C_1$ is $C_1\Delta{V}=4\times10^{-6}\times22.0=88.0\times10^{-6}\,C$ .

Now switch $S_1$ is opened and switch $S_2$ ic closed. Now charge on $C_1$ gets redistributed until potential differences across capacitors $C_1$ and $C_2$  are equal in magnitude. And the net potential difference across the loop consisting of capacitors $C_1$ and $C_2$ is zero. Now let the charge on $C_2$ be q. Then charge on $C_1$ will be $C_1\Delta{V}-q$.

since potential differences are equal $\frac{C_1\Delta{V}-q}{C_1}=\frac{q}{C_2}$

$\frac{88.0\times10^{-6}-q}{4\times10^{-6}}=\frac{q}{9\times10^{-6}}$

charge on capacitor $C_2$ = $q=\frac{9\times88\times10^{-6}}{13}=60.9\times10^{-6}\,C$

charge on capacitor $C_1$ = $(88.0-60.9)\times10^{-6}\,C=27.1\times10^{-6}\,C$