A capacitor $C_1 = 6\ \mu\text{F}$ is charged to a potential difference of $V_0 = 5\ \text{V}$ using a $5\ \text{V}$ battery. The battery is removed and another capacitor $C_2 = 12\ \mu\text{F}$ is inserted in place of the battery. When the switch S is closed, charge flows between the capacitors for some time until equilibrium is reached. What are the charges ($q_1$ and $q_2$) on the capacitors $C_1$ and $C_2$ when equilibrium is reached?
Answer: (A) $q_1 = 10\ \mu\text{C},\ q_2 = 20\ \mu\text{C}$
Initial charge on $C_1$: $6\times5 = 30\ \mu\text{C}$. This charge is shared until both capacitors are at the same voltage:
$$V = \frac{30}{6 + 12} = \frac{5}{3}\ \text{V}$$
$$q_1 = 6\times\frac{5}{3} = 10\ \mu\text{C}, \qquad q_2 = 12\times\frac{5}{3} = 20\ \mu\text{C}$$
Solution by Sreeraj P, M.Sc Physics