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A capacitor $$C_1$$ of capacitance 5 $$\mu$$F is charged to a potential of 30 V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor $$C_2$$ of capacitance 10 $$\mu$$F as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor $$C_2$$ is ______ $$\mu$$C.
Correct Answer: 100
Initial charge on $$C_1$$: $$Q_i = C_1 V_1 = 5\ \mu\text{F} \times 30\ \text{V} = 150\ \mu\text{C}$$
Common potential after closing the switch:
$$V_c = \frac{Q_i}{C_1 + C_2} = \frac{150}{5 + 10} = \frac{150}{15} = 10\ \text{V}$$
Final charge on $$C_2$$: $$Q_2 = C_2 V_c = 10\ \mu\text{F} \times 10\ \text{V} = 100\ \mu\text{C}$$
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