A capacitor C₁ of capacitance 5 μF 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 C₂ of capacitance 10 μF as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor C₂ is __________ μC.

Solution

Physics sir · 3 steps

Charge on C1 has nowhere to go but C2, so share it until both plates sit at one common voltage.

1. Initial charge. With the battery removed, C₁ holds Q = C₁ V = 5 μF × 30 V = 150 μC. C₂ starts with zero charge.
2. Charge is conserved. After the switch closes, the two capacitors are joined top plate to top plate and bottom to bottom, so they are in parallel. The total charge of 150 μC stays the same and spreads until both have the same voltage.

   V_(common) = (C₁ V₁)/(C₁ + C₂) = (150 μC)/(15 μF) = 10 V

3. So. Q₂ = C₂ V_(common) = 10 μF × 10 V = 100 μC. The remaining 50 μC stays on C₁, and 100 + 50 = 150 checks out.

The common slip: 50. Sharing the charge in the inverse ratio of capacitances, as in series, gives 50, which is actually the charge left on C₁.

The idea to keep. When a charged capacitor is joined to an uncharged one, total charge is conserved and the common voltage is (total Q)/(total C). The bigger capacitor ends up with the bigger share.

Answer from the NTA final answer key. Solution by PrepWiser.