Two capacitors having capacitance C₁ and C₂ respectively are connected as shown in figure. Initially, capacitor C₁ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C₁ is now connected to uncharged capacitor C₂ by closing the switch S. The amount of charge on the capacitor C₂, after equilibrium, is :

Solution

Physics sir · 4 steps

No battery, so charge is conserved: let C1 share its charge with C2 until both sit at one voltage.

1. Charge to start. The battery charges C₁ to V, so it holds Q = C₁ V. Once the battery is removed, this charge has nowhere to go except onto C₂, so the total charge stays C₁ V.
2. Common potential. With S closed, both capacitors connect across the same two wires, so they are in parallel. Charge flows until both have the same voltage V', which means C₁ V' + C₂ V' = C₁ V.

   V' = (C₁ V)/(C₁ + C₂)

3. Charge on C2. Q₂ = C₂ V' = (C₁ C₂)/(C₁ + C₂)V. As a check, C₁ keeps (C₁² V)/(C₁ + C₂), and the two charges add back to C₁ V.
4. So. The charge on C₂ is (C₁ C₂)/(C₁ + C₂)V, which is option (a).

The idea to keep. When a charged capacitor is joined to an uncharged one with no battery, conserve total charge and set a common potential: V' = (total charge)/(total capacitance).

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