A 2μF capacitor C₁ is first charged to a potential difference of 10V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor C₂ of 8 μF. The charge in C₂ on equilibrium condition is ____________ μC. (Round off to the Nearest Integer)

Solution

Physics sir · 4 steps

Once the battery is gone, the 20 μC on C1 is all you have. Share it until both voltages match.

1. Charge to share. With S₁ closed, C₁ charges to Q = C₁ V = 2 μF × 10 V = 20 μC. Opening S₁ removes the battery, so this charge cannot change. It can only spread out.
2. Common potential. Closing S₂ puts C₁ and C₂ in parallel. Charge flows until both have the same voltage V_c, and the total charge stays 20 μC.

   V_c = (C₁ V)/(C₁ + C₂) = (20 μC)/(2 μF + 8 μF) = 2 V

3. Charge on C2. Q₂ = C₂ V_c = 8 μF × 2 V = 16 μC. Check: C₁ keeps 2 × 2 = 4 μC, and 16 + 4 = 20 μC.
4. So. The charge on C₂ at equilibrium is 16 μC.

The common slip: 80. Taking C₂ to reach the full 10 V gives 8 × 10 = 80, not 16. The battery is gone, so the voltage falls to the common value of 2 V.

The idea to keep. Battery removed means charge is conserved, not voltage. Total charge divided by total capacitance gives the common voltage, and the charge splits in the ratio of the capacitances.

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