A capacitor, C₁ = 6 μ F is charged to a potential difference of V₀ = 5V using a 5V battery. The battery is removed and another capacitor, C₂ = 12 μ F is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges (q₁ and q₂) on the capacitors C₁ and C₂ when equilibrium condition is reached.

Solution

Physics sir · 4 steps

The battery is gone, so the 30 μC on C1 is all there is. Share it so both capacitors end at one voltage.

1. Charge to share. Before the switch closes, C₁ holds Q = C₁ V₀ = 6 μF × 5 V = 30 μC. C₂ is uncharged. With no battery in the loop, this total charge is conserved.
2. Equilibrium condition. Charge stops flowing when both capacitors are at the same potential difference V. They are effectively in parallel, so the charge splits in the ratio of the capacitances.

   V = (q₁ + q₂)/(C₁ + C₂) = (30 μC)/(18 μF) = 5/3 V

3. Each charge. q₁ = C₁ V = 6 × 5/3 = 10 μC and q₂ = C₂ V = 12 × 5/3 = 20 μC. Check: 10 + 20 = 30 μC, and C₂ has twice the capacitance, so twice the charge.
4. So. At equilibrium q₁ = 10 μC and q₂ = 20 μC, which is option (a).

The idea to keep. When a charged capacitor is joined to an uncharged one with no battery, conserve the total charge and set the voltages equal. The charge then divides in the ratio of the capacitances.

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