A 100 pF capacitor is charged to 60 V and then disconnected from the battery. A second uncharged capacitor is connected in parallel with it. The final common voltage is 20 V. The capacitance of the second capacitor is:
Options

Question

A 100 pF capacitor is charged to 60 V and then disconnected from the battery. A second uncharged capacitor is connected in parallel with it. The final common voltage is 20 V. The capacitance of the second capacitor is:

Options

G1 Physics · Accepted Answer

Initial charge on the first capacitor: Qᵢ = C₁Vᵢ = 100 pF×60 V. After connection, both capacitors are at the common voltage 20 V. Charge is conserved: C₁Vᵢ = (C₁+C₂) V f . V_f. V f ​ . Therefore, 100×60 = (100+C₂)×20. Divide by 20: 300 = 100+C₂. Hence, C₂ = 200 pF. Correct option: B.