JEE Main202622 January 2026Evening ShiftPhysicsCapacitanceActual
A capacitor P with capacitance 10 10⁻⁶ ~F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor P is connected across another capacitor Q with capacitance 20 10⁻⁶ ~F . The charge on capacitor Q when equilibrium is established will be 10⁻⁵ C (assume capacitor Q does not have any charge initially), the value of is _ _ _ _ 。
Correct answer
0
Step-by-step solution
Capacitor P (10 × 10⁻⁶ F) is charged to 6.0 V. Initial charge: Q_P = 10 10⁻⁶ 6 = 60 10⁻⁶ C. When connected to uncharged capacitor Q (20 × 10⁻⁶ F), charge redistributes until both reach same potential. Final voltage: V_f = Q_ total C_ total = 60 10⁻⁶ 30 10⁻⁶ = 2 V. Charge on Q: Q_Q = 20 10⁻⁶ 2 = 40 10⁻⁶ C = 4 10⁻⁵ C. Thus = 4 .