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 α × 10⁻⁵ C (assume capacitor Q does not have any charge initially), the value of α i s _______ .
Physics · Capacitor · connecting of capacitor with battery,connenctiong of two capacitor
JEE Mains 2026 22 January Evening Shift

Question

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 α × 10⁻⁵ C (assume capacitor Q does not have any charge initially), the value of α i s _______ .

Physics · Capacitor · connecting of capacitor with battery,connenctiong of two capacitor

JEE Mains 2026 22 January Evening Shift

G1 Physics · Accepted Answer

Initial charge on capacitor P is Q(initial) = C₁V₁ = (10 × 10⁻⁶)(6) = 60 × 10⁻⁶ C. Capacitor Q is initially uncharged. After the two capacitors are connected in parallel, total charge is conserved. The common final voltage is V = [C₁V₁ + C₂V₂]/(C₁ + C₂). Since V₂ = 0, V = [(10 × 10⁻⁶)(6)]/ [(10 + 20) × 10⁻⁶] = 60/30 = 2 V. Charge on capacitor Q is Q₂ = C₂V = (20 × 10⁻⁶)(2) = 40 × 10⁻⁶ C = 4 × 10⁻⁵ C. Therefore, α = 4.