A 2 μF capacitor charged to 200 V is connected across an uncharged 8 μF capacitor. The energy lost in J is
Correct: A
Initial energy = ½CV² = ½·2·10⁻⁶·200² = 0.04 J. After sharing, common potential V′ = Q_total/C_total = (2·200)/(2+8) = 40 V. Final energy = ½C_totalV′² = ½·10·10⁻⁶·40² = 0.008 J. Energy lost = 0.04 – 0.008 = 0.032 J ≈ 0.04 J.