A 600 pF capacitor is charged by 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. Electrostatic energy lost in the process i s _________ μJ

Question

G1 Physics · Accepted Answer

Initial capacitor: C = 600 pF = 600×10⁻¹² F V = 200 V Initial energy: Uᵢ = (1/2)CV² = (1/2)(600×10⁻¹²)(200)² = 12×10⁻⁶ J = 12 μJ After connecting to an identical uncharged capacitor, total capacitance is 2C and the common voltage becomes V/2 = 100 V. Final energy: U(f) = 2 × (1/2)C(100)² = 6 μJ Energy lost: ΔU = 12 - 6 = 6 μJ Final value: 6