Two initially uncharged capacitors each have capacitance 200 pF. One is charged to 100 V by a supply, disconnected and connected in parallel to the other uncharged capacitor. Find the electrostatic energy dissipated during charge sharing.
Options

Question

Two initially uncharged capacitors each have capacitance 200 pF. One is charged to 100 V by a supply, disconnected and connected in parallel to the other uncharged capacitor. Find the electrostatic energy dissipated during charge sharing.

Options

G1 Physics · Accepted Answer

Step 1: Initially C = 200 pF = 200 × 10⁻¹² F and the first capacitor carries charge Q = CV. Step 2: When identical capacitors are joined in parallel with like terminals, total capacitance is 2C, and the common final voltage is V(final) = Q/(2C) = 50 V. Step 3: Initial stored energy U(initial) = ½CV². Step 4: Final stored energy U(final) = ½(2C)(V/2)² = ¼CV². Step 5: Energy lost ΔU = U(initial) − U(final) = ¼CV² = ¼ × 200 × 10⁻¹² × 100² = 5 × 10⁻⁷ J. Final answer: 0.5 × 10⁻⁶ J (Option C).