Practice question

**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². U = (1/2) C V² = (1/2) × 3 × 10⁻⁶ × (300)² = (1/2) × 3 × 10⁻⁶ × 9 × 10⁴ = 0.135 J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 0.135 J follows, reflecting potential-capacitance relations.