A \( 2 \, \mu\text{F} \) capacitor charged to \( 200 \, \text{V} \) loses how much energy when connected to an uncharged
**Applications** include charge sharing for voltage division, Van de Graaff generator accumulates charge on spherical dome to high potential V = k Q/R, up to MV, using belt to transport charge, capacitance of sphere C=4π ε₀ R ≈10 pF for R=0.1 m, so Q= C V ≈10⁻⁸ C for 1000 V. Initial energy: U_i = (1/2) × 2 × 10⁻⁶ × (200)² = 0.04 J . Charge: Q = 2 × 10⁻⁶ × 200 = 4 × 10⁻⁴ C . Total C = 2 + 3 = 5 μF , V = (4 × 10⁻⁴/5 × 10⁻⁶) = 80 V . Final energy: U_f =
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications