Practice question
Question
A \( 10 \, \mu\text{F} \) capacitor charged to \( 60 \, \text{V} \) is connected to an uncharged \( 30
\, \mu\text{F} \) capacitor. What is the energy lost?
Explanation
**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) × 10 × 10⁻⁶ × (60)² = 0.018 J . Charge: Q = 10 × 10⁻⁶ × 60 = 6 × 10⁻⁴ C . Total C = 10 + 30 = 40 μF , V = (6 × 10⁻⁴/40 × 10⁻⁶) = 15 V . Final energy: U_f =
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