Practice question
Question
A \( 2 \, \mu\text{F} \) capacitor charged to \( 400 \, \text{V} \) is connected to an uncharged \( 6
\, \mu\text{F} \) capacitor. What is the energy lost?
Explanation
**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial energy: U_i = (1/2) × 2 × 10⁻⁶ × (400)² = 0.16 J . Charge: Q = 2 × 10⁻⁶ × 400 = 8 × 10⁻⁴ C . Total C = 2 + 6 = 8 μF , V = (8 × 10⁻⁴/8 × 10⁻⁶) = 100 V . Final energy: U_f = (1/2) × 8 × 10⁻⁶ × (100)² = 0.04 J . Loss: U_i - U_f =
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