NEET2026PhysicsChapterActual
A 4 F capacitor is charged to a potential difference of 50 V . The charging battery is then disconnected, and the capacitor is connected in parallel across an uncharged 6 F capacitor. What is the heat dissipated in the circuit during the redistribution of charge?
Options
- A2 10⁻³ J
- B3 10⁻³ J
- C5 10⁻³ J
- D6 10⁻³ J
Correct answer
B. 3 10⁻³ J
Step-by-step solution
The heat dissipated during the redistribution of charge when two capacitors are connected in parallel is given by the loss in electrostatic potential energy: U = 1 2 C₁ C₂ C₁ + C₂ (V₁ - V₂)^2 Substituting the given values C₁ = 4 F , C₂ = 6 F , V₁ = 50 V , and V₂ = 0 V : U = 1 2 (4 10⁻⁶) (6 10⁻⁶) (4 + 6) 10⁻⁶ (50 - 0)^2 U = 1 2 24 10⁻¹² 10 10⁻⁶ 2500 U = 1.2 10⁻⁶ 2500 U = 3 10⁻³ J Answer: 3 10⁻³ J