COMEDK20269 May 2026Evening ShiftPhysicsCapacitanceActual
A capacitor of capacitance 8 F is fully charged by connecting it to a source of 200V. It is then disconnected from the supply and connected to an uncharged capacitor of capacitance 4 F. The electrostatic energy lost in this sharing is:
Options
- A10.67 10⁻² J
- B5.33 10⁻² J
- C3.53 10⁻³ J
- D21.34 10⁻² J
Correct answer
B. 5.33 10⁻² J
Step-by-step solution
Given: C₁ = 8 F = 8 10⁻⁶ F V₁ = 200 V C₂ = 4 F = 4 10⁻⁶ F V₂ = 0 V The loss of electrostatic energy when two capacitors are connected is given by: U = 1 2 C₁ C₂ C₁ + C₂ (V₁ - V₂)^2 Substituting the values: U = 1 2 8 10⁻⁶ 4 10⁻⁶ (8 + 4) 10⁻⁶ (200 - 0)^2 U = 1 2 32 10⁻¹² 12 10⁻⁶ 40000 U = 16 12 10⁻⁶ 4 10^4 U = 4 3 4 10⁻² U = 16 3 10⁻² = 5.33 10⁻² J Answer: 5.33 10⁻² J