A parallel plate capacitor of capacitance 500 pF charged with 100 ~V supply: It is then disconnected from the supply and connected to another uncharged 500 pF capacitor. The electrostatic energy lost in this process is

Step-by-step solution

Initial stored energy = Energy of capacitor 1 = 1 2 C₁ V₁^2 aligned & = 1 2 500 10⁻¹² (100)^2 ( C₁=500 pF V₁=100 ~V ) & =250 10⁻¹² 10^4=250 10⁻² 10⁻⁶ & =2.5 J aligned When both capacitors are joined, common potential of combination is V_ common =V= Q₁+Q₂ C₁+C₂ or V= C₁ V₁+C₂ V₂ C₁+C₂ or V= 500 10⁻¹² 100 2 500 10⁻¹² =50 ~V Final stored energy = 1 2 (C₁+C₂ ) V^2 aligned & = 1 2 2 500 10⁻¹² (50)^2 & =500 10⁻¹² 2500 & =125 10⁻⁸ & =1.25 10⁻⁶ ~J & =1.25 J aligned Loss of energy, U= Final energy - Initial energy =25-1.25