Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is :

Step-by-step solution

Initial energy of the charged capacitor is given by U_i = 1 2 CV^2 . When it is connected to an identical uncharged capacitor, the common potential becomes V' = CV C+C = V 2 . Final energy of the system is U_f = 1 2 (2C)V'^2 = 1 2 (2C) ( V 2 )^2 = 1 4 CV^2 . Energy lost during the process is U = U_i - U_f = 1 2 CV^2 - 1 4 CV^2 = 1 4 CV^2 . Substituting the given values C = 200 10⁻¹² F and V = 100 V: U = 1 4 200 10⁻¹² (100)^2 U = 50 10⁻¹² 10^4 = 0.5 10⁻⁶ J. Answer: 0.5 10⁻⁶ J