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 :

(A) \(0.5 \times 10^{-6} J\)
Initial energy of the charged capacitor is given by \(U_i=\frac{1}{2} C V^2\).
When it is connected to an identical uncharged capacitor, the common potential becomes \(V^{\prime}=\frac{C V}{C+C}=\frac{V}{2}\).
Final energy of the system is \(U_f=\frac{1}{2}(2 C) V^{\prime 2}=\frac{1}{2}(2 C)\left(\frac{V}{2}\right)^2=\frac{1}{4} C V^2\).
Energy lost during the process is \(\Delta U=U_i-U_f=\frac{1}{2} C V^2-\frac{1}{4} C V^2=\frac{1}{4} C V^2\).
Substituting the given values \(C=200 \times 10^{-12} F\) and \(V=100 V\) :
\(\Delta U=\frac{1}{4} \times 200 \times 10^{-12} \times(100)^2\)
\(\Delta U=50 \times 10^{-12} \times 10^4=0.5 \times 10^{-6} J\).