A 400 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 400 pF capacitor. The electrostatic energy lost in the process is

$4 × 10^{-6} J$

The correct answer is Option (2) → $4 × 10^{-6} J$

Initial energy in first capacitor: $U_i=\frac{1}{2}CV^2=\frac{1}{2}\cdot400\times10^{-12}\cdot(200)^2$

$U_i=0.5\cdot400\times10^{-12}\cdot40000=8\times10^{-6}\ \text{J}$

When connected to another $400\ \text{pF}$ uncharged capacitor, equivalent capacitance:

$C_{\text{eq}}=400+400=800\ \text{pF}=800\times10^{-12}\ \text{F}$

Total charge conserved: $Q=C V=400\times10^{-12}\cdot200=8\times10^{-8}\ \text{C}$

Final voltage across both capacitors: $V_f=\frac{Q}{C_{\text{eq}}}=\frac{8\times10^{-8}}{800\times10^{-12}}=100\ \text{V}$

Final energy: $U_f=\frac{1}{2}C_{\text{eq}}V_f^2=\frac{1}{2}\cdot800\times10^{-12}\cdot(100)^2$

$U_f=4\times10^{-6}\ \text{J}$

Energy lost: $\Delta U=U_i-U_f=8\times10^{-6}-4\times10^{-6}=4\times10^{-6}\ \text{J}$

Answer: $4\times10^{-6}\ \text{J}$