A 4 $\mathrm{\mu }$F capacitor is charged by a 200V battery. It is then disconnected from the supply and is connected  to another uncharged 2 $\mathrm{\mu }$F capacitor. During the process, the loss of energy (in J) is

${\mathrm{C}}_1=4\mathrm{\mu F}=4\times {10}^{-6}\mathrm{F}$; ${\mathrm{V}}_1=200\mathrm{V}$

${\mathrm{E}}_1=\frac{1}{2}{\mathrm{C}}_1{\mathrm{V}}_1^2=\frac{1}{2}\times 4\times {10}^{-6}\times (200{)}^2=8\times {10}^{-2}\mathrm{J}$;

${\mathrm{C}}_2=2\mathrm{\mu F}=2\times {10}^{-6}\mathrm{F}$; 

$\text{ According to the conservation of charge, the initial charge on capacitor }{C}_1$$\text{ is equal to the final charge on capacitors, }{C}_1\text{ and }{C}_2\text{. }$

$$\begin{gathered} \therefore {\mathrm{V}}_2\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)={\mathrm{C}}_1{\mathrm{V}}_1 \\ \Rightarrow {\mathrm{V}}_2=\frac{400}{3}\mathrm{V} \end{gathered}$$

$$\begin{gathered} {\mathrm{E}}_2=\frac{1}{2}\left({\mathrm{C}}_1+{\mathrm{C}}_2\right){\mathrm{V}}_2^2 \\ =\frac{1}{2}(2+4)\times {10}^{-6}\times {\left(\frac{400}{3}\right)}^2 \\ =5.33\times {10}^{-2}\mathrm{J} \end{gathered}$$

$\mathrm{energy} \mathrm{loss}={\mathrm{E}}_1-{\mathrm{E}}_2$

$=8\times {10}^{-2}-5.33\times {10}^{-2}=2.67\times {10}^{-2}\mathrm{J}$