A capacitor of $2\mu f$ is charged to a potential of 4V. using a battery . And then the battery is disconnected and the charged capacitor is connected to an uncharged capacitor of $4\mu \mathrm{f}$ capacitance. When the equilibrium is established the total energy stored in the capacitors is 

The situation is shown in figure, when the switch is closed the charge flows from $\mathbf{2}\mathit{\mu }\mathbf{f}$ capacitor to 4µf capacitor till both acquires the same potential 

Let final common potential of capacitors is V, then from conservation of charge

$\begin{array}{l}2\mu f\times 4V=2\mu f\times V+4\mu f\times V\\ \Rightarrow V=\frac{8}{6}\text{ Volt }=\frac{4}{3}\text{ Volt }\end{array}$

So total final energy stored in capacitors is. 

$\begin{array}{l}\mathrm{\Sigma }{U}_1=\frac{2\mu f\times V^2}{2}+\frac{4\times V^2}{2}\\ \mathrm{\Sigma }{U}_f=\frac{6}{2}\times {\left(\frac{4}{3}\right)}^2=\frac{16}{3}\mu \mathrm{J}\end{array}$