An air filled parallel plate capacitor has capacitance $C_0$. Now the air column in the capacitor is completely replaced by a dielectric slab having a dielectric constant $K$ and the capacitor is connected to a battery of emf $E$ and then the slab is taken out. In the process of taking out the dielectric, it is observed that

$$\begin{gathered} Q_i=C_{0}E and Q_f=k{C}_{0}E \\ △Q=(k-1)C_{0}E \\ {U}_i=\frac{1}{2}{C}_0{E}^2 and U_f=\frac{1}{2}k{C}_0{E}^2 \\ {W}_b=Work done by the battery =△Q\times E =(k-1)C_0{E}^2 = Energy absorbed by the cell. \\ Work done by external agent =W_e \\ Now , W_b+W_e=U_f-U_i+Lost energy . \\ Since there is no resistive element in the circuit and the dielectric slab is removed from the capacitor very slowly , \\ lost energy can be ignored . \\ So W_e=\frac{1}{2}{C}_0(k-1)E^2 \end{gathered}$$