The capacity and the energy stored in a charged parallel plate condenser with air between its plates
are respectively, C₀ and W₀. If the air is replaced by glass (dielectric constant = 5) between the plates, the capacity of the plates and the energy stored in it will respectively be

On introducing dielectric K in a parallel plate capacitor, its capacity becomes,

                                   $\mathrm{C}\text{'}={\mathrm{KC}}_0$

$\therefore \mathrm{C}\text{'}=5{\mathrm{C}}_0$

Also, energy stored,   ${\mathrm{W}}_0=\frac{{\mathrm{q}}^2}{2{\mathrm{C}}_0}$

$\therefore \mathrm{W}\text{'}=\frac{{\mathrm{q}}^2}{2\mathrm{C}\text{'}}=\frac{{\mathrm{q}}^2}{2\times 5{\mathrm{C}}_0}$

$\therefore \frac{{\mathrm{W}}_0}{\mathrm{W}\text{'}}=\frac{5}{1}$

$\text{∴ W'=}\frac{W_0}{5}$

Energy stored in the capacitor will be $\frac{U_0}{K} when dielectric is placed after removing the charging battery from the capacitor$  ( clarity should be given about the battery connection with the capacitor.