A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected
from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect? 

When a parallel plate air capacitor connected to a cell of emf V, then charge stored will be

                                   q = CV

$\Rightarrow$                              $\mathrm{V}=\frac{\mathrm{q}}{\mathrm{C}}$

Also energy stored is $\mathrm{U}=\frac{1}{2}{\mathrm{CV}}^2=\frac{{\mathrm{q}}^2}{2\mathrm{C}}$

As the battery is disconnected from the capacitor, the charge will not be destroyed, i.e. q' = q with the introduction of dielectric in the gap of capacitor, so the new capacitance will be

                                     C' = CK

$\Rightarrow$                                $\mathrm{V}\text{'}=\frac{\mathrm{q}}{\mathrm{C}\text{'}}=\frac{\mathrm{q}}{\mathrm{CK}}$

The new energy stored will be

                                     $\mathrm{U}\text{'}=\frac{{\mathrm{q}}^2}{2\mathrm{CK}}$

                                   $∆\mathrm{U}=\mathrm{U}\text{'}-\mathrm{U}$

                                          $=\frac{{\mathrm{q}}^2}{2\mathrm{C}}\left(\frac{1}{\mathrm{K}}-1\right)=\frac{1}{2}{\mathrm{CV}}^2\left(\frac{1}{\mathrm{K}}-1\right)$

Hence, option (4) is incorrect.