$A$ 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?

Capacitance of the capacitor, $C=\frac{Q}{V}$

After inserting the dielectric, new capacitance ${\mathrm{C}}^{\mathrm{\prime }}=\mathrm{K}.\mathrm{C}$

New potential difference 

$\begin{array}{l}{\mathrm{V}}^1=\frac{\mathrm{V}}{\mathrm{K}}\\ {\mathrm{u}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{CV}}^2=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}\\ {\mathrm{u}}_{\mathrm{f}}=\frac{{\mathrm{Q}}^2}{2\mathrm{f}}=\frac{{\mathrm{Q}}^2}{2\mathrm{K}}=\frac{{\mathrm{C}}^2{\mathrm{V}}^2}{2\mathrm{KC}}=\left(\frac{{\mathrm{u}}_{\mathrm{i}}}{\mathrm{K}}\right)\\ \mathrm{\Delta u}={\mathrm{u}}_{\mathrm{f}}-{\mathrm{u}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{CV}}^2\left\{\frac{1}{\mathrm{K}}-1\right\}\end{array}$

As the capacitor is isolated, so charge will remain conserved p.d. between two plates of the capacitor $=\frac{\mathrm{Q}}{\mathrm{KC}}=\frac{\mathrm{V}}{\mathrm{K}}$