The parallel combination of two air filled parallel plate capacitors of capacitance C and nC  is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is

When parallel combination is fully charged, charge on the combination is

$Q=C_{eq}V=C(1+n)V$

When battery is removed and a dielectric slab is placed between two plates of first capacitor, then charge on the system remains same. Now equivalent capacitance after insertion of dielectric is

$C_{eq}=KC+nC=(n+K)C$

If potential value after insertion of dielectric is V' then charge on system is
$Q^{\mathrm{\prime }}=C_{eq}{V}^{\mathrm{\prime }}=(n+K)C{V}^{\mathrm{\prime }}$
As $Q=Q^{\mathrm{\prime }}$ we have

$\begin{array}{l}C(1+n)V=(n+k)C{V}^{\mathrm{\prime }}\\ \therefore V^{\mathrm{\prime }}=\frac{(1+n)V}{(n+K)}\end{array}$