A dielectric slab fills the lower half of a parallel plate capacitor as shown in figure. (Take plate area as A)
 

Equivalent capacity of the system is $\left({\mathrm{\epsilon }}_0\mathrm{A}/2\mathrm{d}\right)(1+\mathrm{K})$.

The net charge of the lower half of the left hand plate is 1/K times the charge on the upper half of the plate.

Net charges on the lower and upper halves of the left hand plate are different.

Net charge on the lower half of the left hand plate is $\frac{{\mathrm{K\epsilon }}_0\mathrm{A}}{2\mathrm{d}}\times \mathrm{V}$.