A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor of capacitance $C_0$ but has a thickness (3/4)d, where d is the separation of the plates. What is the capacitance of the capacitor when the slab is inserted between the plates?

$1.<mspace\; width="1em"></mspace>\left(\frac{4\mathrm{K}}{\mathrm{K}+3}\right){\mathrm{C}}_0$
$2.<mspace\; width="1em"></mspace>\left(\frac{2\mathrm{K}}{\mathrm{K}+2}\right){\mathrm{C}}_0$
$3.<mspace\; width="1em"></mspace>{\mathrm{C}}_0$
$4.<mspace\; width="1em"></mspace>\frac{3}{4}{\mathrm{C}}_0$