The two plates of a parallel plate capacitor are $4 \mathrm{mm}$ apart. A slab of dielectric constant $3$ and thickness $3 \mathrm{mm}$ is introduced between the plates with its faces parallel to them. The distance between the plates is so adjusted that the capacitance of the capacitor becomes $\frac{2}{3}$ rd of its original value. What is the new distance between the plates?

Figure below shows a parallel plate capacitor of plate area $A$ and plate separation $d$. Its entire space is filled with three different dielectric slabs of the same thickness. Find the equivalent capacitance of the arrangement.

The space between the plates of a parallel plate capacitor of capacitance $C$ is filled with three dielectric slabs of equal thickness, as shown in Figure. If the dielectric constants of the three slabs are $k_1, k_2$ and $k_3$, find the new capacitance.