A parallel plate capacitor with plate area A and plate separation d...

The formula to calculate the capacitance of a parallel plate capacitor having thickness $d$ and with a dielectric of thickness $x$ between the plates is given by

$C = \dfrac{{ε}_{0}A}{\left(d - x\right) + \dfrac{x}{K}} . . . ( 1 )$

Substitute the values of the parameters for first case into equation (1) to obtain the capacitance.

$\begin{cases}{C}_{1} & = & \dfrac{{ε}_{0}A}{\dfrac{2d}{3} + \dfrac{\dfrac{d}{3}}{4}} \\ & = & \dfrac{4{ε}_{0}A}{3d} . . . ( 2 )\end{cases}$

Similarly, substitute the parameters into equation (1) for the second case to obtain the capacitance

$\begin{cases}{C}_{2} & = & \dfrac{{ε}_{0}A}{\dfrac{d}{3} + \dfrac{\dfrac{2d}{3}}{4}} \\ & = & \dfrac{4{ε}_{0}A}{2d} \\ & = & \dfrac{2{ε}_{0}A}{d} . . . ( 3 )\end{cases}$

Divide equation (3) by equation (2) to obtain the capacitance in the second case.

$\begin{cases}\dfrac{{C}_{2}}{{C}_{1}} & = & \dfrac{\dfrac{2{ε}_{0}A}{d}}{\dfrac{4{ε}_{0}A}{3d}} \\ & = & \dfrac{3}{2} \\ {C}_{2} & = & \dfrac{3}{2}{C}_{1} . . . ( 4 )\end{cases}$

Substitute the value of the capacitance in the first case into equation (4) to obtain the value of the capacitance in the second case.

$\begin{cases}{C}_{2} & = & \dfrac{3}{2} \times 2\mu F \\ & = & 3\mu F\end{cases}$