A parallel plate capacitor with plate area A and plate separation d is filled with a dielectric material of dielectric constant K=4. The thickness of…

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

$$C=\frac{{\epsilon }_{0}A}{(d-x)+\frac{x}{K}}...(1)$$C=(d−x)+Kx​ϵ0​A​...(1)

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

$$\begin{matrix}{C}_{1} & = & \frac{{\epsilon }_{0}A}{\frac{2d}{3}+\frac{\frac{d}{3}}{4}} \\ & = & \frac{4{\epsilon }_{0}A}{3d}...(2)\end{matrix}$$C1​​==​32d​+43d​​ϵ0​A​3d4ϵ0​A​...(2)​

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

$$\begin{matrix}{C}_{2} & = & \frac{{\epsilon }_{0}A}{\frac{d}{3}+\frac{\frac{2d}{3}}{4}} \\ & = & \frac{4{\epsilon }_{0}A}{2d} \\ & = & \frac{2{\epsilon }_{0}A}{d}...(3)\end{matrix}$$C2​​===​3d​+432d​​ϵ0​A​2d4ϵ0​A​d2ϵ0​A​...(3)​

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

$$\begin{matrix}\frac{{C}_{2}}{{C}_{1}} & = & \frac{\frac{2{\epsilon }_{0}A}{d}}{\frac{4{\epsilon }_{0}A}{3d}} \\ & = & \frac{3}{2} \\ {C}_{2} & = & \frac{3}{2}{C}_{1}...(4)\end{matrix}$$C1​C2​​C2​​===​3d4ϵ0​A​d2ϵ0​A​​23​23​C1​...(4)​

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

$$\begin{matrix}{C}_{2} & = & \frac{3}{2}\times 2\mu F \\ & = & 3\mu F\end{matrix}$$C2​​==​23​×2μF3μF​