A capacitor having capacitance $1 \mu \mathrm{F}$ with air, is filled with two dielectrics as shown. How many times capacitance will increase?

Key Idea : After filling with dielectrics the two capacitors will be in parallel order. As shown, the two capacitors are connected in parallel. Initially the capacitance of capacitor $C=\frac{\varepsilon_0 A}{d}$ where $A$ is area of each plate and $d$ is the separation between the plates. After filling with dielectrics, we have two capacitors of capacitance. $C_1=\frac{K_1 \varepsilon_0(A / 2)}{d}=\frac{8}{2} \frac{\varepsilon_0 A}{d}=\frac{4 \varepsilon_0 A}{d}=4 C$ and $C_2=\frac{K_2 \varepsilon_0(A / 2)}{d}=\frac{4}{2} \frac{\varepsilon_0 A}{d}=\frac{2 \varepsilon_0 A}{d}=2 C$ Hence, their equivalent capacitance $\begin{aligned} C_{\mathrm{eq}} & =C_1+C_2 \\ & =4 C+2 C=6 C\end{aligned}$ ie, new capacitance will be six times of the original.