A parallel plate capacitor having plates of area $S$ and plate separation $d,$ has capacitance $C_1$ in air. When two dielectrics of different relative permittivities $({\epsilon }_1=2\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}{\epsilon }_2=4)$ are introduced between the two plates as shown in the figure, the capacitance becomes $C_2.$ The ratio $\frac{C_2}{C_1}$ is

 

A capacitance of the capacitor without dielectric.
$C_1=\frac{{\epsilon }_{0}A}{d}$          
Now consider the case with filled dielectric. The given capacitor can be divided into three capacitors as shown.
 
$C^{\text{'}\text{'}}=\frac{2{\epsilon }_0\frac{S}{2}}{d/2}=\frac{2{\epsilon }_{0}S}{d}=2C_1$ 
$C^{\text{'}\text{'}}=\frac{4{\epsilon }_0\frac{S}{2}}{d/2}=\frac{4{\epsilon }_{0}S}{d}=4C_1$ 
${{C^{\text{'}}}^{\text{'}}}^{\text{'}}=\frac{2{\epsilon }_0\frac{S}{2}}{d}=\frac{{\epsilon }_{0}S}{d}=C_1$ 
$C_2=\frac{C^{\text{'}}{C}^{\text{'}\text{'}}}{C^{\text{'}}+C^{\text{'}\text{'}}}+{{C^{\text{'}}}^{\text{'}}}^{\text{'}}=\frac{4}{3}\frac{{\epsilon }_{0}S}{d}+\frac{{\epsilon }_{0}S}{d}=\frac{7}{3}\frac{{\epsilon }_{0}S}{d}$ 
This given : $\frac{C_2}{C_1}=\frac{7}{3}$