A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant \(\varepsilon_1\) and \(\varepsilon_2\), as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are \(C_1\) and \(C_2\) respectively, then \(\frac{C_1}{C_2}\) is :

Area of plate is \(A\). then
\(\begin{aligned}
& \mathrm{C}=\frac{\varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \\
& \mathrm{C}^{\prime}=\frac{\varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}
\end{aligned}\)
Let \(\mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
\(\begin{aligned}
& \mathrm{C}=2 \varepsilon_2 \mathrm{C}_0 \\
& \mathrm{C}^{\prime}=2 \varepsilon_1 \mathrm{C}_0
\end{aligned}\)
\(\mathrm{C} \& \mathrm{C}^{\prime}\) are in series
\(\mathrm{C}_1=\frac{\mathrm{CC}^{\prime}}{\mathrm{C}+\mathrm{C}^{\prime}}=\frac{4 \varepsilon_2 \varepsilon_1 \mathrm{C}_0^2}{2 \mathrm{C}_0\left(\varepsilon_2+\varepsilon_1\right)}\)
\(=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)}\)

Here \(\mathrm{C}=\frac{\varepsilon_1 \varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=\frac{\varepsilon_1 \mathrm{C}_0}{2}\)
\(\mathrm{C}^{\prime}=\frac{\varepsilon_2 \mathrm{C}_0}{2}\)
\(\mathrm{C} \& \mathrm{C}^{\prime}\) are inparallel
\(\mathrm{C}_2=\mathrm{C}^{\prime}+\mathrm{C}=\left(\varepsilon_1+\varepsilon_2\right) \frac{\mathrm{C}_0}{2}\)
Thus \(\frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)} \times \frac{2}{\left(\varepsilon_1+\varepsilon_2\right) \mathrm{C}_0}\)
\(=\frac{4 \varepsilon_2 \varepsilon_1}{\left(\varepsilon_2+\varepsilon_1\right)^2}\)