Two dielectric slabs having dielectric constant ' \(\mathrm{K}_1\) ' and ' \(\mathrm{K}_2\) ' of thickness ' \(\frac{\mathrm{d}}{4}\) and \(\frac{3 \mathrm{~d}}{4}\) are inserted between the plates as shown in figure. The net capacitance between \(A\) and \(B\) is \(\left[\varepsilon_0\right.\) is permittivity of free space]

Capacity of \(1^{\text {st }}\) Capacitor,
\(C_1=\frac{K_1 \varepsilon_0 A}{d / 4}=\frac{4 K_1 \varepsilon_0 A}{d}\)
Capacity of \(2^{\text {nd }}\) Capacitor,
\(\mathrm{C}_2=\frac{\mathrm{K}_2 \varepsilon_0 \mathrm{~A}}{3 \mathrm{~d} / 4}=\frac{4 \mathrm{~K}_2 \varepsilon_0 \mathrm{~A}}{3 \mathrm{~d}}\)
Equivalent capacitance \(\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}\)
\(\begin{aligned}
\frac{1}{\mathrm{C}_1} & =\frac{\mathrm{d}}{4 \mathrm{~K}_1 \varepsilon_0 \mathrm{~A}} ; \frac{1}{\mathrm{C}_2}=\frac{3 \mathrm{~d}}{4 \mathrm{~K}_2 \varepsilon_0 \mathrm{~A}} \\
\frac{1}{\mathrm{C}} & =\frac{\mathrm{d}}{4 \mathrm{~K}_1 \varepsilon_0 \mathrm{~A}}+\frac{3 \mathrm{~d}}{4 \mathrm{~K}_2 \varepsilon_0 \mathrm{~A}} \\
\frac{1}{\mathrm{C}} & =\frac{\mathrm{d}}{4 \varepsilon_0 \mathrm{~A}}\left[\frac{1}{\mathrm{~K}_1}+\frac{3}{\mathrm{~K}_2}\right] \\
\frac{1}{\mathrm{C}} & =\frac{\mathrm{d}}{4 \varepsilon_0 \mathrm{~A}}\left[\frac{\mathrm{K}_2+3 \mathrm{~K}_1}{\mathrm{~K}_1 \mathrm{~K}_2}\right] \\
\therefore \quad \mathrm{C} & =\frac{4 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\left[\frac{\mathrm{K}_1 \mathrm{~K}_2}{3 \mathrm{~K}_1+\mathrm{K}_2}\right]
\end{aligned}\)