A parallel plate air capacitor, with plate separation ' d ' has a capacitance of 9 pF . The space between the plates is now filled with two dielectrics, the first having \(\mathrm{K}_1=3\) and thickness \(\mathrm{d}_1=\mathrm{d} / 3\), while the \(2^{\text {nd }}\) has \(\mathrm{K}_2=6\) and thickness \(\mathrm{d}_2=2 \mathrm{~d} / 3\). The capacitance of the new capacitor is

For a parallel plate capacitor, with dielectric
\(\mathrm{C}=\frac{\mathrm{A} \varepsilon_0 \mathrm{k}}{\mathrm{~d}}\)...(i)
\(C_1=\frac{A \varepsilon_0 k_1}{d_1}=\frac{A \varepsilon_0 3}{d / 3}=9 \frac{A \varepsilon_0}{d}=9 C \quad \ldots[\) From (i) \(]\)
\(C_2=\frac{A \varepsilon_0 k_2}{d_2}=\frac{A \varepsilon_0 6}{2 d / 3}=9 \frac{A \varepsilon_0}{d}=9 C \ldots[\) From (i) \(]\)
As they both are in series,
\(\mathrm{C}_{\text {total }}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{9 \mathrm{C} \times 9 \mathrm{C}}{9 \mathrm{C}+9 \mathrm{C}}=\frac{9}{2} \mathrm{C}=\frac{9}{2} \times 9 \times 10^{-12}\)
\(\therefore \quad \mathrm{C}_{\text {total }}=40.5 \mathrm{pF}\)