Question 119 of 180

A parallel plate capacitor with air between the plates has a capacitance of $9 \mathrm{pF}$. The separation between its plates is ' $d$ '. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_1=3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $\mathrm{k}_2=6$ and thickness $\frac{2 \mathrm{~d}}{3}$. Capacitance of the capacitor is now

A$1.8 \mathrm{pF}$B$45 \mathrm{pF}$C$40.5 \mathrm{pF}$D$20.25 \mathrm{pF}$

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Answer: (C) $40.5 \mathrm{pF}$

$$\begin{aligned} & C^{\prime}=\frac{A \varepsilon_0}{\frac{d_1}{3}+\frac{d_2}{6}}=\frac{A \varepsilon_0}{\frac{d}{9}+\frac{2 d}{18}}=\frac{18 A \varepsilon_0}{4 d} \\ & C^{\prime}=40.5 \mathrm{PF} \end{aligned}$$