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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
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Verified Answer
The correct answer is:
$40.5 \mathrm{pF}$
$40.5 \mathrm{pF}$
$$
\begin{aligned}
& C^{\prime}=\frac{A \varepsilon_0}{\frac{d_1}{3}+\frac{d_2}{6}}=\frac{A \varepsilon_0}{\frac{d}{9}+\frac{2 d}{18}}=\frac{18 A \varepsilon_0}{4 d} \\
& C^{\prime}=40.5 \mathrm{PF}
\end{aligned}
$$
\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}
$$
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