A parallel plate capacitor with air between the plates has capacitance of $9\ pF$. 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 $k_2 = 6$ and thickness $\frac{2d}{3}$ . Capacitance of the capacitor is now.......$pF$
AIEEE 2008, Medium
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The given capacitance is equal to two capacitances connected in series where
$C_{1}=\frac{k_{1} \epsilon_{0} A}{d / 3}=\frac{3 k_{1} \epsilon_{0} A}{d}=\frac{3 \times 3 \epsilon_{0} A}{d}=\frac{9 \epsilon_{0} A}{d}$
and
$C_{2}=\frac{k_{2} \epsilon_{0} A}{2 d / 3}=\frac{3 k_{2} \epsilon_{0} A}{2 d}=\frac{3 \times 6 \epsilon_{0} A}{2 d}=\frac{9 \epsilon_{0} A}{d}$
The equivalent capacitance $C_{\mathrm{eq}}$ is
$\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{d}{9 \epsilon_{0} A}+\frac{d}{9 \epsilon_{0} A}=\frac{2 d}{9 \epsilon_{0} A}$
$\therefore C_{e q}=\frac{9}{2} \frac{\epsilon_{0} A}{d}=\frac{9}{2} \times 9 \,p F=40.5 \,p F$
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