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A parallel plate capacitor with air between the plates has a capacitance of $9\: pF$. The separation between the 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 $ \large\frac{d}{3}$ while the other has dielectric constant $k_2=6$ and thickness $ \large\frac{2d}{3}$. Capacitance of the capacitor is now

$\begin {array} {1 1} (A)\;1.8\: pF & \quad (B)\;45\: pF \\ (C)\;40.5 \: pF & \quad (D)\;20.25\: pF \end {array}$

 

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$ C' = \large\frac{A\varepsilon_0}{\large\frac{d_1}{3}+\large\frac{d_2}{6}} = \large\frac{A\varepsilon_0}{\large\frac{d}{9}+\large\frac{2d}{18}} $ $ = \large\frac{18A \varepsilon_0}{4d}$
$C'=40.5\: pF$
Ans : (C)
answered Feb 14, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1
 

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