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A parallel plate capacitor with air between the plates has a 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 $\large\frac{d}{3}$ while the other one has dielectric constant $K_2 = 6$ and thickness $ \large\frac{2d}{3}$ . Capacitance of the capacitor is now

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$(C) 40.5 pF $
answered Jun 25, 2014 by meena.p
 
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