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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 $\frac{d}{3}$ while the other one has dielectric constant $k_{2} = 6$ and thickness $\frac{2 d}{3}$ .Capacitance of the capacitor is now

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1.8 pF

45 pF

40.5 pF

20.25 pF

answer is C.

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Detailed Solution

$C_{eq} = \frac{C_{1} C_{2}}{C_{1} + C_{2}}; C_{1} = \frac{K_{1} {Aε}_{0}}{(d / 3)} = C_{2} = \frac{K_{2} {Aε}_{0}}{(2 d / 3)}$ $C_{eq} = \frac{3 {Aε}_{0} K_{1} K_{2}}{d (2 K_{1} + K_{2})} = \frac{3 {CK}_{1} K_{2}}{(2 K_{1} + K_{2})} = 40.5 pF$

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