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 dielectric has dielectric constant k1 = 3 and thickness d/3 while the other one has dielectric constant k2 = 6 and thickness 2d/3. Capacitance of the capacitor is now

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20.25 pF

1.8 pF

45 pF

40.5 pF

answer is D.

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Detailed Solution

C=ϵ0Ad=9pF C1=k1ϵ0Ad/3=3k1ϵ0Ad =3×3×9=81pF C2=k2ϵ0A2d/3=3k2ϵ0A2d =3×6×92=81pF

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