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Question

# 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 d3 while the other one has dielectric constant K2=6 and thickness 2d3. Capacitance of the capacitor is now :

A
40.5 pF
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B
20.25 pF
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C
1.8 pF
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D
45 pF
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Solution

## The correct option is C 40.5 pF The capacitance with air is C0=Aϵ0d=9pFAfter inserting dielectric, there are two capacitors C1,C2 and they are in series.here, C1=AK1ϵ0d/3=9Aϵ0d=9×9=81pFand C2=AK2ϵ02d/3=9Aϵ0d=9×9=81pFThus new capacitance Ceq=C1C2C1+C2=81×8181+81=40.5pF

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