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 :
20.25pF
1.8pF
40.5pF
45pF
A
45pF
B
20.25pF
C
1.8pF
D
40.5pF
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Solution
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The capacitance with air is C0=Aϵ0d=9pF
After inserting dielectric, there are two capacitors C1,C2 and they are in series.
here, C1=AK1ϵ0d/3=9Aϵ0d=9×9=81pF
and C2=AK2ϵ02d/3=9Aϵ0d=9×9=81pF
Thus new capacitance Ceq=C1C2C1+C2=81×8181+81=40.5pF
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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 K1=3 and thickness d3 while the other one has dielectric constant K2=6 and thickness 2d3 . Capacitance of the capacitor is now
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