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 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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a
20.25 pF
b
1.8 pF
c
45 pF
d
40.5 pF
answer is D.
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Detailed Solution
Given C = Aε0d=9 pFNow, C'=Aε0d/3k1+2d/3k2=92Aε0d=812pF