A parallel plate capacitor with air between the plates has capacitance of 9pF. 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 k1β=3 and thickness 3dβ while the other one has dielectric constant k2β=6 and thickness 32dβ. Capacitance of the capacitor is now :
A
1.8pF
B
45pF
C
40.5pF
D
20.25pF
Hard
Open in App
Solution
Verified by Toppr
Correct option is C)
The air filled capacitance , C=dAΟ΅0ββ=9pF
According to question there are two capacitor with capacitances C1β and C2β . and they are in series .
Here, C1β=(d/3)Ak1βΟ΅0ββ=9dAΟ΅0ββ=9C Β Β as k1β=3
and C2β=(2d/3)Ak2βΟ΅0ββ=9dAΟ΅0ββ=9C Β Β as k2β=6
The equivalent capacitance, Ceqβ=C1β+C2βC1βC2ββ=9C+9C(9C)(9C)β=29βC=(9/2)9=40.5pF
Solve any question of Electrostatic Potential and Capacitance with:-