A parallel plate capacitor has plate area 100   m 2 and plate separation of 10   m . The space…

A parallel plate capacitor has plate area 100 m2 and plate separation of 10 m. The space between the plates is filled up to a thickness 5 m with a material of dielectric constant of 10 . The resultant capacitance of the system is x pF. The value of ε0=8.85×10-12 F m-1. The value of x to the nearest integer is ______.

Solution

Here, Plate areaA=100 m2
Using, Capacity of parallel plate capacitor is,

C=kε0 A d; where d is separation between the plates and k is dielectric constant.

The capacitance with dielectric will be,
C1=10ϵ0(100)5
=200ϵ0

The capacitance without dielectric or air will be,
C2=ϵ0(100)5=20ϵ0
C1 & C2 are in series so,

 Ceqv. =C1C2C1+C2

=4000ϵ0220
=160.9×10-12161pF

Asked in: JEE Main 2021 (18 Mar Shift 1)

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