A parallel plate capacitor with air between the plates has a capacitance of 8 pF.`(1 pF = 10^(-12)F)` What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 6 ?

Explanation

Capacitance between the parallel plates of the capacitor, C=8 pF
initially, distance betwene the parallel plates was d and it was filled with air. Dielectric constant of air, k=1
capacitance, C is given by the formula.
`C=(kin_(0)A)/d`
`=(in_(0)A)/d....(i)`
Where, A=Area of each plate
`in_(0)`=permitivity of free space
IF distance between the plates is reduced to half , then new distance `d=(d/2)`
dielectric constant of the substance filled in between the paltes `k=6`
Hence, capacitance of the capacitor becomes
`C"=(k in_(0)A)/(d')=(6in_(0)A)/(d/2).....(ii)`
Takin ratios equation (i) and (ii) , we obtain
`C'=2xx6C`
`=12C`
`=12xx8=96 pF`
Therefore, the capacitance between the paltes is 96 pF