Question

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 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?

Answer 1

Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between 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=kε₀A/d=ε₀A/d………………….(1)
Where,
A = Area of each plate
ε₀ = Permittivity of free space
If distance between the plates is reduced to half, then new distance, d₁ = d/2
Dielectric constant of the substance filled in between the plates, k₁= 6
Hence, capacitance of the capacitor becomes