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?

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 = (kε₀ A)/d = ε₀ A/d ———————–Eq-1

Where,

A = Area of each plate

ε₀ = Permittivity of free space

If distance between the plates is reduced to half, then new distance, di = d/2 Dielectric constant of the substance filled in between the plates, k₁= 6 Hence, capacitance of the capacitor becomes

C₁ = k₁ ε₀ A/d₁ = (6 ε₀ A)/(d/2) = (12 ε₀ A)/d………………Eq-2

Taking ratios of equations (1) and (2), we obtain C₁ = 2 x 6 C = 12 C = 12 x 8 pF = 96 pF Therefore, the capacitance between the plates is 96 pF.