Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance, Chapter-2 Exercise 2.5, NCERT Solutions for Class 12 Physics
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?
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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.