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,
A = Area of each plate
`in_0`= Permittivity 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 plates, k' = 6
Hence, capacitance of the capacitor becomes
Taking ratios of equations (i) and (ii), we obtain
=12 x 8=96 pF
Therefore, the capacitance between the plates is 96 pF.