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,
Where, A = Area of each plate
= 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,
= 6
Hence, capacitance of the capacitor becomes
Taking ratios of equations (i) and (ii), we obtain
Therefore, the capacitance between the plates is 96 pF.
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