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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:

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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