Q5.
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.
Student Discussion
between its plates is 10 µf if the distance between the plates is
reduced to half and the medium between plates is replaced by
a medium of dielectric constant k=5, then find the value how
much increase or decrease in the capacitance of the capacitor.
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