NEET-XII-Physics
02: Electrostatic Potential And Capacitance
- #5A 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?
Ans : 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
.gif)
= Permittivity of free space
If distance between the plates is reduced to half, then new distance, d’ =
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.