2.5) A parallel plate capacitor with air between the plates has a capacitance of 8 pF . 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?
As we know,
where A= area of the plate
= permittivity of the free space
d = distance between the plates.
The capacitance between plates initially
Now, Capacitance when the distance is reduced half and filled with the substance of dielectric 6
Hence new capacitance is 96pF.