In a parallel plate air capacitor the distance between plates is reduced to one fourth and the space between them is filled with a dielectric medium of constant 2. If the initial capacity of the capacitor is $4\mu F.$ then its new capacity is

The capacitance of a parallel plate air capacitor is given by
$C_0=\frac{{\epsilon }_{0}A}{d}$
Where, ${\epsilon }_0$ =permittivity of the medium,
A = area of plates
and d = distance between the plates
When the distance between plates is reduced and a dielectric slab is introduced, then the capacitance becomes
$C=\frac{K{\epsilon }_{0}A}{d_1}$ … (ii)
Where, K = dielectric constant of medium.
Here, K = 2, $d_1=\frac{d}{4}$ and $C_0=4\mu F$ = $4\times {10}^{-6}F$
From Eq. (ii), we get
$C=4\left(\frac{K{\epsilon }_{0}A}{d_1}\right)=4K\left(\frac{{\epsilon }_{0}A}{d}\right)$
$=4K{C}_0$ …. (iii) [From Eq. (i)]
Substituting given values in Eq. (iii), we get
$C=4\times 2\times 4=32\mu F$