Question

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

• A. $32\mu F$
• B. $18\mu F$
• C. $8\mu F$
• D. $44\mu F$

Answer 1

Answer: (A)

The capacitance of a parallel plate air capacitor is given by
$C_0=\frac{{\epsilon }_{0}A}{d}....(i)$
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
$C=4\times 2\times 4=32\mu F$