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
$C_{0}=\frac{\varepsilon_{0} A}{d} \,\,\,\,\,\,\,\,....(i)$
where, $\varepsilon_{0}=$ 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
$C=\frac{K \varepsilon_{0} A}{d_{1}}\,\,\,\,\,\, ...(ii)$
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 \varepsilon_{0} A}{d_{1}}\right)=4 K\left(\frac{\varepsilon_{0} A}{d}\right) $
$=4 K C_{0} \,\,\,\,\,\,\,$...(iii) [from Eq. (i)]
Substituting given values in Eq. (iii), we get
$C=4 \times 2 \times 4=32 \mu F$