Question

The capacitance of a parallel plate capacitor with air as medium is $3 \, \mu F$. As a dielectric is introduced between the plates, the capacitance becomes $15 \, \mu F$. The permittivity of the medium in $C^2N^{-1} m^{-2}$ is

• A. $8.15 \times 10^{-11}$
• B. $0.44 \times 10^{-10}$
• C. $15.2 \times 10^{-12}$
• D. $1.6 \times 10^{-14}$

Answer 1

Answer: (B)

The capacitance of air capacitor is given by
$C_{0}=\frac{\varepsilon_{0} A}{d}=3 \mu F .... $(i)
When a dielectric of permittivity $\varepsilon_{r}$ and dielectric constant $K$ is introduced between the plates, then
$C=\frac{K \varepsilon_{0} A}{d}=15 \mu F...$(ii)
Dividing Eq. (ii) by Eg. (i), we get
$\frac{C}{C_{0}}=\frac{\frac{K \varepsilon_{0} A}{d}}{\frac{\varepsilon_{0} A}{d}}=\frac{15}{3} $
$\Rightarrow K=5$
So, the permittivity of the medium
$\varepsilon_{r} =\varepsilon_{0} K$
$=8.85 \times 10^{-12} \times 5 $
$=0.44 \times 10^{-10}$