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 $C^2{N}^{-1}{m}^{-2}$ is

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

Answer 1

Answer: (A)

Capacitance of air capacitor
$C_0=\frac{{\epsilon }_{0}A}{d}=3\mu F$ ......... (i)
When a dielectric of permittivity ${\epsilon }_r$ and dielectric constant K is introduced between the plates, then
Capacitance, $C=\frac{K{\epsilon }_{0}A}{d}=15\mu F$ ..... (ii)
Dividing eq. (ii) by (i), we get
$\frac{C}{C_0}=\frac{\frac{K{ϵ}_{o}A}{d}}{\frac{{\epsilon }_{0}A}{d}}=\frac{15}{3}$
$\Rightarrow K=5$
$\therefore$ permittivity of the medium
${\epsilon }_0={\epsilon }_{0}K$
$=8.85\times {10}^{-12}\times 5=0.44\times {10}^{-10}$