Question

The capacitance of a parallel plate capacitor with air as medium is $3\mu F$. With the introduction of a dielectric medium between the plates, the capacitance becomes $15\mu F$. The permittivity of the medium is

• A. $5C^2{N}^{-1}{m}^{-2}$
• B. $15C^2{N}^{-1}{m}^{-2}$
• C. $0.44\times 10^{-10}{C}^2{N}^{-1}{m}^{-2}$
• D. $8.854\times 10^{-11}{C}^2{N}^{-1}{m}^{-2}$

Answer 1

Answer: (C)

The capacitance of air capacitor
$C_0=\frac{A{\epsilon }_0}{d}=3\mu F\text{...(i) }$
When a dielectric of permittivity ${\epsilon }_r$ or dielectric constant $K$ is introduced between the plates of the capacitor, then its capacitance
$C=\frac{KA{\epsilon }_0}{d}=15\mu F\dots \text{ (ii) }$
Dividing Eq. (ii) by Eq. (i)
$\frac{C}{C_0}=\frac{\frac{KA{\epsilon }_0}{d}}{\frac{A{\epsilon }_0}{d}}=\frac{15}{3}$
$\therefore K=5$
Permittivity of the medium
$\epsilon ={\epsilon }_{0}K$
$=8.854\times 10^{-12}\times 5$
$=44.27\times 10^{-12}$
$=0.44\times 10^{-10}{C}^2{N}^{-1}{m}^{-2}$