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. $10C^2{N}^{-1}{m}^{-2}$
• 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)

$0.44\times {10}^{-10}{C}^2{N}^{-1}{m}^{-2}$

Capacitance of a parallel capacitor with air is $C=\frac{{\epsilon }_{0}A}{d}$
Capacitance of a same parallel plate capacitor with the introduction of a dielectric medium is $C^{\prime }=\frac{{K\epsilon }_{0}A}{d}$
where $K$ is the dielectric constant of a medium
or $\frac{C^{\prime }}{C}=KorK=\frac{15}{3}=5orK=\frac{\epsilon }{{\epsilon }_0}$
or $\epsilon =K{\epsilon }_0=5\times 8.854\times {10}^{-12}=0.44\times {10}^{-10}{C}^2{N}^{-1}{m}^{-2}$