Q. 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) 5
B) 15
C) $0.44 \times 10^{- 10}C^{2}N^{- 1}m^{- 2}$
D) $8.854 \times 10^{- 11}C^{2}N^{- 1}m^{- 2}$
Explanation
(c) Capacitance of a parallel plate capacitor with air is $C = \frac{\varepsilon_{0}A}{d}$
Capacitance of a same parallel plate capacitor with the introduction of a dielectric medium is $C^{'} = \frac{K\varepsilon_{0}A}{d}$ where $K$ is the dielectric constant of a medium
$\Rightarrow \frac{C^{'}}{C} = K$ or $\frac{15}{3} = 5$ or $K = \frac{\varepsilon}{\varepsilon_{0}}$
$\Rightarrow \varepsilon = K\varepsilon_{0} = 5 \times 8.854 \times 10^{- 12} = 0.4427 \times 10^{- 10}C^{2}N^{- 1}m^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.