Q. SI unit of permittivity is
A) $C^{2}m^{2}N^{2}$
B) $C^{2}m^{- 2}N^{- 1}$
C) $C^{2}m^{2}N^{- 1}$
D) $C^{- 1}m^{2}N^{- 2}$
Explanation
From Coulomb’s law $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r^{2}}$ Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \varepsilon_{0} = \frac{q_{1}q_{2}}{4\pi Fr^{2}}$ $\therefore Units\ of\ \varepsilon_{0}\ (permittivity)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{C^{2}}{N - m^{2}} = C^{2}N^{- 1}m^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.