Q. The unit of permittivity of free space $\varepsilon_{0}$ is
A) $Coulomb/newton - metre$
B) $Newton - metre^{2}/coulomb^{2}$
C) $Coulomb^{2}/(newton - metre)^{2}$
D) $Coulomb^{2}/newton - metre^{2}$
Explanation
$F = \frac{1}{4\pi\varepsilon_{0}}.\frac{Q_{1}Q_{2}}{r^{2}}$ $\Rightarrow \varepsilon_{0} \propto \frac{Q^{2}}{F \times r^{2}}$ So $\varepsilon_{0}$ has units of $coulomb^{2}/newton - m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.