Q. The dimensions of permittivity $\varepsilon_{0}$ are
A) $A^{2}T^{2}M^{- 1}L^{- 3}$
B) $A^{2}T^{4}M^{- 1}L^{- 3}$
C) $A^{- 2}T^{- 4}ML^{3}$
D) $A^{2}T^{- 4}M^{- 1}L^{- 3}$
Explanation
$F = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r^{2}}$ $\Rightarrow \ \varepsilon_{0} = \frac{\left| q_{1} \right||q_{2}|}{\lbrack F\rbrack\lbrack r^{2}\rbrack} = \frac{\left\lbrack A^{2}T^{2} \right\rbrack}{\left\lbrack MLT^{- 2} \right\rbrack\lbrack L^{2}\rbrack} = \lbrack A^{2}T^{4}M^{- 1}L^{- 3}\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.