Q. The constant of proportionality $\frac{1}{4\pi\varepsilon_{0}}$ in Coulomb’s law has the following dimensions
A) $C^{- 2}Nm^{2}$
B) $C^{2}N^{- 1}m^{- 2}$
C) $C^{2}Nm^{2}$
D) $C^{- 2}N^{- 1}m^{- 2}$
Explanation
According to Coulombs law $F = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r^{2}}$ $\therefore\frac{1}{4\pi\varepsilon_{0}} = \frac{Fr^{2}}{q_{1}q_{2}} = \frac{(newton)(meter)^{2}}{(coulomb)(coulomb)}$ $= \frac{Nm^{2}}{C^{2}} = C^{- 2}Nm^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.