Q. The constant of proportionality $\frac{1}{4\pi\varepsilon_{0}}\$in Coulomb’s law has the following units

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

From Coulomb’s law $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r^{2}}$ $\Rightarrow \left\lbrack \frac{1}{4\pi\varepsilon_{0}} \right\rbrack = \frac{\lbrack F\ \times r^{2}\rbrack}{\lbrack q\rbrack^{2}}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{\lbrack newton\rbrack\lbrack metre\rbrack^{2}}{\lbrack coulomb\rbrack^{2}}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = {Nm}^{2}C^{- 2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.