Question

The unit of electrical permittivity of free space ${\epsilon }_0$ is

• A. $C{N}^{-1}{m}^{-1}$
• B. $N{m}^2{C}^{-2}$
• C. $C^2{N}^{-2}{m}^{-2}$
• D. $C^2{N}^{-1}{m}^{-2}$

Answer 1

Answer: (D)

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$
$\left[{\epsilon }_0\right]=\frac{\left[q_1\right]\left[q_2\right]}{[4\pi ][F]\left[r^2\right]}=\frac{C^2}{N{m}^2}=C^2{N}^{-1}{m}^{-2}$