Question

The S.I unit of permittivity of free space $\left({\epsilon }_0\right)$ is

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

Answer 1

The correct option is D

$C^2{N}^{-1}{m}^{-2}$

Formula of permittivity of free space: ${\epsilon }_0=\frac{q_1{q}_2}{4{\mathrm{\pi Fr}}^2}$,where $q_1,q_2$ are charges, $F$ is the electric force and $r$ is the distance between the charges.

S.I. unit of permittivity of free space:

$$\begin{gathered} {\epsilon }_0=\frac{C\times C}{N\times m^2} \\ {\epsilon }_0=\frac{C^2}{N{m}^2} \\ {\epsilon }_0=C^2{N}^{-1}{m}^{-2} \end{gathered}$$

Thus, the S.I. unit of permittivity of free space is $C^2{N}^{-1}{m}^{-2}$.

So, option D is the correct answer.