Question

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

• A. coulomb/newton-metre
• B. Newton metre${}^2$ / Coulomb${}^2$
• C. Coulomb${}^2$ /Newton metre${}^2$
• D. Newton metre)${}^2$

Answer 1

Answer: (C)

Key Idea : Substitute the units for all the quantities involved in an expression written for permittivity of free space.
By Coulomb's law, the electrostatic force
$F=\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1{q}_2}{r^2}$
$\Rightarrow {\epsilon }_0=\frac{1}{4\pi }\times \frac{q_1{q}_2}{r^{2}F}$
Substituting the units for $q,r$ and $F$, we obtain unit of
${\epsilon }_0=\frac{\text{ coulomb }\times \text{ coulomb }}{\text{ newton }-(\text{ metre }{)}^2}$
$=\frac{(\text{ coulomb }{)}^2}{\text{ newton }-(\text{ metre }{)}^2}$
$=C^2/N-m^2$