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. coulomb${}^2$/(newton-metre)${}^2$

Answer 1

Answer: (C)

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
${\epsilon }_0=\frac{\text{coulomb}\times \text{coulomb}}{\text{newton}-{\left(\text{metre}\right)}^2}=\frac{{\left(\text{coulomb}\right)}^2}{\text{newton}-{\left(\text{metre}\right)}^2}$
$=C^2/N-m^2$
Method 2: $F=\frac{1}{4\pi {\epsilon }_0}.\frac{Q_1{Q}_2}{r^2}\Rightarrow {\epsilon }_0\propto \frac{Q^2}{F\times r^2}$
So ${\epsilon }_0$ has units of coulomb${}^2$/ newton-$m^2$