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The unit of permittivity of free space $\varepsilon_0$ is :-
A
Newton metre$^2$/ Coulomb$^2$
B
Coulomb$^2$ /Newton metre$^2$
C
Coulomb$^2$/ (Newton metre)$^2$
D
Coulomb/Newton metre
Detailed Solution
Coulomb's law: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$
Rearranging: $\varepsilon_0 = \dfrac{q_1 q_2}{4\pi F r^2}$
$4\pi$ is a pure number, so the unit of $\varepsilon_0 = \dfrac{\text{Coulomb} \times \text{Coulomb}}{\text{Newton} \times \text{metre}^2}$
$[\varepsilon_0] = \dfrac{\text{Coulomb}^2}{\text{Newton metre}^2}$, i.e. $C^2 N^{-1} m^{-2}$
So the unit of permittivity of free space is Coulomb$^2$/Newton metre$^2$.
Rearranging: $\varepsilon_0 = \dfrac{q_1 q_2}{4\pi F r^2}$
$4\pi$ is a pure number, so the unit of $\varepsilon_0 = \dfrac{\text{Coulomb} \times \text{Coulomb}}{\text{Newton} \times \text{metre}^2}$
$[\varepsilon_0] = \dfrac{\text{Coulomb}^2}{\text{Newton metre}^2}$, i.e. $C^2 N^{-1} m^{-2}$
So the unit of permittivity of free space is Coulomb$^2$/Newton metre$^2$.