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Q.
The unit of permittivity of free space, $\varepsilon_{0}$ is:
Physical World, Units and Measurements
Solution:
By Coulomb’s law, the electrostatic force
$F=\frac{1}{4\pi\varepsilon_{0}}\times\frac{q_{1}q_{2}}{r^{2}} \Rightarrow \varepsilon_{0}=\frac{1}{4\pi}\times\frac{q_{1}q_{2}}{r^{2}F}$
Substituting the units for $q, r$ and $F,$ we obtain unit of
$\varepsilon_{0}=\frac{\text{coulomb} \times \text{coulomb}}{\text{newton}-\left(\text{metre}\right)^{2}}=\frac{\left(\text{coulomb}\right)^{2}}{\text{newton}-\left(\text{metre}\right)^{2}}$
$=C^{2}/N-m^{2}$ Method 2: $F=\frac{1}{4\pi\varepsilon_{0}}.\frac{Q_{1}Q_{2}}{r^{2}}\Rightarrow \varepsilon_{0}\propto\frac{Q^{2}}{F\times r^{2}}$
So $\varepsilon_{0}$ has units of coulomb$^{2}$/ newton-$m^{2}$