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Q. The unit of electrical permittivity of free space $\varepsilon_{0}$ is

Physical World, Units and Measurements

Solution:

$ F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$
$\left[\varepsilon_{0}\right]=\frac{\left[q_{1}\right]\left[q_{2}\right]}{[4 \pi][F]\left[r^{2}\right]}=\frac{C^{2}}{N m^{2}}=C^{2} N^{-1} m^{-2}$