Question
The unit of permittivity of free space, $${\varepsilon _0}$$ is
A.
coulomb/newton-metre
B.
$${\text{newton - metr}}{{\text{e}}^2}/{\text{coulom}}{{\text{b}}^2}$$
C.
$${\text{coulom}}{{\text{b}}^2}/{\text{newton - metr}}{{\text{e}}^2}$$
D.
$${\text{coulom}}{{\text{b}}^2}/{\left( {{\text{newton - metre}}} \right)^2}$$
Answer :
$${\text{coulom}}{{\text{b}}^2}/{\text{newton - metr}}{{\text{e}}^2}$$
Solution :
$$\eqalign{
& {\text{According to Coulomb's law, the electrostatic force}} \cr
& F = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{{{q_1}{q_2}}}{{{r^2}}} \cr
& {q_1}\,{\text{and }}{q_2}\, = {\text{charges, }}r = {\text{distance between charges}} \cr
& {\text{and}}\,{\varepsilon _0} = \,{\text{permittivity of free space}} \cr
& \Rightarrow {\varepsilon _0} = \frac{1}{{4\pi }} \times \frac{{{q_1}{q_2}}}{{{r^2}F}} \cr
& {\text{Substituting the units for }}q,r{\text{ and }}F,\,{\text{we obtain unit}}\,{\text{of}}\,{\varepsilon _0} \cr
& = \frac{{{\text{coulomb}} \times {\text{coulomb}}}}{{{\text{newton}} - {{\left( {{\text{metre}}} \right)}^2}}} \cr
& = \frac{{{{\left( {{\text{coulomb}}} \right)}^2}}}{{{\text{newton}} - {{\left( {{\text{metre}}} \right)}^2}}} \cr} $$