The dimensions of permittivity {varepsilon _0 | vedclass

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(B) According to Coulomb's Law,the force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by: $F = \frac{1}{{4\pi {\vareps

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${A^2}{T^4}{M^{ - 1}}{L^{ - 3}}$

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(B) According to Coulomb's Law,the force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by: $F = \frac{1}{{4\pi {\varepsilon _0}}}\,\frac{{{q_1}{q_2}}}{{{r^2}}}$Rearranging the formula to solve for permittivity ${\varepsilon _0}$:${\varepsilon _0} = \frac{{{q_1}{q_2}}}{{4\pi F{r^2}}}$Substituting the dimensions of each quantity:$[q] = [AT]$$[F] = [MLT^{-2}]$$[r] = [L]$Therefore,the dimensions of ${\varepsilon _0}$ are:$[{\varepsilon _0}] = \frac{[AT][AT]}{[MLT^{-2}][L^2]}$$[{\varepsilon _0}] = \frac{[A^2T^2]}{[ML^3T^{-2}]}$$[{\varepsilon _0}] = [A^2T^4M^{-1}L^{-3}]$Thus,the correct option is $B$.

The dimensions of permittivity ${\varepsilon _0}$ are

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