Let [varepsilon_0] be the dimensional formula | 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 \varepsil

Vedclass · Accepted Answer

$[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$

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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 for $\varepsilon_0$: $\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$Substituting the dimensions: $[q] = [AT]$,$[F] = [MLT^{-2}]$,$[r] = [L]$$[\varepsilon_0] = \frac{[AT][AT]}{[MLT^{-2}][L^2]} = \frac{[A^2 T^2]}{[ML^3 T^{-2}]}$$[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$

Let $[\varepsilon_0]$ be the dimensional formula of the permittivity of free space. If $M = \text{mass}$,$L = \text{length}$,$T = \text{time}$,and $A = \text{electric current}$,then which of the following is correct?

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