Single Choice

Let $$[\varepsilon _{0}]$$ denote the dimensional formula of the permittivity of vacuum. If $$M=$$ mass, $$L=$$ length, $$T=$$ time and $$A=$$ electric current, then:

A$$[\varepsilon _{0}]=[M^{-1}L^{-3}T^{4}A^{2}]$$
Correct Answer
B$$[\varepsilon _{0}]=[M^{-1}L^{2}T^{-1}A^{-2}]$$
C$$[\varepsilon _{0}]=[M^{-1}L^{2}T^{-1}A]$$
D$$[\varepsilon _{0}]=[M^{-1}L^{-3}T^{2}A]$$

Solution


1
4πε0

q2
r2

=F
ε0=
[A2T2]
[MLT−2L2]

=[M−1L−3A2T4]

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