Let denote the dimensional formula of the permittivity of vacuum

PHYSICS · Single correct · CBSE · Physics · Units and Measurements · Units and Measurements

Problem

Let [ϵ0][\epsilon_0] denote the dimensional formula of the permittivity of vacuum. If MM = mass, LL = length, TT = time and AA = electric current, then: (1) ϵ0=[M−1L−3T2A]\epsilon_0 = [M^{-1}L^{-3}T^2A] (2) ϵ0=[M−1L−3T4A2]\epsilon_0 = [M^{-1}L^{-3}T^4A^2] (3) ϵ0=[M1L2T1A2]\epsilon_0 = [M^1L^2T^1A^2] (4) ϵ0=[M1L2T1A]\epsilon_0 = [M^1L^2T^1A]

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