PhysicsEasy156×since 2002Q8220Let [ε0{\varepsilon _0}ε0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:Aε0=[M−1L−3T2A]{\varepsilon _0} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^2}A} \right]ε0=[M−1L−3T2A]Bε0={\varepsilon _0} =ε0=[M−1L−3T4A2]\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right][M−1L−3T4A2]Cε0=[M1L2T1A2]{\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}{A^2}} \right]ε0=[M1L2T1A2]Dε0=[M1L2T1A]{\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}A} \right]ε0=[M1L2T1A]Check answerSkip