Let [_0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current , then: - Physics MCQ JEE Main 2026 | ExamTest.live

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

Solution & Step-by-step Explanation

From Coulomb's Law: F = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r ^{2}} \Rightarrow ϵ_{0} = \frac{q ^{2}}{F r ^{2}} .Dimensions of Charge [q] = [A T] . So [q^{2}] = [A^{2} T^{2}] .Dimensions of Force [F] = [M L T^{- 2}] .Dimensions of r^{2} = [L^{2}] . [ϵ_{0}] = \frac{[ A ^{2} T ^{2} ]}{[ M L T ^{- 2} ] [ L ^{2} ]} = \frac{[ A ^{2} T ^{2} ]}{[ M L ^{3} T ^{- 2} ]} = [M^{- 1} L^{- 3} T^{4} A^{2}]

Try it yourself before checking the explanation above.

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

[\epsilon_0] = [M^{-1}L^{-3}T^4A^2]

[\epsilon_0] = [M^{-1}L^2T^{-1}A^{-2}]

[\epsilon_0] = [M^{-1}L^2T^{-1}A]

[\epsilon_0] = [M^{-1}L^{-3}T^2A]

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