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The dimensional formula of $\frac{1}{2} ε_{0} E^{2} is (ε_{0}$ is permittivity of free space and E is electric field)

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${ML}^{2} T^{- 2}$

${MLT}^{- 2}$

${ML}^{- 1} T^{- 2}$

${ML}^{- 2} T^{- 1}$

answer is C.

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Detailed Solution

$\frac{1}{2} ε_{0} E^{2} = [M^{- 1} L^{- 3} T^{4} I^{2}] {[{MLT}^{- 3} I^{- 1}]}^{2} = [M^{- 1} L^{- 3} T^{4} I^{2}] {[M^{2} L^{2} T^{- 6} I^{- 2}]}^{=} [{ML}^{- 1} T^{- 2}]$

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