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

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

MLT^–2

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

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

answer is C.

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Detailed Solution

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

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