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Q.

The dimension of $\frac{1}{2} ε_{0} E^{2}$ . Where $ε_{0}$ is permittivity of free space and E is electric field is

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a

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

b

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

c

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

d

$[{MLT}^{- 1}]$

answer is B.

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Detailed Solution

$u = \frac{1}{2} ε_{0} E^{2} = \frac{U}{V}$

$[\frac{1}{2} ε_{0} E^{2}] = \frac{[{ML}^{2} T^{- 2}]}{[L^{3}]} = [{ML}^{- 1} T^{- 2}]$

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