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Question

Physics Question on Dimensional Analysis

The dimension of 12ε0E2\frac{1}{2}\varepsilon _0 E^2, where ε0\varepsilon _0 is permittivity of free space and EE is electric field, is

A

ML2T−2ML^2T^{-2}

B

ML−1T−2ML^{-1}T^{-2}

C

ML2T−1ML^2T^{-1}

D

MLT−1MLT^{-1}

Answer

ML−1T−2ML^{-1}T^{-2}

Explanation

Solution

Energy density of an electric field EE is
uE=12ε0E2u_E=\frac{1}{2}\varepsilon _0 E^2
where ε0\varepsilon _0 is permittivity of free space
uE=EnergyVolumeu_E=\frac{\text{Energy}}{\text{Volume}}
=ML2T−2L3=\frac{ML^2T^{-2}}{L^3}
=ML−1T−2=ML^{-1}T^{-2}
Hence, the dimension of 12ε0E2 is ML−1T−2\frac{1}{2}\varepsilon _0 E^2 \, is \, ML^{-1}T^{-2}