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Question: The dimensions of \(\frac { 1 } { 2 }\) \(\varepsilon_{0}E^{2}\) (\(\varepsilon_{0}\)= permittivity...

The dimensions of 12\frac { 1 } { 2 } ε0E2\varepsilon_{0}E^{2} (ε0\varepsilon_{0}= permittivity of free space ; E = electric field ) is

A

MLT−1MLT^{- 1}

B

ML22T−2- 2

C

ML−1- 1T−2- 2

D

ML2T−1ML^{2}T^{- 1}

Answer

ML−1- 1T−2- 2

Explanation

Solution

Energy density = 12ε0E2=EnergyVolume\frac{1}{2}\varepsilon_{0}E^{2} = \frac{\text{Energy}}{\text{Volume}}

=[ML2T−2L3]=[ML−1T−2]= \left\lbrack \frac{ML^{2}T^{- 2}}{L^{3}} \right\rbrack = \lbrack ML^{- 1}T^{- 2}\rbrack