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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

[ML2T2][M L^2 T^{-2}]

B

[ML1T2][M L^{-1} T^{-2}]

C

[ML2T1][M L^2 T^{-1}]

D

[MLT1][M L T^{-1}]

Step-by-Step Solution

The quantity 12ε0E2\frac{1}{2}\varepsilon_0 E^2 represents the energy density (energy per unit volume) of an electric field. The dimensional formula of energy is [ML2T2][M L^2 T^{-2}] and that of volume is [L3][L^3]. Therefore, the dimensional formula of energy density is [ML2T2][L3]=[ML1T2]\frac{[M L^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}]. Alternatively, you can derive it by substituting the individual dimensions: Dimension of permittivity of free space, ε0=[M1L3T4A2]\varepsilon_0 = [M^{-1} L^{-3} T^4 A^2] Dimension of electric field, E=ForceCharge=[MLT2][AT]=[MLT3A1]E = \frac{\text{Force}}{\text{Charge}} = \frac{[M L T^{-2}]}{[A T]} = [M L T^{-3} A^{-1}] Dimension of ε0E2=[M1L3T4A2]×[MLT3A1]2=[M1L3T4A2]×[M2L2T6A2]=[ML1T2]\varepsilon_0 E^2 = [M^{-1} L^{-3} T^4 A^2] \times [M L T^{-3} A^{-1}]^2 = [M^{-1} L^{-3} T^4 A^2] \times [M^2 L^2 T^{-6} A^{-2}] = [M L^{-1} T^{-2}].

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