Dimension of 1/2 ∈0E^2 (Where ∈0 = permittivity of vaccum and E = electric field) is ​

Question

Dimension of \(\frac12\epsilon_0E^2\) (Where ∈₀ = permittivity of vaccum and E = electric field) is 

(A) [MLT^-1]

(B) [ML²T^-2]

(C) [ML^-1T^-2]

(D) [ML²T^-1]

Answer 1

Correct option is (C) [ML^-1T^-2]

Dimension of ∈₀ = M⁻¹L⁻³T⁴A²

Dimension of E(Electric field) = MLT⁻²/AT

= MLT⁻¹A⁻¹

Dimension of \(\frac 1{2\epsilon_0E^2} = \frac 1{(M^{-1}L^{-3}T^4A^2)(MLT^{-1}A^{-1})^2}\)

\(=\frac 1{ML^{-1}T^2}\)

\(= ML^{-1}T^{-2}\)