JEE MainPhysicsUnits and Dimensions
If E represents the electric field, ₀ is the permeability of free space, and c is the speed of light in vacuum, then the dimensional formula of the expression E ^2 ₀ c is :
Options
- A[ M L ⁻¹ T ⁻²]
- B[ M L ^0 T ⁻³]
- C[ M L ^2 T ⁻³]
- D[ M L ⁻² T ⁻¹]
Correct answer
B. [ M L ^0 T ⁻³]
Step-by-step solution
We know the relation between the speed of light c , permeability ₀ , and permittivity ₀ of free space is: c = 1 ₀ ₀ ₀ = 1 ₀ c^2 Substitute this into the given expression: E ^2 ₀ c = E ^2 ( 1 ₀ c^2 ) c = ₀ E ^2 c The term ₀ E ^2 is proportional to the energy density of an electric field, which has the dimensions of Energy / Volume: [ ₀ E ^2] = [ M L ^2 T ⁻²] [ L ^3] = [ M L ⁻¹ T ⁻²] The dimension of the speed of light c is [ L T ⁻¹] . Multiplying these gives the dimensions of the complete expression: [ ₀ E ^2 c] = [