JEE Main 2024 Physics question

Physics Units and Measurement JEE Main 2024

If ε₀ is the permittivity of free space and E is the electric field, then ε₀E² has the dimensions:

(A)[M⁰L⁻²TA]
(B)[ML⁻¹T⁻²]Correct
(C)[M⁻¹L⁻³T⁴A²]
(D)[ML²T⁻²]

Answer: (B)

Solution

½ε₀E² is the energy density (energy per unit volume), with dimensions ML⁻¹T⁻².

Question ID VK-012390