Q 11-01-113JEE MainJEE Main 2024 (8 Apr, Shift 2)Easy
If $\epsilon_0$ is the permittivity of free space and $E$ is the electric field, then $\epsilon_0E^2$ has the dimensions:
Answer: (D) $[\text{ML}^{-1}\text{T}^{-2}]$
$\frac12\epsilon_0E^2$ is the energy density of an electric field (energy per unit volume):
$$[\epsilon_0E^2] = \frac{[\text{ML}^2\text{T}^{-2}]}{[\text{L}^3]} = [\text{ML}^{-1}\text{T}^{-2}]$$
Solution by Sreeraj P, M.Sc Physics