Q 12-08-237JEE MainJEE Main 2025 (7 Apr, Shift 1)Easy
If $\epsilon_0$ denotes the permittivity of free space and $\Phi_E$ is the flux of the electric field through the area bounded by a closed surface, then the dimensions of $\left(\epsilon_0\dfrac{d\Phi_E}{dt}\right)$ are that of:
Answer: (D) Electric current
$\epsilon_0\dfrac{d\Phi_E}{dt}$ is Maxwell's displacement current $i_d$, which enters the Ampere–Maxwell law alongside the conduction current. So it has the dimensions of electric current.
Solution by Sreeraj P, M.Sc Physics