Electrostatics Part 3

By Gauss's law, the electric flux through a closed surface is related to the charge enclosed by it as$\varepsilon_0 \phi_E = q_{\text{in}}$.Differentiating both sides with respect to time,$\varepsilon_0 \dfrac{d \phi_E}{d t} = \dfrac{d q_{\text{in}}}{d t}$.The rate of flow of charge, $\dfrac{d q_{\text{in}}}{d t}$, is electric current.Hence $\left[ \varepsilon_0 \dfrac{d \phi_E}{d t} \right] = [I] = A$ (ampere).This quantity is exactly the displacement current introduced in Maxwell's equations, so its dimensions are those of electric current.Therefore the correct option is: electric current.