The dimensions of the quantity \( \epsilon_0 \frac{d\Phi_E}{dt} \) are to be determined. \( \epsilon_0 \) represents the permittivity of free space, and \( \Phi_E \) denotes the electric flux.- The dimensions of \( \epsilon_0 \) are derived as: \[ [\epsilon_0] = \frac{\text{C}^2}{\text{Nm}^2} = \frac{\text{A}^2 \cdot \text{s}^4}{\text{kg} \cdot \text{m}^3} \]- The dimensions of electric flux \( \Phi_E \) are established as: \[ [\Phi_E] = \text{Electric field} \times \text{Area} = \left[\frac{\text{N}}{\text{C}}\right] \times \text{m}^2 = \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^2} \]The dimensions of the rate of change of electric flux, \( \frac{d\Phi_E}{dt} \), are calculated as follows:- The dimensions of \( \frac{d\Phi_E}{dt} \) are: \[ \left[\frac{d\Phi_E}{dt}\right] = \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^3} \]Consequently, the dimensions of \( \epsilon_0 \frac{d\Phi_E}{dt} \) are computed by multiplying the individual dimensions:\[[\epsilon_0 \frac{d\Phi_E}{dt}] = [\epsilon_0] \cdot \left[\frac{d\Phi_E}{dt}\right] = \frac{\text{A}^2 \cdot \text{s}^4}{\text{kg} \cdot \text{m}^3} \cdot \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^3} = \text{A} \cdot \text{s}^{-2}\]This resultant dimension corresponds to that of electric current.Therefore, the correct classification is (1) Electric current.
