If $\epsilon_0$ denotes the permittivity of free space and $\Phi_E$ is the flux of the electric field through the area bounded by the closed surface, then dimensions of $\left(\epsilon_0 \frac{d \phi_E}{d t}\right)$ are that of :

<p>We can use Gauss’s law to analyze the dimensions. Gauss’s law states that</p> <p>$\Phi_E = \frac{q}{\epsilon_0},$</p> <p>where</p> <p><p>$\Phi_E$ is the electric flux,</p></p> <p><p>$q$ is the electric charge,</p></p> <p><p>$\epsilon_0$ is the permittivity of free space.</p></p> <p>To find the dimensions of $\epsilon_0 \frac{d\Phi_E}{dt}$, follow these steps:</p> <p><p>Differentiate Gauss’s law with respect to time:</p> <p>$\frac{d\Phi_E}{dt} = \frac{1}{\epsilon_0}\frac{dq}{dt}.$</p></p> <p><p>Multiply the equation by $\epsilon_0$:</p> <p>$\epsilon_0 \frac{d\Phi_E}{dt} = \frac{dq}{dt}.$</p></p> <p><p>Recognize that the time rate of change of charge, $\frac{dq}{dt}$, is by definition the electric current.</p></p> <p>Thus, the dimensions of $\epsilon_0 \frac{d\Phi_E}{dt}$ are those of electric current.</p> <p>The correct answer is: Option C, electric current.</p>