Explanation

If E is the intensity of electric field over a small area element dS and $\theta$ is angle between E and outdrawn normal to area element. Therefore, electric flux through this element is $\ \ \ \ \ \ \ \ \ \ \ \ d\phi_{E} = (dS)\left( E\cos\theta \right)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = E\ dS\cos\theta = \mathbf{E.dS}$ Hence, $\phi_{E} = E\ .S$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{V}{d}\ .S$ $\therefore Unit\ of\ \phi_{E} = \frac{volt\ \times {metre}^{2}}{metre}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = volt - metre$