Explain electric flux.

If a small planar element of area $\overrightarrow{\Delta S}$ is placed normal to $\vec{E}$ at a point, the number of field lines crossing it is proportional to $\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\Delta \mathrm{S}}$.

Suppose we tilt the area element by angle $\theta$, the number of field lines crossing $\Delta \mathrm{S}$ is proportional to $\mathrm{E} \Delta \mathrm{S} \cos \theta$.

When $\theta=90^{\circ}$, field lines will be parallel to surface and will not cross it at all.

When $\theta=0^{\circ}$, field lines will be normal to surface as shown in figure.

The vector associated with every area element of a closed surface is taken to be in the direction of the outward normal. Thus, the area element vector $\overrightarrow{\Delta S}$ at a point on a closed surface equals $\Delta \mathrm{S} \hat{n}$ where $\Delta \mathrm{S}$ is the magnitude of the area element and $\hat{n}$ is a unit vector in the direction of outward normal at that point.

Electric flux is no. of electric field lines passing through or associated with the surface placed in electric field.

$\therefore$ Electric flux $\Delta \phi$ through an area element $\Delta \overrightarrow{\mathrm{S}}$ is $\Delta \phi=\overrightarrow{\mathrm{E}} \cdot \Delta \overrightarrow{\mathrm{S}}=\mathrm{E} \Delta \mathrm{S} \cos \theta$ the angle $\theta$ here is the angle between $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{S}}$.

$\phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\Delta \mathrm{S}}$

$=\mathrm{E} \Delta \mathrm{S} \cos \theta$

where $\theta$ is angle between $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\Delta \mathrm{S}}$.

$SI$ unit of electric flux is $\mathrm{Nm}^{2} \mathrm{C}^{-1}$ or $\mathrm{Vm}$ and it is a scalar quantity.

Definition of electric flux : "Electric flux associated with any area is areal integration of vector electric field on that area".