We consider a Gaussian surface of a pill box of extremely small length and extremely small cross-section as $\sigma$.

A fraction of it is inside the surface and the remaining part is outside the surface.

The total charge enclosed by this pill box is $q=\sigma d s$

where $\sigma=$ surface charge density of conductor.

At every point on the surface of the conductor $\vec{E}$ is perpendicular to the surface. Hence, it is parallel to the surface vector $\overrightarrow{\mathrm{E}} \| d \vec{s}$

But inside the surface $\vec{E}=0$. Hence, the flux coming out from the cross-section of pill box inside the surface $=0$.

The flux coming out from the cross-section of pill box outside the surface, $\phi=\overrightarrow{\mathrm{E}} \cdot d \vec{s}=\mathrm{E} d s \cos 0^{\circ}=\mathrm{E} d s$

According to Gauss's theorem,

$\phi=\mathrm{E} d s$

$\therefore \frac{q}{\varepsilon_{0}}=\mathrm{E} d s$

$\therefore \frac{\sigma d s}{\varepsilon_{0}}=\mathrm{E} d s$

$\therefore \mathrm{E}=\frac{\sigma}{\varepsilon_{0}}$

In the vector form $\overrightarrow{\mathrm{E}}=\frac{\sigma}{\varepsilon_{0}} \cdot \hat{n}$

If $\sigma$ is positive $\overrightarrow{\mathrm{E}}$ is in the direction of normal coming out from the surface. If $\sigma$ is negative $\overrightarrow{\mathrm{E}}$ is in direction of normal entering into the surface.