Obtain an expression for the electric field at the surface of a charged conductor.

(N/A) We consider a Gaussian surface in the form of a pillbox of extremely small length and extremely small cross-sectional area $dS$.
$A$ fraction of this pillbox is inside the conductor,and the remaining part is outside the surface.
The total charge enclosed by this pillbox is $q = \sigma dS$,where $\sigma$ is the surface charge density of the conductor.
At every point on the surface of the conductor,the electric field $\vec{E}$ is perpendicular to the surface. Hence,it is parallel to the area vector $d\vec{S}$,so $\vec{E} \parallel d\vec{S}$.
Inside the conductor,the electric field $\vec{E} = 0$. Therefore,the flux coming out from the cross-section of the pillbox inside the surface is $0$.
The flux coming out from the cross-section of the pillbox outside the surface is $\phi = \vec{E} \cdot d\vec{S} = E dS \cos 0^{\circ} = E dS$.
According to Gauss's theorem,the total flux $\phi = \frac{q}{\varepsilon_{0}}$.
Substituting the values,we get $E dS = \frac{\sigma dS}{\varepsilon_{0}}$.
Therefore,$E = \frac{\sigma}{\varepsilon_{0}}$.
In vector form,$\vec{E} = \frac{\sigma}{\varepsilon_{0}} \hat{n}$,where $\hat{n}$ is the unit vector normal to the surface.
If $\sigma$ is positive,$\vec{E}$ is in the direction of the normal pointing outward from the surface. If $\sigma$ is negative,$\vec{E}$ is in the direction of the normal pointing inward toward the surface.