The electric field intensity near a conducting surface having uniform surface charge density ' $\sigma$ ' is given by ( $\epsilon_0=$ permittivity of free space)

Electric field near a conducting surface

A Gaussian pillbox with one face inside the conductor and the other just outside provides the appropriate surface for applying Gauss's law. The electric field inside the conductor is zero, while the field outside is perpendicular to the surface. The flux through the curved surface is zero due to the parallel alignment with the field lines.

Only the external end face contributes to the flux, giving $\Phi = EA$ for area $A$. The enclosed charge is $q_{\text{enc}} = \sigma A$. Applying Gauss's law:

$EA = \frac{\sigma A}{\epsilon_0}$

Simplifying yields the magnitude $E = \frac{\sigma}{\epsilon_0}$, directed normal to the surface with orientation determined by the sign of $\sigma$.

The electric field intensity near the conducting surface is therefore $\frac{\sigma}{\epsilon_0}$.