boundary condition of perpendicular component of electric field of a thin sheet

Question

This question is in reference to Introduction to Electrodynamics by David Griffith

By Gauss's law:

$\oint_{S} \vec{E}.d\vec{a}=\frac{Q_{enclosed}}{\epsilon_{0} }$

where

$Q_{enclosed}$ is the charge enclosed by a Gaussian surface.

If we imagine a Gaussian pillbox of area A with a finite depth on the "top" face of the sheet, we take the sum electric field of an eclosed charge through the surface A and the electric field of the same enclosed charge through the sides of finite depth.

But, by the dot product, only the component of the electric field perpendicular to the surface $d\vec{a}$ of the Gaussian pillbox exists.

So, $\vec{E} = \frac{\sigma}{\epsilon_{0} }\hat{n}$

where $\sigma$ is the surface charge density.

Griffith's explanation that $E^{\perp }_{above}-E^{\perp}_{below}=\frac{\sigma}{\epsilon_{0} }$ does not enable me to fully understand how this expression is arrive at.

Perhaps, someone may be able to assist me with the schematic set up so that I may have a better insight.

Answer 1

$\oint_{S} \vec{E}.d\vec{l}=\frac{Q_{enclosed}}{\epsilon_{0} }$.

It is not a line integral on the left hand side but a surface integral which states that the net electric flux out of an enclosed surface is equal to the net charge enclosed by the surface divided by the permittivity.

So in this case

$$E^{\perp }_{above}A-E^{\perp}_{below}A=\frac{\sigma A}{\epsilon_{0} }$$

which leads to Griffith’s result.

The expression $\vec{E} = \frac{\sigma}{\epsilon_{0} }\hat{n}$ is true for the electric field at the surface of a conductor where $E^{\perp}_{below}$ inside the conductor is zero.