The formula for the discontinuity of the electric field across a boundary is given by$$\vec{E}_{\text{above}} – \vec{E}_{\text{below}} = \frac{\sigma}{\epsilon_0} \hat{n}.$$ In the derivation of this formula we use the Gaussian box and hence Gauss's Law. What I don't understand is, at which point in the proof does the result become generally applicable to any electric field above and below any surface rather than just the electric field produced by the charge enclosed in the small Gaussian box?
I am asking since for any conductor it states in literature I am using (Griffiths Introduction to Electrodynamics), that the electric field immediately outside the conductor is "$\vec{E} = \frac{\sigma}{\epsilon_0} \hat{n}$." This formula was obtained from the above formula and the fact that the electric field inside a conductor is zero.
Best Answer
Gauß law for electrodynamics states that $$\nabla \vec{E}= \frac{\rho}{\varepsilon_0} $$ at any point and no matter by what the electric field $\vec{E}$ is produced. $\vec{E}$ here stands for the total electric field at a point produced by all sources that are present, not just any selected few. So the whole derivation is valid for any surface and any electric field. The assumption in the derivation is just that $\rho$ is only nonzero on the surface.