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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.

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  • $\begingroup$ The idea is that when you zoom in enough into any surface, it looks flat, hence for every point on any arbitrary surface, you can use the Gaussian box argument. $\endgroup$
    – Omar Nagib
    Commented Jul 25, 2016 at 14:29

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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.

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  • $\begingroup$ Could you elaborate a bit on your reasoning why the whole derivation is valid for any surface and any electric field? Not very clear. $\endgroup$
    – Alex
    Commented Jul 25, 2016 at 14:11
  • $\begingroup$ @Alex: Gauß law is valid everywhere with $\vec{E}$ being the total electric field (not just a part of the field produced by something special). It can be integrated over any volume. If I happen to take the volume in the proof, I can use the argument made in the proof. $\endgroup$
    – Sanya
    Commented Jul 25, 2016 at 14:17
  • $\begingroup$ @Sanya "Gauß law is valid everywhere with E⃗ being the total electric field"...provided the field vanishes at infinity. $\endgroup$
    – gented
    Commented Jul 25, 2016 at 16:16
  • $\begingroup$ @GennaroTedesco could you elaborate on why that is necessary, especially in the context of the question? $\endgroup$
    – Sanya
    Commented Jul 25, 2016 at 17:00
  • $\begingroup$ @Sanya simply because Gauß law is derived assuming that the fields vanish at infinity (in particular when applying Stoke's theorem on its integral form). Otherwise, it just doesn't hold true. $\endgroup$
    – gented
    Commented Jul 25, 2016 at 17:12
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I think the point is that the electric field contribution from the patch (which is the surface charge inside the Gaussian box) is the significant contributing one in the formula for the following reason:

In the proof after we form our Gaussian box we let the sides of the box (of length say $\epsilon$) tend to zero. Then we arrive at the result that the difference in the electric field produced by charges in the box are $\vec{E}_{\text{above box}} - \vec{E}_{\text{below box}} = \frac{\sigma}{\epsilon_0} \hat{n}$. Your concern is about why the other surface charge contribution outside the Gaussian box does not seem to be present in the derivation of the general formula. The outside contribution simply cancel, since $$\vec{E}_{above} = \vec{E}_{other} + \vec{E}_{\text{above box}}$$ and similarly $$\vec{E}_{below} = \vec{E}_{other} + \vec{E}_{\text{below box}}$$ hence $$\vec{E}_{above} - \vec{E}_{below} = \frac{\sigma}{\epsilon_0} \hat{n}.$$ This is expected since the discontinuity is only from the patch covered by the Gaussian surface, if we removed that patch we would have a continuous electric field in this space. It is also clear to see for cases where we have symmetry, like an infinite charged plane (with uniform surface charge), that $\vec{E}_{above} = \vec{E}_{\text{above box}}$ and $\vec{E}_{below} = \vec{E}_{\text{ below box}}$ (also the formula from the proof clearly holds).

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