# Why does electric field undergo a discontinuity when we cross $any$ surface charge $σ$?

According to Griffith's book on electrodynamics, electric field always undergoes a discontinuity when crossing a surface charge $$σ$$. I do understand that in certain cases like the surface of a spherical charge shell, electric field undergoes discontinuity. However I have no clue why there is discontinuity on crossing any kind of surface charge? Please explain.

• Because $\nabla\cdot{\bf E}=\rho/\epsilon$. – The Photon Dec 23 '18 at 6:40

It just comes from Gauss's law. That is, Gauss's law states that

$$\oint _S \mathbf{E\cdot } d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0}$$

Suppose now we apply this to a plane. WOLOG, suppose that it is oriented so that it has normal vector $$\hat{z}$$ on the top and $$-\hat{z}$$ on the bottom, then it follows that, for surface charge enclosed by a region of area $$dA$$, that

$$\mathbf{E\cdot}dA\hat{z} - \mathbf{E}\cdot{dA\hat{z}} = \frac{\sigma dA}{\epsilon_0}.$$

It immediately follows that

$$E_1^\perp - E_2^\perp = \frac{\sigma}{\epsilon_0}.$$

Griffiths refers to these as $$E^{\perp}_{abv.}$$ and $$E^{\perp}_{bel.}$$respectively.