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.
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$\begingroup$ Because $\nabla\cdot{\bf E}=\rho/\epsilon$. $\endgroup$– The PhotonCommented Dec 23, 2018 at 6:40
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1 Answer
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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.
answered Dec 23, 2018 at 6:42