''electric field always undergoes a discontinuity when you cross a surface charge $\sigma$'' GRIFFITHS
In the derivation; Suppose we draw a wafer-thin Gaussian Pillbox, extended just barely over the edge in each direction. Gauss law states that:
$$\int_{S} E \cdot A = Q_\text{enc}/ \epsilon$$
and so $$E_{\perp\;\text{above}} - E_{\perp\;\text{below} } = \sigma/ \epsilon . $$
My question is why not $2A$? $$\int_{S} E \cdot A = 2EA$$ because the top area of pillbox and the bottom area of pillbox, just as because the 2 parts of the flux...
SO.. WHY NOT : $$E_{\perp\;\text{above}}- E_{\perp\;\text{below} } = \sigma/ 2\epsilon ~? $$
And why there is tangencial component of electric field; not just perpendicular to the surface, which can be seen as flat just looking very close to the surface.
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