Understanding electric field discontinuity

From divergence theorem/Gauss's law: $$\oint_S \vec{E} . d\vec{a}=\frac{1}{\epsilon_0}\sigma A.$$ From what I understand, Gauss's law is where the electric field discontinuity (at a boundary) is derived from (mathematically):

$$\vec{E}_{above} - \vec{E}_{below} =\frac{\sigma}{\epsilon_0} \hat{n}.$$ However, what I don't understand, is what causes this discontinuity physically.

It has been confusing me for several months and the explanations that my teachers have given me, haven't resolved my problem. If anyone could clear this up for me that'd be great. Thanks.

• what is confusing me, is why the discontinuity is always $\frac{1}{\epsilon_0}\sigma A$ specifically – George Dec 24 '17 at 16:04
• I think you mean $\sigma/\epsilon_0$ ... no $A$. It is sensible that the discontinuity depends on $\sigma$. That it is linear in $\sigma$ is due to the fact that electric fields add linearly. The $\epsilon_0$ is there only to make the units come out right. In certain systems of units other than the S.I. system, there is no $\epsilon_0$ – garyp Dec 24 '17 at 16:19