# Deriving the Electrostatic boundary conditions

When deriving the electrostatic boundary conditions for any charge distribution (to my knowledge at least), Griffiths in his textbook references this illustration:

So, when considering the boundary conditions for any distribution with a surface charge density painted on it $$\sigma$$, it looks like we can consider this image. Is this because if we zoom in super super close to any volume, locally, we can approximate this surface with charge as a plane of charge? If not, how else is this approximation justifiable to derive the electrostatic boundary conditions generally?

In addition, if we took this idea of a plane of charge seriously, I'd expect the perpendicular component of $$E$$ below to be pointing in the opposite direction befitting of a plane of charge.

We seem to be, for any surface, assuming the geometry of the surface and the direction of the electric field (which is either due to the surface charge or the volume charge density this surface is covering) through it.

What's going on here?

In fact, in the analysis of the parallel component of the field, this illustration is used:

Here, it makes sense that the two non-trivial components of the line integral are of opposite sign, but only because Griffiths elected to have $$\mathbf E$$ look this way. Why does he have justification in doing so while still stating this applies generally?

I think my main question is:

Basically, why is $$E_{below}$$ and $$E_{above}$$ parallel, as opposed to, for instance, antiparallel or at some arbitrary angle from eachother?

Why is it also the case for this (which was also used by Griffiths for deriving the magnetostatic boundary conditions):

The $$B_{above}$$ and $$B_{below}$$ certainly don't come from the surface current, as they can't diverge from a plane like that from a cursory glance at Maxwell's equations, but why are they oriented this way?

• Basically, why is $E_{below}$ and $E_{above}$ parallel, as opposed to, for instance, antiparallel or neither? May 5, 2019 at 10:52
The electric field shown as "above" and "below" is not the field due to the surface charge only but the total field, including the field produced by other charge distribution. The point of this diagram is not to show the field of the surface charge alone (this is indeed pointing in opposite directions on the two sides of the surface) but how a pre-existing field (taken for example pointing up) is modified by the surface chage. Assuming that the field of the surface charge is smaller than the external field (in absence of the surface) the direction of the external field does not change but its magnitude changes. You can assume (for example) that the surface charge is positive and then its field will add to the external field in the above region and subtract in the below region. As the field of the surface charge is $$\sigma/2\epsilon_0$$ the difference between the above and below fields will be $$\sigma/\epsilon_0$$. Same result you obtain by applying Gauss' law to the total field, the one including both external and surface fields (which is represented on the diagram).