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

enter image description here

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:

enter image description here

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):

enter image description here

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?

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Thankfully the continuity relations at the boundary are local so yes it’s legit to make it a local argument by considering the surface to be locally flat. The direction is as shown because one defines the positive direction to be in the direction of a normal vector to the surface, and this normal vector is same for both sides of the surface.

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  • $\begingroup$ The normal vector is the same for both sides of the surface? Are you implying it's pointing into the surface at one side? That's the only way I can imagine it if it's the same for both sides of the surface. I thought it was just because the field will always be piercing the surface like this locally, but it was just a guess. $\endgroup$ – sangstar May 4 '19 at 9:12
  • $\begingroup$ I've edited my answer, which may make my confusion a bit more clear. $\endgroup$ – sangstar May 4 '19 at 10:01
  • $\begingroup$ Basically, why is $E_{below}$ and $E_{above}$ parallel, as opposed to, for instance, antiparallel or neither? $\endgroup$ – sangstar May 5 '19 at 10:52
  • $\begingroup$ @sangstar away on travel back later. $\endgroup$ – ZeroTheHero May 5 '19 at 11:35

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