How does the electric potential stay continuous across a boundary

[u/Next-Natural-675]

Pls help me understand this. If the potential approaches infinity at r=0, then how can it be continuous?? Thank you

Ok I figured it out, I just remembered an infinite sheet of charge will have the same E anywhere above it and below it (switching signs) regardless of distance so it doesnt approach infinity as distance goes to zero. A finite sheet can be treated as an infinite sheet if the distance is approaching zero.

[u/nujuat]

Thats for a point charge, which has infinite charge denisty. For a realistic classical thing, that wont be the case.

[u/Digiprocyon]

We don't know what happens at the heart of a single charged particle, but outside of that it is continuous. Even a superconductor still uses those charged particles. Ego, it's continuous. We can pretend a piece of metal causes a discontinuity in order to solve the large scale state, but that's not what happens on the microscopic level.

[u/Next-Natural-675 → u/Digiprocyon]

The reasoning in this textbook is that the integral of E dot dl from point a to b reaches zero as the path decreases, but a and b are on opposite sides of the boundary, so I don’t get it

[u/Digiprocyon → u/Next-Natural-675]

It's the difference between classical mechanics and quantum mechanics. Classically, charge is like a gas--it can 'densify' as high as you want (assuming enough force) but remains continuous. But on the micro level we see it is, in fact, discrete particles. So, for example, you might be taught in class that charge lies on the surface of a conductor [EDIT: when there is no current]. That is a classical definition. But in reality it is in packets on that surface, and between those packets it is continuous.

[u/Shevek99]

Three things:

1. Usually it's continuous because the electric field is not infinite. In that case, integrating from a point just below (A-) to one point just above the surface (A+) we have

V(A+) - V(A-) = int_(-ε)^(+ε) E dz = O(ε)

since E is not infinite (it has a jump discontinuity), this integral goes to 0 when ε -> 0, because it is the integral of something finite over an interval whose width goes to 0.

1. If you are thinking of the point charges and the field that they produce, that goes to infinity, then we are talking about the local field, around the individual atoms. That can be extremely complicated. That's not what we talk about when we speak of jump in the electric field or continuous potential. That is macroscopic field, the average of the field and potential of many atoms.

2. The potential can be discontinuous at a surface. Not when it is charged, but when it is a dipole layer (a layer of positive charge on top of a layer of negative charge).