In electromagnetism, we say that any conservative electric field $\vec{E}(\vec{r})$ is associated to a scalar potential $V(\vec{r})$ such that $\vec{E}(\vec{r}) = -\nabla V(\vec{r})$. If the electric field is continuous, the respective electric potential must be differentiable because, if not, its gradient could not be calculated everywhere.

There are some cases, though, in which the electric field is discontinuous, leading to a non-differentiable electric potential. The latter is, however, still continuous.

Why is this? Why is it that even when the electric field is discontinuous, the electric potential is not? Must the electric potential always be continuous everywhere? A mathematical approach (i.e. not just a qualitative insight) is what I'm looking for.


No. For example, the potential of a point charge is discontinuous at the location of the point charge, where the potential becomes infinite.

Since all charges in nature seem to be point charges (elementary particles such as electrons and quarks), electric potential always has discontinuities somewhere. When we work with continuous charge distributions, we are simply using an approximation that averages over lots of point charges and smears out the discontinuities in their charge density, potential, field, field energy density, etc.

  • $\begingroup$ And what about the first question? "Why is it that even when the electric field is discontinuous, the electric potential is not?" $\endgroup$ – Tendero Sep 7 '19 at 18:10
  • $\begingroup$ That isn’t true. In the case of the point charge that I mentioned, both the field and the potential are discontinuous. $\endgroup$ – G. Smith Sep 7 '19 at 21:44
  • $\begingroup$ Isn't the electric field infinite in that point? What I want to understand is why the electric potential is continuous when the electric field shows finite discontinuities. $\endgroup$ – Tendero Sep 8 '19 at 2:43
  • $\begingroup$ The change in potential between two points is the line integral of the field between those points. Suppose the field has a finite discontinuity somewhere. Take two points $\epsilon$ away from the discontinuity and calculate the line integral, which will have two finite pieces. As $\epsilon\to 0$, this line integral goes to zero, so the change in potential across the finite discontinuity is zero, so the potential is continuous there. $\endgroup$ – G. Smith Sep 8 '19 at 4:03

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