Is electric potential always continuous?

Question

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.

Answer 1

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.

Answer 2

Yes, it is. In electromagnetism, the electric field is defined in terms of the electric potential through the equation $\mathbf{E}=-\nabla \phi$, where $\nabla$ is the gradient operator. If the electric potential were to have a discontinuity or singularity at a certain point, it would lead to a divergence in the electric field at that point, which is physically unrealistic. In classical electromagnetism, such singularities or discontinuities are not allowed within the domain of valid solutions.

As for point charges, they are not physical objects, just mathematical models. The Maxwell equations don't have solutions that are discontinuous functions. On the other hand, singular potentials play a crucial role in defining Green's functions and simplifying various concepts through the introduction of analyticity. They provide formal solutions for non-homogeneous equations in integral form and serve as valuable mathematical models.

Answer 3

A real discontinuity in the mathematical sense (not a divergence) of the electric potential is only possible in the presence of a surface dipole distribution (see, for instance, this page ). It can be obtained starting with the expression of the potential due to a continuous dipole density ${\bf P(r)}$: $$ \phi({\bf r})=\frac{1}{4 \pi \epsilon_0}\int {\bf P(r')}\cdot\nabla'\left( \frac{1}{| {\bf r} - {\bf r'} |} \right) dV' $$ Analyzing the effect of a dipole density confined on a surface, it can be shown that the limit values of the potential on the two faces of the surface must be different.