# Continuity of Potential of an electric field

Consider a point charge $q$ at $(0,0,0)$, the potential at $\bf{r}$ is given by $V(\bf{r})$ $= \frac{q}{4\pi\epsilon_0r}$. If you consider a path through $(0,0,0)$, you encounter a discontinuity in electric field and the direction of field changes. Is the potential for the field of point charge discontinuous at the location of the point charge?

I'm facing this confusion because, in the book by Griffiths, it is mentioned that the electric field is discontinuous at the surface of charge, with charge density $\sigma$ i.e $E_{above}-E_{below}=\frac{\sigma}{\epsilon_0}\hat{n}$. It is also mentioned that the potential is continuous across a boundary like that. Is my analogy between point and surface charge wrong?

Also, is potential for an electric field continuous everywhere? Or is it continuous across field discontinuities?

Consider a point charge $q$ at $(0,0,0)$, the potential at $\bf{r}$ is given by $V(\bf{r})$ $= \frac{q}{4\pi\epsilon_0r}$. If you consider a path through $(0,0,0)$, you encounter a discontinuity in electric field and the direction of field changes. Is the potential for the field of point charge discontinuous at the location of the point charge?

If you specify the potential as a function from real numbers to real numbers then the function is discontinuous at zero by virtue of the fact that it is not defined at zero. I.e., the definition of continuity $$\lim_{r\to 0}V(r)\to V(0)$$ does not hold because V(0) is not defined (it is not a real number).

I'm facing this confusion because, in the book by Griffiths, it is mentioned that the electric field is discontinuous at the surface of charge, with charge density $\sigma$ i.e $E_{above}-E_{below}=\frac{\sigma}{\epsilon_0}\hat{n}$. It is also mentioned that the potential is continuous across a boundary like that. Is my analogy between point and surface charge wrong?

The analogy seems somewhat imperfect to me because both $E$ and $V$ go to infinity in the first example, but neither does in the second example. If E remains finite, phi will be continuous (as discussed below).

Also, is potential for an electric field continuous everywhere? Or is it continuous across field discontinuities?

Typically $\phi$ is continuous since, for example, if $\phi$ had a discontinuity $\Delta\phi$ at some point $x_0$ then $$\Delta\phi=\lim_{\delta\to 0}\phi(x_0+\delta)-\phi(x_0)=\lim_{\delta\to 0}\int_{x_0}^{x_0+\delta}\frac{d\phi}{dx}dx=\lim_{\delta\to 0}\int_{x_0}^{x_0+\delta}E_j(x)dx =\lim_{\delta\to 0}E_j(x_0)\delta\;,$$ which is zero unless E(x_0) blows up. I.e., for finite electric fields (even discontinuous finite electric fields) the potential remains continuous.

But, remember, in general, the potential does not have to be continuous everywhere (as evidenced by the case mentioned above of a point charge). This is because the governing equation for $\phi$ is $$\nabla^2 \phi=-\rho/\epsilon_0\;,$$ where $\rho$ is the charge density. And, I'll bet that you can come up with some wacky charge densities (a single point charge is one case) where $\phi$ is discontinuous.

A point charge is given at a single, specific location in space. Think of it as an idealized source of charge. At the point where r = (0, 0, 0), electric field is zero and V(r) = $\infty$. As the value of r approaches 0 from any direction, V increases towards infinity but it is never discontinuous.

For any other surface, the concept is well-explained by Hyperphysics. At the surface of, say, a charged sphere of radius R, we have $E = \frac{Q}{4\pi\epsilon R^2}$. Anywhere inside the sphere, the sum of the surface charges from all directions will be equal to zero, so electric field is discontinuous. Because the electric field is zero, the electric potential remains constant inside the sphere.

If you draw (or visualize) electric field and potential graphs as functions of distance from the center of both a point charge and a charged sphere (as seen in the Hyperphysics link), it should all make more sense.

For both a point charge and a charged sphere, electric field is discontinuous at the "surface" (in the case of a point charge, the surface IS the location of the charge) and the electric potential is continuous.