"This is because of abrupt discontinuity of fields"

I have read this or similar sentences in many papers. I am bit puzzled. How and under what conditions electric field can be discontinous? In my opinion this is unphysical, field lines start at one charge and end at opposite charge. Then how can fields be discontinous midway?


4 Answers 4


Yes, the electric field can be discontinuous, and in fact the discontinuity is in a region of space where there are charges. You can see this already from Coulomb's law $E(r) = kq/r^2$, where the function $E(r)$ is discontinuous at $r=0$. Another remarkable example is an infinitely large charged plane with surface charge density $\sigma$. If the plane has equation $z=0$, then the electric field points in a certain direction in the region $z>0$, and in the opposite direction in the region $z<0$, while the intensity is uniformly $\frac{\sigma}{2\varepsilon_0}$, so the $z$ component of the electric field has a discontinuity jump of $\frac{\sigma}{\varepsilon_0}$.

As @Andrew Steane pointed out in the comment, there are subtleties: technically purely 2 dimensional charge distributions don't really exist, in general the charges are distributed in a 3 dimensional arrangement with one characteristic length much smaller than the other two. In this case, a mathematical discontinuity is not present, but the field component changes continuously, yet abrouptly. The answer above was just meant to show how in standard electromagnetism it is not so strange to have situations where a discontinuous field is a very good approximation to solve standard problems.

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    $\begingroup$ For a question like this you need to keep in mind the degree of approximation. The point charge solution cannot be valid all the way to $r=0$ because it involves infinite fields and energies and the theory is then outside its regime of validity. For physically possible cases it is debatable whether electric field is ever strictly discontinuous. For a charged plane the charge is not infinitely thin. Etc. $\endgroup$ Commented Jul 23, 2023 at 10:55
  • $\begingroup$ Yeah, I totally agree. I will modify my answer adding this consideration $\endgroup$
    – Matteo
    Commented Jul 23, 2023 at 11:05
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    $\begingroup$ Technically, a point where a function is not defined is not a point where it is discontinuous. Even using a wider definition, based on the behavior of the limit at an accumulation point, $1/r^n$ is not discontinuous. $\endgroup$ Commented Jul 23, 2023 at 14:31

I think it is useful to complement @Matteo's answer (for the part related to the surface charge density) with a discussion of the physical conditions justifying the existence of a discontinuous electric field.

It is essential to distinguish between microscopic and macroscopic fields. The microscopic fields are controlled by all the physical point-like sources in the system and may vary quite a lot with time and over microscopic distances. Still, there are always continuous (of course, they are not defined at each point where a charge is located, but that is not a discontinuity, according to the mathematical definitions).

Macroscopic fields are a different story. They can be obtained from the microscopic fields through time and spatial averages, over times and spatial scales large with respect to the atomic scale.

Such averaging process has profound mathematical consequences. Spatial and time variations generally become smoother but with an important exception. The interfaces between different homogeneous media are usually a few atomic layers wide. Suppose there is some pile-up of charge in such interfacial region at the macroscopic level. In that case, it has to be described as a charge distribution confined to the separation surface between the two media.

It is a consequence of Gauss law that such two-dimensional charge density introduces a real discontinuity in the normal values of the electric displacement field at the interface. I.e., introducing the field on the two sides of the surface (${\bf D}_1$ and ${\bf D}_2$) $$ ({\bf D}_1 - {\bf D}_2)\cdot {\bf \hat n} = \sigma. $$

Notice that a similar discontinuity appears in the tangential component of the magnetic field ${\bf H}$ at a surface with a confined surface current density.

In both cases, the origin of the macroscopic discontinuities can be traced back to the need to describe sources at the interfaces in terms of surface densities.

In terms of field lines, the discontinuities do not introduce any inconsistency. Field lines start or end at a surface density, not only at point-like charges. However, in correspondence with the surface charge, the field has a finite value (does not diverge), but it is different on the two sides of the surface.


Since there seems to be no mention of dielectrics, let me add something. A capacitor figure is taken from an article by Jeremy Tatum. Jeremy Tatum figure

As you can see from the figure, for a capacitor consisting of two types of dielectrics, the electric flux density $D$ is continuous at the dielectric boundary, but the electric field $E$ is discontinuous. Since the electric charge is zero at the dielectric boundary, $\text{div}\vec{D}=0$ holds, and so on.

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    $\begingroup$ But it may still be continuous at the microscopic level, right? $\endgroup$ Commented Jul 24, 2023 at 11:56
  • $\begingroup$ @user253751, I think you are probably right. However, I think it is still insufficient to explain how the macroscopic laws dealing with many-particle systems are connected from the microscopic fundamental laws for few-particle systems. I would like to know how to derive $D$ from $E$ alone, how to derive $H$ from $B$ alone, how to derive $\sigma$(stress) from $\epsilon$(strain) alone. $\endgroup$
    – HEMMI
    Commented Jul 25, 2023 at 0:25
  • $\begingroup$ the question is not about connecting microscopic to macroscopic; it's about discontinuity $\endgroup$ Commented Jul 25, 2023 at 5:42

Another point to consider:

imagine a point charge sitting still. Electrostatic field lines diverge outward from it in all directions, to terminate at infinity.

Now we suddenly apply a large force to it and it begins to accelerate. Once it is set in motion, we cut off that force and it coasts. The field lines have to follow the accelerating charge while at the same time remaining connected to the field lines that existed before it began to move. They do this by getting kinked. a gentle acceleration produces a smoothly curved kink. a strong acceleration will produce a sharper kink. Make the acceleration infinite and the field lines will kink with zero radius and the E-field becomes discontinuous (no derivative at the kink).

You can think of the kinked portions of the electric field lines as mapping a magnetic field which does not diverge but runs in a circle where the kink is, around the charge in a plane perpendicular to the particle's direction of motion.

the kink self-propels itself outwards along the field lines and zooms off to infinity as an electromagnetic wave with an electric and a magnetic field component.


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