I was solving a problem in which i came across a radially symmetric discontinuous electric field, which puzzled me for a moment but then I figured that there must be surface charge density at the point of discontinuity causing this.

My question is :

Is such discontinuity in 3D always the result of a singularity like a surface charge distribution, line charge or point charge? By singularity I mean infinite volume charge density. If so, why?


2 Answers 2


Yes, a discontinuity in the electric field is always associated with a charge distribution with infinite density. That's a necessary implication of Gauss' law. To think of it qualitatively, where there is a discontinuity in the field, there must be field lines either starting or ending. (If it's not obvious why that is, think about the fact that the electric field magnitude is different on one side of the discontinuity than on the other. There will be more field lines on the side with the stronger side, or else lines pointing in different directions on the different sides if the amplitude is positive on one side and negative on the other.) But electric field lines start and end only on charges, or at infinity.

To think about it even more physically, consider the force on a positive test charge. It will be different on either side of the discontinuity. But electric forces are exerted by charges, so the only way to produce a jump in the force across the discontinuity is to have some charge there.

In fact, it turns out that if the electric field amplitude jumps by $\Delta E$ across a discontinuity, there must be a surface charge there (charge per unit area) of magnitude $\epsilon_\circ \Delta E$.


To include the mathematics, this can be seen directly from the differential form of Gauss's law:

$$\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}$$

If divergence isn't intuitive to you, you can at least consider that it is something like a derivative. So this law is saying the "derivative" of the field is proportional to a finite quantity (charge), and thus the field is continuous. Only a singularity in that quanatity can produce an "infinite" derivative of the field, and thus a discontinuity in the field itself.


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