At a superconductor is what I know of, but what are some other cases?

  • $\begingroup$ Do you mean like textbook discontinuities in EM field? $\endgroup$ Commented Oct 19, 2022 at 4:06
  • $\begingroup$ A superconductor doesn't cause a discontinuity in an em field. $\endgroup$ Commented Oct 19, 2022 at 4:12

1 Answer 1


The Maxwell equations,

\begin{align} \nabla \cdot \mathbf{E}& = \frac {\rho} {\varepsilon_0} & \nabla \cdot \mathbf{B}& = 0 \\ \nabla \times \mathbf{E}& = -\frac{\partial \mathbf{B}} {\partial t} & \nabla \times \mathbf{B}& = \mu_0\left(\mathbf{J} + \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t} \right) \end{align}

define the derivatives of the fields. A discontinuity in a function corresponds to a point where its rate of change is infinite. So you can have field discontinuities where the right-hand sides of Maxwell’s equations are infinite. The classic examples are when the charge density $\rho$ or the current density $\mathbf J$ represent a nonzero charge or current confined to zero volume, such as at a point charge, at a line charge or line current, or at a surface charge or a surface current. Your favorite textbook will discuss the conditions the fields must obey across such a boundary.


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