Electrostatics – Using Gauss’s Law When Point Charges Lie Exactly on the Gaussian Surface

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Suppose you place a point charge $+Q$ at the corner of an imaginary cube.

Since electric field lines are radial, there is no flux through the three adjacent (adjacent to the charge) sides of the cube. However there is some amount of flux passing through the other three sides of the cube (flowing out of the cube).

We can estimate that the flux through these three surfaces combined is equal to $Q/(8\epsilon)$. As, if you consider $7$ other cubes having the charge at the corner, each of them would have equal flux flowing out by symmetry and since the total flux through the $8$ cubes is $Q/\epsilon$, each cube would have a flux of $Q/(8\epsilon)$.

Now apply Gauss' law to the cube, and we find that the cube encloses a charge of $Q/8$.

This means that 1/8th of the charge belongs to this cube.

But the charge we placed was a point charge with no dimensions. It cannot be split into parts.

What is wrong?

Best Answer

Gauss's law applies to situations where there is charge contained within a surface, but it doesn't cover situations where there is a finite amount of charge actually on the surface - in other words, where the charge density has a singularity at a point that lies on the surface. For that, you need the "Generalized Gauss's Theorem" [PDF], which was published in 2011 in the conference proceedings of the Electrostatics Society of America. (I found out about this paper from Wikipedia.)

The Generalized Gauss's Theorem as published in that paper says that $$\iint_S \vec{E}\cdot\mathrm{d}\vec{A} = \frac{1}{\epsilon_0}\biggl(Q_{\text{enc}} + \frac{1}{2}Q_{\text{con}} + \sum_{i}\frac{\Omega_i}{4\pi}q_{i}\biggr)$$ where

  • $Q_{\text{enc}}$ is the amount of charge fully enclosed by the surface $S$ and not located on $S$
  • $Q_{\text{con}}$ is the amount of charge that lies on the surface $S$ at points where $S$ is smooth
  • $q_i$ for each $i$ represents a point charge that is located on $S$ at a point where $S$ is not smooth (i.e. on a corner), and $\Omega_i$ represents the amount of solid angle around that that point charge that is directed into the region enclosed by $S$.

There are a few edge cases (haha) not handled by this formulation (although it should be straightforward to tweak the argument in the paper to cover those), but fortunately it does cover the case you're asking about, where a point charge is located at a corner of a cube. In that case, the amount of solid angle around the corner that is directed into the interior of the cube is $\Omega_0 = \frac{\pi}{2}$. Plugging in that along with $q_0 = Q$ (the magnitude of the charge), you find that $$\iint_S \vec{E}\cdot\mathrm{d}\vec{A} = \frac{1}{\epsilon_0}\biggl(0 + \frac{1}{2}(0) + \frac{\pi/2}{4\pi}(Q)\biggr) = \frac{Q}{8\epsilon_0}$$ which agrees with what you've found intuitively.

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