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?

  • 5
    $\begingroup$ Please call it a point charge. "Having no dimensions" sounds like something else. $\endgroup$
    – J.G.
    Apr 17, 2020 at 19:13
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    $\begingroup$ I would take your point charge as the limit of a spherical charged insulator as the radius goes to zero. Before the radius gets to zero, exactly $\frac 18$ of the charge is inside the cube. There is no contradiction at all. $\endgroup$ Apr 17, 2020 at 20:33
  • 1
    $\begingroup$ @RossMilikan But what if I want to model the point charge as the limit of an octopus shaped charge? Multipole expansion gives me the full right to do so. $\endgroup$
    – ACat
    May 13, 2020 at 23:51

4 Answers 4


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.

  • 9
    $\begingroup$ The Generalised Gauss' Theorem should be published in the Exercise section of Physics textbooks (without solution, of course) rather than on proceedings, for that's what it really is. $\endgroup$
    – Phoenix87
    Apr 17, 2020 at 8:29
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    $\begingroup$ To be clear Gauss' Law does NOT require the Q to be enclosed in S. If it is outside the Flux = 0 as expected. The interesting behavior here is in taking the limit as Q approaches S from outside/inside and seeing the difference. $\endgroup$
    – user196418
    Apr 17, 2020 at 19:44
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    $\begingroup$ How is the argument given in the paper any better than the intuitive expectation? I mean it's just handwaving, right? Shouldn't the correct answer simply be that the flux diverges when you put a finite charge on the Gaussian surface? It seems like the author (and whoever writes these silly questions for students) forgets the lesson of physics that you can't always calculate what you want but if you do renormalize, you might what you need. ;) $\endgroup$
    – ACat
    May 13, 2020 at 23:36
  • $\begingroup$ Very interesting, could you please point out what happens if we have an non isotropic medium? $\endgroup$
    – carloc
    May 14, 2020 at 6:03
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    $\begingroup$ @AdilMohammed It's the same as the integral in the regular version of Gauss's law, i.e. it represents the total flux through the surface. $\endgroup$
    – David Z
    May 22, 2021 at 18:44

How do you define a point charge? Let us add some formality to that: Consider a spherical charge of radius $r$ centered at the cube's vertex, with uniform charge density $\rho_v=\frac{3Q}{4\pi r^2}$ (so that the total charge is $Q$ and the electric field is the same as a point charge's at a distance $d > r$). We can define our point charge as the limit of this spherical charge as $r\rightarrow 0$. The amount of charge enclosed by the cube for any $r>0$ is, by the same symmetry argument you used, $Q/8$. so the charge "enclosed" for all purposes of Gauss' law is: $$Q_{enc} = \lim_{r\rightarrow 0}{\frac{Q}{8}} = \frac{Q}{8}$$ Now, why should we use a sphere for the limit and not another shape that could give a different result? That's because only a uniform sphere can replicate the electric field of a unit charge over all space outside its body, no matter the size (radius), and thus converge to a unit charge on the limit for all electrical purposes.

  • $\begingroup$ "Now, why should we use a sphere for the limit and not another shape that could give a different result? That's because only a uniform sphere can replicate the electric field of a unit charge over all space" This is not true. An octopus shaped charge also produces the exact same electric field in the limit when the size of the octopus goes to zero. That's what we learn from the multipole expansion. $\endgroup$
    – ACat
    May 13, 2020 at 23:45
  • $\begingroup$ @DvijD.C. By octopus shaped charge do you mean an octupole? the idea behind my answer is to define a point charge in terms of a non-singular charge distribution. A definition of a point charge in terms of an octupole which in turn is defined in terms of point charges is inconsistent. If you do not refer to an octupole then I would like, if possible, that you explain to me that distribution with more detail please. $\endgroup$ May 14, 2020 at 2:41
  • $\begingroup$ Haha, no, I actually meant an octopus 🐙. I just meant that I can model the point charge as the limit of any arbitrarily shaped object, there is no reason to say that only a sphere would work. My octopus is supposed to be completely non-singular (just like your sphere). I now take the limit where the size of the octopus goes to zero while keeping the total charge constant, thus getting a singular point charge in the limit. As I take this limit, only the $Q/r$ term would survive in the multipole expansion of its potential and it would produce the same electric field as that of a point charge. $\endgroup$
    – ACat
    May 14, 2020 at 2:50
  • $\begingroup$ @DvijD.C. But then that would mean your octopus charge does not produce the exactly same field of a point charge for positive sizes, no matter how small. I chose a sphere because it produces the same field even for positive sizes. The idea is that in a real-world experiment, a small spherical charge would be a "fair enough" point charge, and in such experiment OP's conclusion would hold. I will edit my answer to clarify this. $\endgroup$ May 14, 2020 at 5:40
  • $\begingroup$ The electric field of the finite sized object is not the relevant thing. We have no obligation to have the same kind of electric field as that of a point charge before we take the limit. The only requirement is that it should produce the electric field of the point charge in the limit. That's the whole point. And a small octopus shaped charge would also be a "fair enough" point charge. Again, that's the point of multipole expansion. $\endgroup$
    – ACat
    May 14, 2020 at 10:03

Gauss’ law says the net flux through a closed surface equals the net charge enclosed by the surface divided by the electrical permitivity of the space.

I don’t see how a point charge at the corner of a cube can be considered as enclosed by the surfaces of the cube. Gauss’ law applies to a closed surface. The cube minus three surfaces does not constitute a closed surface. Moreover, as @ZeroTheHero points out, it does not make sense to divide a charge having no dimensions into an eighth.

In summary, I see no Gauss’ law paradox.

Hope this helps

  • 3
    $\begingroup$ The OP is still proposing a closed surface. They are just reasoning that flux through some parts are $0$. The OP's issue is that the net flux is some fraction of $Q$, so by Gauss's law that much charge is inside the closed surface. $\endgroup$ Apr 16, 2020 at 23:09
  • $\begingroup$ @AaronStevens I think DavidZ's answer compliments this one. The original Gauss's Law cannot be applied (as this answer says) and it requires the generalized theorem as described in the other answer. $\endgroup$ Apr 16, 2020 at 23:23
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    $\begingroup$ @StephenG Certainly. I'm just saying the OP is considering a closed surface and explaining why they are confused. $\endgroup$ Apr 17, 2020 at 0:45
  • $\begingroup$ @StephenG True. But the original Gauss' law can be applied if you put 8 of such cubes together and calculate the net flux through the bigger cube. $\endgroup$
    – Vivek
    Apr 20, 2020 at 1:38
  • $\begingroup$ @Vivek yes but the eight cubes is a different surface. The charge is ON one of the surfaces of the original cube. As such how can the original surface “enclose” the charge? That’s the point David Z and I are trying to make. $\endgroup$
    – Bob D
    Apr 20, 2020 at 1:59

Gauss' law requires charges enclosed to be completely enclosed within the volume that you're considering (i.e. the charge must be contained in an open subset in the usual topology of $\mathbb{R}^3$ that is totally inside the compact domain under consideration). If not, you can argue using symmetry considerations. For example if your charge is at the boundary of a smooth surface, it would yield half a solid angle contribution because it's half inside/half outside etc. For a proper argument, imagine a charge at the boundary of a smooth volume, then reflect the volume about the tangent plane and consider the limit of a surface that encloses the union of both these volumes from the outside...by symmetry we would have $Q/\epsilon_0$ flux for both and $Q/2\epsilon_0$ through each.

The generalized divergence theorem in the answer of @DavidZ seems to have generalized this. I wasn't aware of the generalized result, but now it's easy to imagine it would hold for $4\pi/N$ solid angle, if $N$ is an integer - just cover the $4\pi$ solid angle with $N$ of these volumes....from this one can extend to rational fractions of $4\pi$...and then by continuity to all arbitrary solid angles.

In your case, you can imagine enclosing the central charge inside 8 symmetrical cubes joined at the vertex where the point charge resides....then you'd get $Q/\epsilon_0$ flux through all of them and $Q/8\epsilon_0$ through each of them by symmetry.


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