Using Gauss's law when point charges lie exactly on the Gaussian surface

Question

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?

Answer 1

Gauss's law applies to situations where there is charge contained within a surface (or entirely outside of it, of course), 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 $\Omega_0 = \frac{\pi}{2}$ (because it's one eighth of a full sphere, which has a solid angle of $4\pi$ steradians) . 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.

Answer 2

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.

Answer 3

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

Answer 4

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.

Answer 5

Gauss's Law does not apply if there is a point charge on the Gaussian surface. That is the case even if we try to adopt a convention to treat the charge as being 'inside' or 'outside' the surface - either way the law will not apply.

To see this we can calculate the total flux $\Phi$ such a charge produces through a sphere $S$. This flux is not directly well-defined since the electric field at the charge itself is infinite, however $\Phi$ does have a well-defined value if we calculate it as a limit. For any $\delta \in (0, \pi/2)$ we can calculate the total flux $\Phi_{\delta}$ through an approximation $S_{\delta}$ to $S$ which is defined by the spherical coordinates (using the Physics convention [SPH]) : $\theta \in [\delta, \pi - \delta]$ and $\varphi \in [0, 2\pi]$, since the electric field does not become infinite at any point of $S_{\delta}$. Then we can define $\Phi = \lim_{\delta \rightarrow 0^+} \Phi_{\delta}$.

There is a second reason for using this limiting process namely to be able to apply the spherical formula for surface integration [MSE] which requires the surface not to intersect the $z$-axis so that a division by $0$ doesn't occur in that formula when $\sin \theta = 0$.

Let $S$ have radius $R$ and set up $xyz$-axes so that the point charge $q$ lies on the positive $z$-axis at a distance $R$ from $O$. We will calculate the total flux $\Phi$ through $S_{\delta}$ due to $q$.

Let $\underline{\mathbf{\hat{r}}} = \sin \theta \cos \varphi \: \underline{\mathbf{i}} + \sin \theta \sin \varphi \: \underline{\mathbf{j}} + \cos \theta \: \underline{\mathbf{k}}$ be the radial unit vector, so the position vector $\overrightarrow{QP}$ of the general point $P$ on $S_{\delta}$ wrt $q$ is given by $\overrightarrow{QP} = -R \, \underline{\mathbf{k}} + R \, \underline{\mathbf{\hat{r}}}$. Then at $P$ we have : $$ \underline{\mathbf{E}} = \frac{q}{4 \pi \epsilon_0} \cdot \frac{-R \, \underline{\mathbf{k}} + R \, \underline{\mathbf{\hat{r}}}}{| {-R} \, \underline{\mathbf{k}} + R \, \underline{\mathbf{\hat{r}}} \: |^3} $$ where $$ |\underline{\mathbf{\hat{r}}} - \underline{\mathbf{k}}|^2 = (1 - \cos \theta)^2 + \sin^2 \theta = 2 - 2 \cos \theta. $$ Since a unit outward normal to $S_{\delta}$ at $P$ is $\underline{\mathbf{n}} = \underline{\mathbf{\hat{r}}}$ and since $\underline{\mathbf{k}} \cdot \underline{\mathbf{\hat{r}}} = \cos \theta$ we have : $$ \underline{\mathbf{E}} \cdot \underline{\mathbf{n}} = \frac{q}{4 \pi \epsilon_0 R^2} \cdot \frac{1}{2 \sqrt{2}} \cdot \frac{1}{(1 - \cos \theta)^{1/2}} $$ Using the formula for surface integration in spherical coordinates described in [MSE], with $r(\theta, \varphi) = R, \; \forall \; (\theta, \varphi) \in D_{\delta} = [\delta, \pi - \delta] \times [0, 2 \pi]$ and $\frac{\partial r}{\partial \theta} = \frac{\partial r}{\partial \varphi} = 0$ throughout $S_{\delta}$ we have : \begin{eqnarray*} \Phi_{\delta} = \int_{S_{\delta}} \underline{\mathbf{E}} \cdot \underline{\mathbf{n}} & = & \int_{D_{\delta}} \; \frac{q}{8 \pi \epsilon_0 R^2 \sqrt{2}} \cdot \frac{1}{(1 - \cos \theta)^{1/2}} \cdot R^2 \sin \theta \\ & = & \int_{\delta}^{\pi - \delta} \int_0^{2 \pi} \frac{q}{8 \pi \epsilon_0 \sqrt{2}} \cdot \frac{\sin \theta}{(1 - \cos \theta)^{1/2}} \; d\varphi \; d\theta \\ & = & \int_{\delta}^{\pi - \delta} \frac{q}{4 \sqrt{2} \epsilon_0} \cdot \frac{\sin \theta}{(1 - \cos \theta)^{1/2}} \; d\theta = \frac{q}{4 \sqrt{2} \epsilon_0} \cdot \left[ 2(1 - \cos \theta)^{1/2} \right]_{\delta}^{\pi - \delta} \\ \mbox{and hence} \hspace{2em} \Phi & = & \lim_{\delta \rightarrow 0^+} \Phi_{\delta} = \frac{q}{2 \sqrt{2} \epsilon_0} ( \sqrt{2} - 0 ) = \frac{q}{2 \epsilon_0} \end{eqnarray*}

We can calculate the above integral for $q$ at any distance $R_2 \geq 0$ from $O$ : \begin{eqnarray*} \underline{\mathbf{E}} & = & \frac{q}{4 \pi \epsilon_0} \cdot \frac{-R_2 \, \underline{\mathbf{k}} + R \, \underline{\mathbf{\hat{r}}}}{| {-R_2} \, \underline{\mathbf{k}} + R \, \underline{\mathbf{\hat{r}}} \: |^3} = \frac{q}{4 \pi \epsilon_0 R^2} \cdot \frac{-\frac{R_2}{R}\underline{\mathbf{k}} + \underline{\mathbf{\hat{r}}}}{|-\frac{R_2}{R}\underline{\mathbf{k}} + \underline{\mathbf{\hat{r}}}|^3} \\ \mbox{where} \hspace{2em} \left|-\frac{R_2}{R}\underline{\mathbf{k}} + \underline{\mathbf{\hat{r}}}\right|^2 & = & \left(\frac{R_2}{R} - \cos \theta\right)^2 + \sin^2 \theta = 1 + \frac{R_2^2}{R^2} - 2\frac{R_2}{R} \cos \theta. \end{eqnarray*} Then we obtain : \begin{eqnarray*} \Phi_{\delta} = \int_{S_{\delta}} \underline{\mathbf{E}} \cdot \underline{\mathbf{n}} & = & \frac{q}{4 \pi \epsilon_0} \int_{D_{\delta}} \frac{-(R_2/R)\sin \theta \cos \theta + \sin \theta}{\left( 1 + (R_2/R)^2 -2 (R_2/R) \cos \theta \right)^{3/2}} \\ & = & \frac{q}{2 \epsilon_0} \cdot \left[ \left(\cos \theta - \frac{R}{R_2}\right) \left(1 + \frac{R_2^2}{R^2} -2\frac{R_2}{R}\cos \theta\right)^{-1/2} + \frac{R}{R_2}\left(1 + \frac{R_2^2}{R^2} -2\frac{R_2}{R} \cos \theta \right)^{1/2} \right]_{\delta}^{\pi - \delta}, \\ & & \mbox{using integration by parts for the first part of the integral.} \end{eqnarray*} Then we obtain : $$ \Phi = \lim_{\delta \rightarrow 0^+} \Phi_{\delta} = \left\{ \begin{array}{ll} q/\epsilon_0, & \mbox{if } 0 \leq R_2 < R \\[1ex] q/2\epsilon_0, & \mbox{if } R_2 = R \\[1ex] 0, & \mbox{if } R_2 > R \end{array} \right. $$ in keeping with Gauss's Law in the case $R_2 \neq R$. The electric flux $\Phi$ is a discontinuous function of the distance $R_2$ of $q$ from the centre of the sphere. If we had multiple point charges on the surface of $S$, and/or a continuous surface charge distribution $\sigma$ across $S$ then because of additivity of the electric field the above formula for $\Phi$ would still apply but with $q$ replaced by $Q$, the total surface charge. If $S$ was exposed only to a volume charge distribution $\rho$ then the total surface charge would be $Q = 0$.

A similar process of taking a limit can be used to calculate the average electric potential $V_{\mathrm{avg}}$ over a sphere $S$ (eg see [DJG, p117-118]) due to a point charge $q$ on its surface. The potential will be infinite at the charge itself so $V_{\mathrm{avg}}$ is not well-defined, but as we did with the flux $\Phi$ above we can use a limit to give it a well-defined value : \begin{eqnarray*} V_{\mathrm{avg}} & = & \lim_{\delta \rightarrow 0^+} \, \frac{1}{A_{\delta}} \, \int_{S_{\delta}} V \; dS, \hspace{2em} \mbox{where $A_{\delta}$ is the area of $S_{\delta}$} \\ & = & \lim_{\delta \rightarrow 0^+} \, \frac{1}{A_{\delta}} \, \int_{\delta}^{\pi - \delta} \int_0^{2 \pi} \; \frac{q}{4 \pi \epsilon_0} \cdot \frac{R \sin \theta}{\left( 2 - 2 \cos \theta \right)^{1/2}} \; d\varphi \; d\theta = \frac{q}{4 \pi \epsilon_0 R} \end{eqnarray*} and for $q$ not on $S$ at distance $R_2 \geq 0$ from $O$ : \begin{eqnarray*} V_{\mathrm{avg}} & = & \lim_{\delta \rightarrow 0^+} \frac{1}{A_{\delta}} \int_{\delta}^{\pi - \delta} \int_0^{2 \pi} \; \frac{q}{4 \pi \epsilon_0} \cdot \frac{R^2 \sin \theta}{\left( R^2 + R_2^2 - 2R R_2 \cos \theta \right)^{1/2}} \; d\varphi \; d\theta = \frac{q}{8 \pi \epsilon_0 R R_2} \left[ (R + R_2) - |R - R_2| \right], \end{eqnarray*} where in the latter integral the only reason we have used the limiting process is so the spherical surface integral formula [MSE] is valid, ie. to avoid a division by $0$ in that formula. Then we have : $$ V_{\mathrm{avg}} = \left\{ \begin{array}{ll} \displaystyle \frac{q}{4 \pi \epsilon_0 R}, & \mbox{if $0 \leq R_2 < R$, (ie. as if $q$ at centre) } \\[3ex] \displaystyle \frac{q}{4 \pi \epsilon_0 R}, & \mbox{if $R_2 = R$, (ie. as if $q$ at centre)} \\[3ex] \displaystyle \frac{q}{4 \pi \epsilon_0 R_2}, & \mbox{if $R_2 > R$, (ie. potential at centre due to $q$)} \end{array} \right. $$ In contrast to the flux $\Phi$, $V_{\mathrm{avg}}$ is a continuous function of $R_2$.

Another situation where the limiting process applies is in calculating the potential at a point $P$ due to a uniformly charged spherical shell of total charge $Q$ where by performing a similar integration as those above we obtain : $$ V = \left\{ \begin{array}{ll} \displaystyle \frac{Q}{4 \pi \epsilon_0 R}, & \mbox{if } r < R \\[3ex] \displaystyle \frac{Q}{4 \pi \epsilon_0 R}, & \mbox{if } r = R \\[3ex] \displaystyle \frac{Q}{4 \pi \epsilon_0 r}, & \mbox{if } r > R \end{array} \right. $$

where $R$ is the radius of the shell and $P$ is at distance $r$ from its centre. In each of the 3 cases the limit process avoids the division by $0$ in the spherical surface integral formula [MSE]. On the shell's surface it secondly avoids the problem of an infinite potential due to a non-zero charge density at $P$, which also produces a division by $0$. From Gauss's Law and the calculation above, the total flux $\Phi$ of this charge distribution through a Gaussian surface of radius $r$ and concentric with the shell is given by : $$ \Phi = \left\{ \begin{array}{ll} \displaystyle 0, & \mbox{if } r < R \\[1ex] \displaystyle Q/2\epsilon_0, & \mbox{if } r = R \\[1ex] \displaystyle Q/\epsilon_0, & \mbox{if } r > R \end{array} \right. $$ For the case $r = R$ the flux can't be calculated from Gauss's Law, but it can be calculated from the Generalized Gauss Theorem discussed in the answer [PSE] to the present question. Because of the symmetry, the electric field of the uniformly charged spherical shell must be radial and a function of $r$ only. From the latter formula its magnitude $E$ is given by : $$ E = \left\{ \begin{array}{ll} 0, & \mbox{if } r < R \\[2ex] \displaystyle \frac{Q}{8 \pi \epsilon_0 R^2}, & \mbox{if } r = R \\[2ex] \displaystyle \frac{Q}{4 \pi \epsilon_0 r^2}, & \mbox{if } r > R \end{array} \right. $$ However the value of $E$ for $r = R$ is purely a value calculated within the mathematical model which arises from the physical model of electrostatics and this physical model reaches the limits of its applicability when considering electric field right on a surface of zero thickness - it would be unlikely to be possible to measure such a field value in an experiment or to detect it in any way.

In practice the limiting process used above is not normally written out, and the limits $0$ and $\pi$ are just applied to the integration and the correct answer is arrived at. A square root term in the denominator which goes to zero becomes a square root term in the numerator once integrated and then going to zero is not a problem. But a check should be made mentally that a problem is not arising due to a division by $0$.

References

[SPH] Spherical coordinate system, https://en.wikipedia.org/wiki/Spherical_coordinate_system

[MSE] Maths StackExchange Answer : Surface integrals in spherical coordinates, https://math.stackexchange.com/a/4867164/417024

[DJG] David J. Griffiths (2013), Introduction To Electrodynamics, 4th Edition, Pearson Education Inc.

[PSE] Physics StackExchange Answer : Using Gauss's law when point charges lie exactly on the Gaussian surface, https://physics.stackexchange.com/a/544481/111652