# Flux through side of a cube

I am looking at Griffiths introduction to Electrodynamics 3rd ED. Problem 2.10 asks for the flux of $E$ through the right face of the cube, when a charge $q$ is in the back left corner of the cube. The solution manual says it should be $\frac q {24\epsilon _0}$ but I think it should be $\frac q {6\epsilon _0}$ because each side should experience the same flux, and the face of the cube represents 1/6th of the cube.

• Nov 7, 2017 at 21:50

Contrary to what queueoverflow says, you do not actually need to perform any integration here; a pretty cool symmetry argument will give you the answer.

Let the cube we are considering in the problem have side length $\ell$.

The trick is to consider putting the charge at the center of an imaginary cube of side length $2\ell$. The flux through the surface of this cube is just $q/\epsilon_0$ by Gauss's Law since it is a closed surface containing the charge. Now imagine dividing each face of this larger cube into four squares of side length $\ell$. By symmetry, the flux through each of these squares is the same, but there are a total of 24 such squares since there are six faces, so the flux through each of these squares is $q/(24\epsilon_0$).

Now, simply notice that if we were to cut the larger cube into eight cubes of side length $\ell$, then each of these squares mentioned in the last paragraph would be a face of one of these cubes at which the charge is at a corner. QED.

Regarding your confusion. Notice that a bunch of the flux coming from the charge does not even pass through the faces of the cube when the charge is at its corner. Only one eighth of its total flux goes through the faces of the cube as illustrated by the argument above. Then, each of the three faces opposite the charge get one third of this flux because the other three faces next to the charge are parallel to the electric field, so there is no flux through them.

• Oh, that is cool. $1/24 = 1/8 \cdot 1/3$. Sep 13, 2013 at 11:04

I had a similar argument which also failed, curiously:

In the cube shown, the three sides touching the charge contribute no flux because the electric field is parallel to the surface. The other 3 are symmetrically opposed to the charge so they contribute the same to the flux. Thus the total flux is 3 times the flux of the desired side, and you get

\begin{equation} \frac{1}{3} \frac{q}{\epsilon_0} \end{equation}

The issue with this argument is subtle: does the charge count as "enclosed" or not, since it is on the boundary? A closer look at the divergence theorem specifies that the vector field $\vec{E}$ would have to be differentiably continuous for for the theorem to work, but there's a discontinuity at $r=0$, so the use of Gauss' Law isn't even legitimate, mathematically speaking. I can only see the "$2l$" argument working because at least the discontinuity isn't on the boundary. I don't know any legitimate resolution to this and I would be curious to hear it.

In general the flux through an oriented open or closed surface $$\:\mathrm{S}\:$$ due to a point charge $$\:Q\:$$ is
$$\begin{equation} \Phi_{\mathrm{S}}=\dfrac{\Theta}{4\pi}\dfrac{Q}{\epsilon_{0}} \tag{01} \end{equation}$$ where $$\:\Theta\:$$ the solid angle by which the charge $$\:Q\:$$ sees the surface. In our case this solid angle is $$\:\pi/6\:$$, as proved in the Figures, so
$$\begin{equation} \Phi_{\mathrm{S}}=\dfrac{\pi/6}{4\pi}\dfrac{Q}{\epsilon_{0}}=\dfrac{1}{24}\dfrac{Q}{\epsilon_{0}} \tag{02} \end{equation}$$   