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A charge $q$ sits at the back corner of a cube, as shown in Figure. What is the flux of $E$ through the shaded side?

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One of the solution stated that.

Looking at the figure, we notice two things: (1) only one eighth of the charge is actually contained in the box; and (2) there is no flux through the three sides touching the charge. Where it is situated, the charge only generates flux through the three opposite sides, a third of which goes through the shaded side.

My question is how the three sides don't contribute any flux whereas we can see that a small fraction amount of flux can pass through the three sides. Please explain a bit more.

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1) One eighth of the charge is contained in the cube because you can place eight cubes such that they have a common vertex, the one your charge is at. If your charge was on the middle of an edge, it would be shared by four cubes. If your charge was in the middle of one of the faces, it would be shared by two cubes and so on.

2) The three adjacent faces have no flux through them because flux lines are radial. Every line inside the cube from the vertex of the charge can only intersect one of the three opposite faces. Draw straight lines from the vertex and you'll see that they only pass through the three opposite faces (one of which is your shaded face).

3) The three opposite faces get an equal amount of flux (by symmetry). Hence your flux through the shaded face is $1/8\times 1/3 = 1/24$ of the total flux. The total flux is related to the magnitude of the charge through Gauss' law.

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My question is how the three sides don't contribute any flux whereas we can see that a small fraction amount of flux can pass through the three sides.

Not really. You see, the electrostatic field $\vec{E}$ of the charge is always radially outwards. If the charge is situated at the exact corner of the cube, then the field is exactly coplanar with the three faces passing through that corner. What I mean, is that the field doesn't cross any of these 3 faces. And that means, the flux of the electric field through these 3 surfaces is 0. (Because, $\vec{E}$ and $\vec{dA}$ are perpendicular and so $\vec{E}\dot{}\vec{dA} = 0$)

Now, consider a bigger cube, with each edge twice as long as the given cube's, with its center at the charge itself. It is not difficult to realise that this cube can be decomposed into 8 cubes, of smaller size, each with the charge at one corner. Also, all of these cubes are perfectly symmetric, in the sense that the flux through each of them is the same. On the other hand, the flux through the whole of the bigger cube is $\frac{q}{\epsilon_0}$, as provided by Gauss's Law. Hence, we can conclude that the flux through the given cube is exactly $\frac{1}{8}$-th of the total flux of the field through the bigger cube, id est, $\frac{q}{8\epsilon_0}$. Because flux through the adjacent 3 faces are 0, the flux through each of the remaining has to be $\frac{1}{3}$-rd of this flux, because these sides are also perfectly symmetric.

The answer to the question comes out to be $\frac{q}{24\epsilon_0}$.

If the charge were not exactly at the corner, the situation would have differed. If it were slightly outside the cube but nearly at the corner, the flux through the distant 3 faces would remain unchanged ($\frac{q}{8\epsilon_0}$) whereas the flux through the 3 close-by faces would be the negative of that because the net flux through the cube has to be 0. (There is no charge in it now.)

If the charge were just inside the cube but nearly at the corner, then too the flux through the 3 distant faces would remain unchanged, but now the net flux through the cube has to be $\frac{q}{\epsilon_0}$ so the flux through the three adjacent sides would be $\frac{7q}{8\epsilon_0}$.

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Hint: You can always construct imaginary cubes (7 more, in this case.) such that your charge is completely enclosed inside your multi-cube system. You can find the flux through this system via Gauss' Law. The total flux through each side (of the new configuration) would be equal via symmetry.

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Suppose you have a charge in a plane. The electric field from the charge is spherically symmetric, and that means every field line from the charge that intersects the plane has an equal and opposite field line intersecting the plane. So when we integrate $\mathbf{E}\cdot\text{d}\mathbf{A}$ the two field lines will cancel out and the net flux will be zero.

Flux through plane

Incidentally, this is a classic question and is best treated by imagining the charge at the centre of eight cubes. Then by symmetry the flux through any single cube is one eight of the total.

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Well for the three sides touching the particle, the vector $\bf E$ coming from the particle, is always contained in the surface. This means that there is no component in the direction normal to any of the surfaces. So this means $\bf E$$d \bf S$ and therefore the flux through them, $\int \bf{E}$ $d \bf S$ will result zero.

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