A charge q sits at the back corner of a cube as shown what is the flux of e through the shaded side
Answers
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.
Answer:
q/24ϵ0
Explanation:
Not really. You see, the electrostatic field E⃗ 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, E⃗ and dA→ are perpendicular and so E⃗ ˙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 qϵ0, as provided by Gauss's Law. Hence, we can conclude that the flux through the given cube is exactly 18-th of the total flux of the field through the bigger cube, id est, q8ϵ0. Because flux through the adjacent 3 faces are 0, the flux through each of the remaining has to be 13-rd of this flux, because these sides are also perfectly symmetric.
The answer to the question comes out to be q24ϵ0