In figure, a charge q is placed at a distance `delta rarr 0` near one of the corners of a cube of edge I on a line of symmetry along diagonal.

1. What is the flux through each of the faces containing the point a?
2. What is the flux through the other three faces?

Explanation

Use of symmetry consideration may be useful in problems of flux calculation.

`*` We can imagine a charged particle placed at the center of a cube of side `2l`.
`*` The flux enclosed with the cube is `phi = q//epsilon_0`
`*` The flux passing through on of the faces of the cube `phi' = phi//6 = q//6epsilon_0`
`*` Hence, the flux passing through the face bcgf is `phi'//4 = q//24epsilon_0`
`*` Each of the faces `(efgh), (bcgh)`, and `(dcgh)` is symmetrical with respect to charge. Hence, the flux passing through each of the face is `q//24epsilon_0`
The electric field lines for the faces `(efgh),(bcgf)` and `(dcgh)` are away from the faces. Hence, the flux associated with each of the faces will be positive `(i.e., +q//24epsilon_0)`. Hence, the total flux through these sides is
`(3 xxq)/(24epsilon_0) = q/(8epsilon_0)`
`*` As `delta rarr 0`, we can say the faces `(abcd),(avfe)` and `(adhe)` are also symmetrical about charge. Charge is slightly outside the cube.
`*` The number of electric field lines passing through the faces, which do not contain the point a, is same as the number of electric field lines passing through the faces containing the point a.
`*` Hence, the same amount of flux will pass through the faces containing the point a, i.e., `q//8epsilon_0`.
`*` The electric field lines are toward the faces containing the point a. Hence, the flux will be negative, i.e., `phi" = -q/8epsilon_0`.
Hence, the flux through each of the faces containing the point 'a' will be `phi"//3 = q//24epsilon_0`.