If a point charge q is placed at one corner of a cube, what is the flux linked with the cube?

Explanation

By the same procedure you can see that total eight cubes are required to enclose the Q completely and symmetrically. So, the flux through one of the cubes (that is the thing asked in question) is `Q//8epsilon_(0)` (because total electric flux from Q as per Gauss.s law `Q//epsilon_(0)`)

Extension of the problem : When the charge is put at one of the corners of a cube, there are three faces which touch the charge and other three faces do not touch the charge. Now, if we ask you what is the flux through one of the faces not touching the charge Q? The correct answer is `Q//24epsilon_(0)`. Because the total flux through the cube is `Q//8epsilon_(0)` and it is being shared only by those faces not touching the charge (because the faces which are touching that charge have theri surface area vector perpendicular to the electric field emanating form the charge Q and hence flux through them is zero because `cos90^(@)=0`) equally, hence we divided `Q//8epsilon_(0)` further by 3.