Find flux through the hemispherical surface

Explanation

Flux means number of electric lines being shared by the surface. You can understand as if the charge is point source of light and by flux we have to calculate the light shared by the given surface. Total flux a point charge Q happens to have is `Q//epsilon_(0)`.

Sometimes you are so much lost in the Gauss.s law that you happen to think that unless the surface is a closed one, the flux through the open bodies is zero. Let us remind you that the flux is an independent concept. Gauss.s law talks about the maximum flux a charge can produce. But if the surface is not a closed one, it never means that the flux through it is zero.
Note that : If the `vec(E)` line from the charge cuts the surface once, there will be net contribution to the electric flux. In such type of question try to enclose the charge completely and symmetrically by as many bodies required as that of the given body. Here a hemisphere is given, so you known that if another hemisphere is placed below it will enclose the charge completely by a sphere. So, two hemispheres are enclosing the charge Q completely, which will share the total electric flux (`Q//epsilon_(0)` as per Gauss.s law) equally So, the flux through one hemisphere is Ql `2epsilon_(0)`.
Note that in all the following cases the flux through the body is `Q//2epsilon_(0)`
(a) A charge Q is placed at the centre of the base of a cone.

Then the flux through the cone is `Q//2epsilon_(0)` with the came logic.
(b) A charge Q is placed at the centre of one of the faces of a cube.

Flux through the cube will be `Q//2epsilon_(0)`.