Question

show with the help of a diagram that outward flux due to point charge q in vacuum within a gaussian surface is independant of shape and size

Answer 1

Since, the result is independent of radius of sphere. Thus, electric flux through gaussian surfaces I, II and III will be the same. Think of electric flux as number of electric field lines entering (in the case of negative charge) and exiting (in the case of positive charge) a charge. Thus, electric flux through any Gaussian surface (closed 3D surface) is equal to charge enclosed in that Gaussian surface divided by permittivity of the free space.

Hope, this helps...