Gaussian surface for a uniformly charge thin disc? Is it possible to find a Gaussian surface to apply Gauss's law for a uniformly charge thin disc?
Many methods use summation (integration) of fields on charged rings.
 A: It is not doable.
The reason is that you need to find a geometrical surface on which $\vec E\cdot d\vec S= \vert \vec E\vert dS\cos(\theta)$ so you can “pull out” the constant factor $\vert \vec E\vert$ outside the integral for the flux.
There is no such surface for a disk.  To use a cylinder you would need to have not a thin disk but an infinitely long one since the cylinder is invariant under translation about the symmetry axis passing through the center and parallel to the sides of the cylinder.
Moreover, in the limit where you are very far from the disk, the field is that of a point particle, and for that you’d use a sphere.  As a result you’d need a surface that changes its shape as you increase the distance from your disk: i.e. a sphere at large distance but something else at short distance.  Nobody has discovered such a useful shape to find the field of a disk by using Gauss’ law.
A: If you want something in context of the given problem, you can think of the circular disk with the given restrictions to be an infinitely large uniformly charged sheet and by using symmetry you can say that the field at 'z' is a uniform outgoing field and solve it by taking a cylinder with the top and bottom above and below the disk.
As for a gaussian surface for a disk in general, I don't think there is any specific shape for an equipotential surface around the disk. moreover it will be much easier to calculate the field using coulombs law.
Well if you want a shape you can always bactrack it using the field derived from coulumb's law
