Final answer:
The electric field E above the center of a flat circular disk of radius r with uniform surface charge density (σ) is given by E = {σ} / {2e_0} (1 - z / z^2 + r^2}^1/2), and in the limit (r), it approaches the electric field above a point charge, while for z gg r, it tends towards the electric field above an infinite plane.
Step-by-step explanation:
The question is asking to find the electric field (E-field) a distance z above the center of a flat circular disk with radius r that carries a uniform surface charge density. To derive the formula for this scenario, one needs to integrate the contributions of all the infinitesimal charge elements on the disk to the electric field at point z.
When r goes to infinity the disk resembles an infinitely large sheet of charge and the field becomes uniform. In the limit where z is much greater than r (z >> r), the disk can be considered a point charge, and the electric field will drop off according to the inverse square law of distance.
For a disk of finite radius r, the electric field at a distance z above the center can be described using an integration of rings of charge which result in a more complex expression. It involves an integral that takes into account the varying distance of each infinitesimal charge element from the point z where the electric field is being calculated. This involves an integration over the area of the disk, taking advantage of the symmetry of the problem.
In the case that r approaches infinity, or the disk is considered a portion of an infinite plane of charge, the electric field is perpendicular to the plane and is given by E = σ / (2ε_0), where σ is the surface charge density and ε_0 is the vacuum permittivity.