Elektrisch veld van een geladen schijf

Consider a uniformly charged thin disk of radius R. The electric field at a point above its center at a perpendicular distance d is given by the surface integral over the disk.

Choose a ring of charge at a radius r, of width dr. Its area is the product of its circumference and width.

Let the electric field produced by any small charge element on the ring make an angle θ with the z-axis. Resolve this into two components: parallel and perpendicular to the z-axis.

There is a charge element whose electric field's perpendicular component is equal and opposite. So, the perpendicular components cancel. The disk's symmetry along its plane implies that it's the same for any such pair.

Since the z-components reinforce, the electric field points away from the disk. Integrating over r, the resultant field is obtained.

At large distances, the expression reduces to that of a point charge equal to the disk's total charge.

On the other hand, at small distances, the disk looks like an infinite plane whose electric field is constant.