The Electric Field Due to a Charged Disk — Question 1

Question 1

A disk field is a sum of ring contributions with different radii

A thin insulating disk of radius R has fixed uniform positive surface charge density $\sigma$. Find its field on the positive z axis perpendicular to its center, using Coulomb constant k.

1. Derive the field by integrating circular rings of radius $\rho$ and width $d\rho$.

2. Find the one-sided limit just above the center and the far-field limit, expressing the latter in terms of total charge Q.

Question 1 - Solution

A disk field is a sum of ring contributions with different radii

(a) Method and calculation

$$dq=2\pi\sigma\rho\,d\rho,\quad dE_z=\frac{kz\,dq}{(z^2+\rho^2)^{3/2}}.$$$$E_z=2\pi k\sigma z\int_0^R\frac{\rho\,d\rho}{(z^2+\rho^2)^{3/2}}=2\pi k\sigma\left(1-\frac z{\sqrt{z^2+R^2}}\right).$$

(b) Interpretation and further calculation

$$z\to0^+:\ E_z\to2\pi k\sigma.$$$$z\gg R:\quad E_z\simeq\frac{\pi k\sigma R^2}{z^2}=\frac{kQ}{z^2},\quad Q=\pi R^2\sigma.$$Near the surface the charged area appears broad; far away the finite total charge controls the leading field.

Physical check

The limit is taken from above the ideal sheet, where the field is well defined; it is not a claim about a charge element acting on itself.