Potential Due to a Continuous Charge Distribution — Question 5

Question 5

A disk potential follows by summing concentric rings

A thin disk of radius R has uniform positive surface density $\sigma$. Observe on its axis at z>0. Use infinity reference.

1. Derive V(z) from rings of radius s.

2. Find V at the disk center and show that the large-z limit agrees with a point charge of total charge Q.

Question 5 - Solution

A disk potential follows by summing concentric rings

(a) Method and calculation

$$dq=2\pi\sigma s\,ds,\quad V(z)=2\pi k\sigma\int_0^R\frac{s\,ds}{\sqrt{s^2+z^2}}.$$$$V(z)=2\pi k\sigma(\sqrt{R^2+z^2}-z).$$

(b) Interpretation and further calculation

$$V(0)=2\pi k\sigma R.$$$$z\gg R:\quad V\simeq2\pi k\sigma\frac{R^2}{2z}=\frac{kQ}{z},\quad Q=\pi R^2\sigma.$$

Physical check

The integrable central ring contribution gives a finite disk-center potential despite the infinitesimal kernel singularity.