Potential Due to a Continuous Charge Distribution — Question 9

Question 9

A finite cylinder center potential combines ring and axial integration

A solid cylinder of radius R and half-length L has uniform volume density $\rho>0$. Its axis is z and its center is the origin. Use infinity reference.

1. Write and evaluate the double integral for V at the center.

2. Find the thin-cylinder limit L much smaller than R and show agreement with a disk of surface density $\sigma=2L\rho$.

Question 9 - Solution

A finite cylinder center potential combines ring and axial integration

(a) Method and calculation

$$V(0)=2\pi k\rho\int_{-L}^{L}\int_0^R\frac{s\,ds\,dz}{\sqrt{s^2+z^2}}=4\pi k\rho\int_0^L(\sqrt{R^2+z^2}-z)dz.$$$$V(0)=2\pi k\rho\left[L\sqrt{R^2+L^2}+R^2\operatorname{arsinh}(L/R)-L^2\right].$$

(b) Interpretation and further calculation

$$L\ll R:\quad V(0)=4\pi k\rho LR-2\pi k\rho L^2+O(k\rho L^3/R).$$$$V_{\mathrm{leading}}=2\pi k(2L\rho)R=2\pi k\sigma R.$$

Physical check

The limit refers to a finite cylinder becoming thin; no infinity reference is claimed for an infinite cylinder.