Applying Gauss' Law: Cylindrical Symmetry — Question 5

Question 5

A declining radial density can place the maximum field inside a cylinder

An infinite insulating cylinder of radius R has density $\rho(r)=\rho_0(1-r^2/R^2)$ for $r<R$ and zero outside, with $\rho_0>0$.

1. Derive its interior field and surface value.

2. Locate the maximum field over all radii and calculate its ratio to the surface field.

Question 5 - Solution

A declining radial density can place the maximum field inside a cylinder

(a) Method and calculation

$$\lambda_{\mathrm{enc}}=2\pi\rho_0\int_0^r\left(1-\frac{s^2}{R^2}\right)s\,ds=\pi\rho_0\left(r^2-\frac{r^4}{2R^2}\right).$$$$E_r=\frac{\rho_0}{2\epsilon_0}\left(r-\frac{r^3}{2R^2}\right),\quad E_R=\frac{\rho_0R}{4\epsilon_0}.$$

(b) Interpretation and further calculation

$$\frac{dE_r}{dr}=\frac{\rho_0}{2\epsilon_0}\left(1-\frac{3r^2}{2R^2}\right)=0\Rightarrow r_*=R\sqrt{2/3}.$$$$E_{\max}=\frac{\rho_0R}{3\epsilon_0}\sqrt{2/3},\quad\frac{E_{\max}}{E_R}=\frac43\sqrt{2/3}\approx1.08866.$$Beyond R the field decreases as $1/r$, so the interior maximum is global.

Physical check

Positive charge density alone does not require the field to increase with radius; cylindrical area also grows.