Applying Gauss' Law: Spherical Symmetry — Question 4

Question 4

A sphere with vanishing edge density reaches peak field beneath its surface

A spherical insulator of radius R has density $\rho(r)=\rho_0(1-r^2/R^2)$ for $r<R$, with $\rho_0>0$. No surface sheet or exterior charge is present.

1. Derive the interior field and its value at R.

2. Locate the global maximum field and compare it with the surface value.

Question 4 - Solution

A sphere with vanishing edge density reaches peak field beneath its surface

(a) Method and calculation

$$Q_{\mathrm{enc}}=4\pi\rho_0\left(\frac{r^3}{3}-\frac{r^5}{5R^2}\right).$$$$E_r=\frac{\rho_0}{\epsilon_0}\left(\frac r3-\frac{r^3}{5R^2}\right),\quad E_R=\frac{2\rho_0R}{15\epsilon_0}.$$

(b) Interpretation and further calculation

$$\frac{dE_r}{dr}=\frac{\rho_0}{\epsilon_0}\left(\frac13-\frac{3r^2}{5R^2}\right)=0\Rightarrow r_*=\frac{\sqrt5}{3}R.$$$$E_{\max}=\frac{2\sqrt5\rho_0R}{27\epsilon_0},\quad\frac{E_{\max}}{E_R}=\frac{5\sqrt5}{9}\approx1.24226.$$The exterior decreases as $1/r^2$, confirming the interior extremum is the global maximum.

Physical check

The field and its derivative match the exterior at R because the volume density itself tends to zero there.