Applying Gauss' Law: Spherical Symmetry — Question 3

Question 3

A quadratic radial density produces a cubic interior field

A spherical insulator has density $\rho(r)=\rho_0r^2/R^2$ for $r<R$, with $\rho_0>0$, and no charge outside. Compare it with a uniformly charged sphere of the same R and same total charge.

1. Derive its enclosed charge, interior field and total charge.

2. Find the density of the uniform comparison sphere and the ratio of interior fields at the same nonzero radius.

Question 3 - Solution

A quadratic radial density produces a cubic interior field

(a) Method and calculation

$$Q_{\mathrm{enc}}=4\pi\int_0^r\frac{\rho_0u^2}{R^2}u^2\,du=\frac{4\pi\rho_0r^5}{5R^2}.$$$$E_{\mathrm{graded}}=\frac{\rho_0r^3}{5\epsilon_0R^2},\quad Q=\frac{4\pi\rho_0R^3}{5}.$$

(b) Interpretation and further calculation

$$\rho_{\mathrm{uniform}}=\frac Q{4\pi R^3/3}=\frac{3\rho_0}{5},\quad E_{\mathrm{uniform}}=\frac{\rho_0r}{5\epsilon_0}.$$$$E_{\mathrm{graded}}/E_{\mathrm{uniform}}=(r/R)^2.$$The stronger outward weighting reduces the enclosed fraction more severely at small r than a uniform sphere does.

Physical check

Both profiles have the same exterior field, so exterior measurements alone cannot distinguish their radial density distributions.