Applying Gauss' Law: Spherical Symmetry — Question 6

Question 6

A neutral core-and-coating sphere can confine a nonzero field to its interior

A sphere has density $\rho_1>0$ for $r<a$ and uniform density $\rho_2$ in $a<r<b$. Choose $\rho_2$ so the total charge is zero. There are no separate surface sheets or external sources.

1. Find $\rho_2$ and the field in the core and exterior.

2. Derive the coating-region field, determine its sign and show where the largest field occurs.

Question 6 - Solution

A neutral core-and-coating sphere can confine a nonzero field to its interior

(a) Method and calculation

$$\frac{4\pi}{3}[\rho_1a^3+\rho_2(b^3-a^3)]=0\Rightarrow\rho_2=-\frac{\rho_1a^3}{b^3-a^3}.$$$$E_r=\frac{\rho_1r}{3\epsilon_0}\quad(r<a),\qquad E_r=0\quad(r>b).$$

(b) Interpretation and further calculation

$$E_r=\frac{\rho_1a^3}{3\epsilon_0(b^3-a^3)}\left(\frac{b^3}{r^2}-r\right)>0\quad(a<r<b).$$$$\frac{dE_r}{dr}=\frac{\rho_1a^3}{3\epsilon_0(b^3-a^3)}\left(-\frac{2b^3}{r^3}-1\right)<0.$$The core field rises to $E_{\max}=\rho_1a/(3\epsilon_0)$ at a, then decreases to zero through the coating.

Physical check

Total neutrality determines the exterior field here only because the entire distribution is spherically symmetric.