Discussion
A solid sphere of radius R with its center at origin has a volume charge density \(\rho = A\sqrt{r}\), where A is a constant. The magnitude of electric field, \(\bar{E}\), at \(r \lt R\), varies with r as ........ .
Answer: C
Inside the sphere, \(E \propto r^{3/2}\).
By Gauss's law, the field at radius r depends on the charge enclosed:
\[ \begin{aligned} q(r) &= \int_{0}^{r} A\sqrt{r'}\, 4\pi r'^{2}\, dr' \\ &= \frac{8\pi A}{7} r^{7/2} \\ E &= \frac{q(r)}{4\pi\epsilon_{0}r^{2}} = \frac{2A}{7\epsilon_{0}} r^{3/2} \end{aligned} \]
For a uniform density the same steps give \(E \propto r\); the extra factor \(\sqrt{r}\) in the density raises it to \(r^{3/2}\). \(r^{-2}\) is the law outside the sphere, and \(r^{\pm 1/2}\) would come from forgetting the \(r^{2}\) of the volume element.
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