Question
For the electric field E, in a region of space where a non-uniform, but spherically symmetric distribution of charge has a charge density ρ as
ρ = ρ0(1 − r/R) for r ≤ R
ρ = 0 for r ≥ R
one can say that
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Solution
[Key Concept: Gauss's law with the enclosed charge of a radially varying density gives E as a function of r; its maximum follows from dE/dr = 0.]
$$\begin{matrix} & E_{\text{in~}}4\pi r^{2} = \frac{\int_{0}^{r}\mspace{2mu}\mspace{2mu}\rho_{0}\left( 1 - \frac{r}{R} \right)4\pi r^{2}dr}{\varepsilon_{0}} \\ & E_{\text{in~}} = \frac{\rho_{0}}{\varepsilon_{0}r^{2}}\left\lbrack \frac{r^{3}}{3} - \frac{r^{4}}{4R} \right\rbrack \\ & E_{\text{in~}} = \frac{\rho}{\varepsilon_{0}}\left\lbrack \frac{r}{3} - \frac{r^{2}}{4R} \right\rbrack \end{matrix}$$
$$\begin{matrix} & \frac{dE}{dR} = 0 \Rightarrow \frac{1}{3} - \frac{r}{2R} \Rightarrow r = \frac{2R}{3} \\ & \frac{d^{2}E}{dR^{2}} < 0\text{~at~}r = \frac{2R}{3} \\ & E_{\max} = \frac{\rho}{\varepsilon_{0}}\left\lbrack \frac{2R}{9} - \frac{4R^{2}}{9 \times 4R} \right\rbrack \\ & \frac{\rho}{\varepsilon_{0}}\frac{R}{9} \end{matrix}$$
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