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Consider a sphere of radius $R$ with charge density distributed as :
$\rho(r) =k r$, $r \leq R $
$=0$ for $r> R$.
$(a)$ Find the electric field at all points $r$.
$(b)$ Suppose the total charge on the sphere is $2e$ where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.
Solution
$(a)$ Let us consider a sphere $\mathrm{S}$ of radius $\mathrm{R}$ and two hypothetic spheres of radius $r<\mathrm{R}$ and $r>\mathrm{R}$. Electric field intensity for point $r<\mathrm{R}$,
$\oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{S}}=\frac{1}{\epsilon_{0}} \int \rho d \mathrm{~V}$
But, volume $\mathrm{V}=\frac{4}{3} \pi r^{3}$
$d \mathrm{~V} =\frac{4}{3} \pi \times 3 r^{2} d r$
$d \mathrm{~V} =4 \pi r^{2} d r$
$\therefore \oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{S}}=\frac{1}{\epsilon_{0}} 4 \pi k \int_{0}^{r} r^{3} d r \quad[\rho=k r]$ $\therefore(\mathrm{E}) 4 \pi r^{2}=\frac{4 \pi k}{\epsilon_{0}} \cdot \frac{r^{4}}{4}$ $\therefore \mathrm{E}=\frac{1}{4 \epsilon_{0}} \cdot k r^{2}$ As charge density is positive it means the direction of $\overrightarrow{\mathrm{E}}$ is radially outwards.
