Consider a sphere of radius $R$ with charge density distributed as:
$\rho(r) = kr$ for $r \leq R$
$\rho(r) = 0$ 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 elementary charge. Where can two protons be embedded such that the force on each of them is zero? Assume that the introduction of the protons does not alter the charge distribution.

(N/A) Let us consider a Gaussian sphere of radius $r$. Using Gauss's Law: $\oint \vec{E} \cdot d\vec{S} = \frac{q_{enclosed}}{\epsilon_0}$.
For $r \leq R$,the enclosed charge is $q(r) = \int_0^r \rho(r) 4\pi r^2 dr = \int_0^r (kr) 4\pi r^2 dr = 4\pi k \int_0^r r^3 dr = \pi k r^4$.
Applying Gauss's Law: $E(4\pi r^2) = \frac{\pi k r^4}{\epsilon_0} \implies E = \frac{kr^2}{4\epsilon_0}$.
For $r > R$,the total charge is $Q = \pi k R^4 = 2e$. Thus,$E(4\pi r^2) = \frac{2e}{\epsilon_0} \implies E = \frac{2e}{4\pi \epsilon_0 r^2}$.
$(b)$ For the force on a proton to be zero,the electric field due to the sphere must be cancelled by the electric field due to the other proton. Let the protons be at distance $d$ from the center on opposite sides. The field from the sphere at distance $d$ is $E_s = \frac{q(d)}{4\pi \epsilon_0 d^2} = \frac{\pi k d^4}{4\pi \epsilon_0 d^2} = \frac{k d^2}{4\epsilon_0}$.
Since $Q = \pi k R^4 = 2e$,we have $k = \frac{2e}{\pi R^4}$. So $E_s = \frac{2e d^2}{4\pi \epsilon_0 R^4}$.
The field from the other proton at distance $2d$ is $E_p = \frac{e}{4\pi \epsilon_0 (2d)^2} = \frac{e}{16\pi \epsilon_0 d^2}$.
Equating $E_s = E_p$: $\frac{2e d^2}{4\pi \epsilon_0 R^4} = \frac{e}{16\pi \epsilon_0 d^2} \implies d^4 = \frac{R^4}{8} \implies d = \frac{R}{8^{1/4}} = \frac{R}{\sqrt[4]{8}}$.