Consider a sphere of radius R with charge density distributed as

(a)

The symmetry of the problem suggests that the electric field is radial.

For points $r

(a)

Consider Gaussian surfaces as shown in figure given below.

For points at $r

$\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}}=\frac{1}{\varepsilon_{0}} \int p d V$

Now, $V=\frac{4}{3} \pi r^{3} \Rightarrow d V=3 \times \frac{4}{3} \pi r^{3} d r$ or $d V=4 \pi r^{2} d r$

$\Rightarrow \oint \mathrm{E} \cdot \mathrm{dS}=\frac{1}{\varepsilon_{0}} 4 \pi K \int_{0}^{r} r^{3} d r[\because p(r)=K r]$

$\Rightarrow(E) 4 \pi r^{2}=\frac{4 \pi K}{\varepsilon_{0}} \frac{r^{4}}{4}$

$\Rightarrow E=\frac{1}{4 \varepsilon_{0}} \mathrm{Kr}^{2}$

Here, charge density is positive. Therefore, direction of $\mathbf{E}$ is radially outwards.

For points $r>R$,

$\oint \mathrm{E} \cdot \mathrm{dS}=\frac{1}{\varepsilon_{0}} \int p \cdot d V$

$\Rightarrow E\left(4 \pi r^{2}\right)=\frac{4 \pi K}{\varepsilon_{0}} \int_{0}^{R} r^{3} d r=\frac{4 \pi K}{\varepsilon_{0}} \frac{R^{4}}{4}$

$\Rightarrow E=\frac{K}{4 \varepsilon_{0}} \frac{R^{4}}{r^{2}}$

Now charge density is again positive. So, the direction of $E$ is radially outward.

(b)

The two protons must be on the opposite sides of the centre along a diameter. Suppose the protons are at a distance $r$ from the centre.

$q=\int_{0}^{\pi} p d V=\int_{0}^{8}(K r) 4 \pi r^{2} d r$

$q=4 \pi K \frac{R^{4}}{4}=2 e$

$\therefore k=\frac{2 e}{\pi R^{4}}$

Consider the forces on proton 1 . The attractive force due to the charge distribution is

$F_{1}=-e E=\frac{-e K r^{2}}{4 \varepsilon_{0}}$

Repulsive force on proton 1 due to proton 2 is

$F_{2}=\frac{e^{2^{2}}}{4 \pi \varepsilon_{0}(2 r)^{2}}$

So, $F_{N n}=\left[\frac{-e r^{2}}{4 \varepsilon_{0}} \frac{Z e}{\pi R^{4}}+\frac{e^{2}}{16 \pi \varepsilon_{0} r^{4}}\right]$

Thus, net force on proton 1 will be zero. when

$\frac{e r^{2} 2^{*}}{4 \varepsilon_{0} \pi R^{4}}=\frac{e^{2}}{16 \pi \varepsilon_{0} r}$

$\Rightarrow r^{4}=\frac{R^{4}}{8}$

$\Rightarrow r=\frac{R}{(8)^{1 / 4}}$

Therefore, distance of both the protons from the centre must be $r=\frac{R}{(8)^{1 / 4}}$.