Question

$1)$ Consider a sphere of radius $R$ with charge density distributed as $\rho \left(r\right)=kr$ for $r\le R$

$=0$ for $r>R$

$A)$ Find the electric field at all points $r.$

$2)$ Consider a sphere of radius $R$ with charge density distributed as $\rho \left(r\right)=kr$ for $r\le R$

$=0$ for $r>R$

$B)$ 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.

Answer 1

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

Step 1: For points $r<R$

Given, $\rho \left(r\right)=kr$ for $r\le R$

consider a spherical Gaussian surface. Then on the surface

$\oint \overrightarrow{E}.\overrightarrow{dS}=\frac{q_{in}}{{\epsilon }_o}$

$\oint \overrightarrow{E_r}.\overrightarrow{dS}=\frac{1}{{\epsilon }_o}{\int }_v\rho dv$

$E_r\oint dS=\frac{1}{{\epsilon }_o}{\int }_0^{r}kr\left(4\pi r^{2}dr\right)$ 

$E_{r}4\pi r^2=\frac{1}{{\epsilon }_o}4\pi k{\int }_0^r{r}^{3}dr$

$E_r=\frac{1}{4{\epsilon }_o}k{r}^2$

$\overrightarrow{E}\left(r\right)=\frac{1}{4{\epsilon }_o}k{r}^2\hat{r}$

Step 2: For points $r>R$

Given,$\rho \left(r\right)=0$ for $r>R$

consider a spherical Gaussian surface of radius $r$



$\oint \overrightarrow{E}.\overrightarrow{ds}=\frac{q_{in}}{{\epsilon }_0}$

$\oint \overrightarrow{E_r}.\overrightarrow{ds}=\frac{1}{{\epsilon }_0}{\int }_v\rho dv$

$E_r\oint ds=\frac{1}{{\epsilon }_0}\left({\int }_0^R\rho \right(r)4\pi r^{2}dr+{\int }_R^r\rho (r\left)4\pi r^{2}dr\right)$

$E_{r}4\pi r^2=\frac{1}{{\epsilon }_0}{\int }_0^{R}kr\left(4\pi r^{2}dr\right)$

$E_r=\frac{k}{4{\epsilon }_0}\frac{R^4}{r^2}$

$\overrightarrow{E}\left(r\right)=\frac{k}{4{\epsilon }_0}\frac{R^4}{r^1}\hat{r}$

Final Answer:

For $r\le R,\overrightarrow{E}\left(r\right)=\frac{1}{4{\epsilon }_0}k{r}^2\hat{r}$

For $r>R,\overrightarrow{E}\left(r\right)=\frac{k}{4{\epsilon }_0}\frac{R^4}{r^2}\hat{r}$

$B)$ Assumption: Introduction of the proton does not alter the negative charge distribution.



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
Now,

${\int }_0^R\rho dv=2e$

$4\pi {\int }_0^{R}k{r}^{3}dr=2e$

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

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

For $r\le R$

$\overrightarrow{E}\left(r\right)=\frac{1}{4{\epsilon }_0}k{r}^2\hat{r}$

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

$-e\overrightarrow{E_r}=-\frac{e}{4{\epsilon }_0}k{r}^2\hat{r}=-\frac{2e^2}{4\pi {\epsilon }_0}\frac{r^2}{R^4}\hat{r}$

The repulsive force from coulomb's force is $\frac{e^2}{4\pi {\epsilon }_0}\frac{1}{(2r{)}^2}\hat{r}.$

Net force is

$\overrightarrow{F_{net}}=(\frac{e^2}{4\pi {\epsilon }_{0}4r^2}-\frac{2e^2}{4\pi {\epsilon }_0}\frac{r^2}{R^4})\hat{r}$

Given, $\overrightarrow{F_{net}}=0$

$\frac{e^2}{16\pi {\epsilon }_0{r}^2}=\frac{2e^2}{4\pi {\epsilon }_0{r}^2}\frac{r^2}{R^4}$

$r^4=\frac{4r^4}{32}=\frac{R^4}{8}$

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

Thus, the protons must be at a distance $r=\frac{R}{\sqrt[4]{48}}$ from the centre.

Final Answer: $r=\frac{R}{\sqrt[4]{48}}$