Consider a sphere of radius R with charge density distributed as ρ (r ) = kr for r ≤ R 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.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation-

(i) when r

$?E.ds$   = $\frac{1}{{\epsilon }_0}\int \mathrm{\rho }\mathrm{d}\mathrm{V}$ and we know that V = $\frac{4}{3}\pi r^3$

$dV=3*\frac{4}{3}\pi r^{3}dr$ = 4 $\pi r^{2}dr$

  $?E.ds$ = $\frac{1}{{\epsilon }_0}4\pi K\int r^{3}dr$

  E(4 ) $\pi r^2$ = $\frac{4\pi K{r}^4}{{\epsilon }_{0}4}$ = $\frac{k{r}^2}{4{\epsilon }_0}$ so it is clear that E is radially outwards.

  When r>R

$?E.ds$ = $\frac{1}{{\epsilon }_0}\int \mathrm{\rho }\mathrm{d}\mathrm{V}$

              E= $\frac{k{R}^4}{4{\epsilon }_0{r}^2}$ again field is outwards

(ii) When two protons are there then they must

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation-

(i) when r

$?E.ds$   = $\frac{1}{{?}_0}?\mathrm{?}\mathrm{d}\mathrm{V}$ and we know that V = $\frac{4}{3}?r^3$

$dV=3*\frac{4}{3}?r^{3}dr$ = 4 $?r^{2}dr$

  $?E.ds$ = $\frac{1}{{?}_0}4?K?r^{3}dr$

  E(4 ) $?r^2$ = $\frac{4?K{r}^4}{{?}_{0}4}$ = $\frac{k{r}^2}{4{?}_0}$ so it is clear that E is radially outwards.

  When r>R

$?E.ds$ = $\frac{1}{{?}_0}?\mathrm{?}\mathrm{d}\mathrm{V}$

              E= $\frac{k{R}^4}{4{?}_0{r}^2}$ again field is outwards

(ii) When two protons are there then they must