Answer
(i) $E= 0$
(ii)$ E= \frac{k}{ε_{0}}(\frac{r - a}{r^2})\hat r$
(iii)$ E= \frac{k}{ε_{0}}(\frac{b - a}{r^2})\hat r$
Work Step by Step
(i) By guasses law
$ ∫E.ds =\frac{q_{en}}{ε_{0}}$
but $q_{en} = 0 $ for $r\lt a$
Hence $E =0$
(ii) for $a\lt r \lt b$
$ ∫E.ds =\frac{q_{en}}{ε_{0}}$
lets compute $q_{en}$ first
Consider a volume element $dτ$ at a distance x from the center of the shell with a thickness of $dx$, such that $ a\lt x \lt b$
$q_{en} = ∫_{a}^{r}ρdτ $
$q_{en} = ∫_{a}^{r} \frac{k}{x^{2}} dτ = ∫_{a}^{r} \frac{k}{x^{2}} 4πx^{2}dx$
$q_{en} = 4πk∫_{a}^{r}dx = 4πk|x|_{a}^{r} = 4πk(r-a)$
By gausses law
$ ∫E.ds =\frac{q_{en}}{ε_{0}}$
$ |E| 4πr^{2}=\frac{q_{en}}{ε_{0}}$
$ |E| 4πr^{2}=\frac{4πk(r-a)}{ε_{0}}$
$|E| =\frac{k}{ε_{0}}\frac{(r-a)}{r^{2}}$
since $E $ is in the direction of $\vec r$
$\vec E =\frac{k}{ε_{0}}\frac{(r-a)}{r^{2}}\hat r$
(iii) For $ r\gt b$
$ q_{en} =q_{total} =∫_{a}^{b}ρdτ = 4πk|x|_{a}^{b} = 4πk(b-a)$
By gausses law
$ ∫E.ds =\frac{q_{en}}{ε_{0}}$
$ |E| 4πr^{2}=\frac{4πk(b-a)}{ε_{0}}$
$|E| =\frac{k}{ε_{0}}\frac{(b-a)}{r^{2}}$
or $\vec E =\frac{k}{ε_{0}}\frac{(b-a)}{r^{2}}\hat r$
