Question

Find the maximum intensity E${}_{max}$ and the corresponding distance r${}_m$

Answer 1

Charge within radius r is

$q={\int }_0^r{\rho }_0[1-\frac{r}{R}]4\pi r^{2}dr$

$=4\pi {\rho }_0{\int }_0^r(r^2-\frac{r^3}{R})dr$

$=4\pi {\rho }_0(\frac{r^3}{3}-\frac{r^4}{4R})$ (i)

For r < R,

$E=\frac{kq}{r^2}=\frac{{\rho }_{0}r}{3{\epsilon }_0}[1-\frac{3r}{4R}]$ (ii)

For total charge, put $r=R$ in Eq. (i)

$Q=4\pi {\rho }_0[\frac{R^3}{3}-\frac{R^3}{4}]=\frac{4\pi {\rho }_0{R}^3}{12}$

For r > R,

$E=\frac{kQ}{r^2}=\frac{{\rho }_0{R}^3}{12{\epsilon }_0{r}^2}$ (iii)

At surface, from both Eqs. (ii) and (iii), put $r=R$.

$E_0=\frac{{\rho }_{0}R}{12{\epsilon }_0}$

For maximum, $dE/dr=0$ from eq. (ii).

$r=\frac{2}{3}R$

$E=\frac{{\rho }_{0}R}{9{\epsilon }_0}$

Also, at $r=R/3$,

$E=\frac{{\rho }_{0}R}{12{\epsilon }_0}$