Let three be a spherically symmetric charge distribution with charge density varying as $\rho(r) = \rho_{0}\left( \frac{5}{4} - \frac{r}{R} \right)$ upto$\ r = R,\ and\ \rho(r) = 0\ \ for\ r R,\ where\ r$ is the distance from the origin. The electric field at a distance $r(r

Question

Let three be a spherically symmetric charge distribution with charge density varying as $\rho(r) = \rho_{0}\left( \frac{5}{4} - \frac{r}{R} \right)$ upto$\ r = R,\ and\ \rho(r) = 0\ \ for\ r > R,\ where\ r$ is the distance from the origin. The electric field at a distance $r(r < R)$ from the origin is given by

Dr. Abubacker · Accepted Answer

(b) Apply shell theorem, the total charge upto distance $r$ can be calculated as followed
$dq = 4\pi r^{2}.dr.ho$
$\  = 4\pi r^{2}.dr.ho_{0}\left\lbrack \frac{5}{4} - \frac{r}{R} ightbrack$
$\  = 4\piho_{0}\left\lbrack \frac{5}{4}r^{2}dr - \frac{r^{3}}{R}dr ightbrack$
$\int_{}^{}{dq = q = 4\piho_{0}\int_{0}^{r}\left( \frac{5}{4}r^{2}dr - \frac{r^{3}}{R}dr ight)}$
$= 4\piho_{0}\left\lbrack \frac{5}{4}\frac{r^{3}}{3} - \frac{1}{R}\frac{r^{4}}{4	ext{ }} ightbrack$
$E = \frac{kq}{r^{2}}$
$= \frac{1}{4\pi\varepsilon_{0}}\frac{1}{r^{2}}.4\piho_{0}\left\lbrack \frac{5}{4}\left( \frac{r^{3}}{3} ight) - \frac{r^{4}}{4R} ightbrack$
$E = \frac{ho_{0}r}{4\varepsilon_{0}}\left( \frac{5}{3} - \frac{r}{R} ight)$

Explanation

Reviewed by Dr. Abubacker

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