Question 324: Electrostatics & Capacitance

Question

A spherically symmetric charge distribution is considered with charge density varying as
$ho(r)= \begin{cases}ho_{0}\left(\frac{3}{4}-\frac{r}{R}ight) & 	ext { for } r \leq R \ 	ext { zero } & 	ext { for } r>R\end{cases}$
Where, $r(r < R)$ is the distance from the centre O (as shown in figure). The electric field at point P will be:
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l6sahgfm/2e63886c-b02c-41f9-963c-c78b197ed17b/dd281520-1b3d-11ed-91de-637bbeffe542/file-1l6sahgfn.png?format=png)

Accepted Answer

$\left( {4\pi {r^2}} \right){E_\rho } = {{{Q_{in}}} \over {{\varepsilon _0}}}$
$ = {{\int_0^r {{\rho _0}\left( {{3 \over 4} - {r \over 4}} \right)4\pi {r^2}dr} } \over {{\varepsilon _0}}}$
$ = {{{\rho _0}\pi 4} \over {{\varepsilon _0}}}\left( {{{{r^3}} \over 4} - {{{r^4}} \over {4R}}} \right)$
${E_\rho } = {{{\rho _0}} \over {4{\varepsilon _0}}}\left( {r - {{{r^2}} \over R}} \right)$
$ = {{{\rho _0}r} \over {4{\varepsilon _0}}}\left( {1 - {r \over R}} \right)$

Answer

$\frac{\rho_{0} \mathrm{r}}{4 \varepsilon_{0}}\left(\frac{3}{4}-\frac{r}{R}\right)$

Answer

$\frac{\rho_{0} r}{3 \varepsilon_{0}}\left(\frac{3}{4}-\frac{r}{R}\right)$

Answer

$\frac{\rho_{0} r}{4 \varepsilon_{0}}\left(1-\frac{r}{R}\right)$

Answer

$
\frac{\rho_{0} r}{5 \varepsilon_{0}}\left(1-\frac{r}{R}\right)
$

Electrostatics & Capacitance

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