MCQ
A spherically symmetric charge distribution is considered with charge density varying as

$\rho(r)=\left\{\begin{array}{ll}\rho_{0}\left(\frac{3}{4}-\frac{r}{R}\right) & \text { for } r \leq R \\ \text { Zero } & \text { for } r>R\end{array}\right.$

Where, $r ( r < R )$ is the distance from the centre $O$ (as shown in figure). The electric field at point $P$ will be.

  • A
    $\frac{\rho_{0} r}{4 \varepsilon_{0}}\left(\frac{3}{4}-\frac{r}{R}\right)$
  • B
    $\frac{\rho_{0} r}{3 \varepsilon_{0}}\left(\frac{3}{4}-\frac{r}{R}\right)$
  • $\frac{\rho_{0} r}{4 \varepsilon_{0}}\left(1-\frac{r}{R}\right)$
  • D
    $\frac{\rho_{0} r}{5 \varepsilon_{0}}\left(1-\frac{r}{R}\right)$

Answer

Correct option: C.
$\frac{\rho_{0} r}{4 \varepsilon_{0}}\left(1-\frac{r}{R}\right)$
c
$\oint \overrightarrow{ E } \cdot d \overrightarrow{ s }=\frac{ Q _{\text {in }}}{\varepsilon_{ o }}$

$E .4 \pi r ^{2}=\frac{\int_{0}^{ r } \rho_{ o }\left(\frac{3}{4}-\frac{ r }{ R }\right) 4 \pi r ^{2} dr }{\varepsilon_{0}}$

$E 4 \pi r ^{2}=\frac{\rho_{ o } 4 \pi}{\varepsilon_{ o }}\left(\frac{3}{4} \frac{ r ^{3}}{3}-\frac{ r ^{4}}{4 R }\right)$

$Er { }^{2}=\frac{\rho_{ o } r ^{3}}{4 \varepsilon_{ o }}\left\{1-\frac{ r }{ R }\right\}$

$E =\frac{\rho_{0} r }{4 \varepsilon_{ o }}\left\{1-\frac{ r }{ R }\right\}$

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