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  4. A spherically symmetric charge distribution is considered with charge density varying as ρ(r)= begincasesρ0((3/4)-(r/R)) text for r ≤ R text Zero text for r>R endcases Where, r ( r < R ) is the distance from the centre O (as shown in figure). The electric field at point P will be :

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Q. A spherically symmetric charge distribution is considered with charge density varying as $\rho(r)= \begin{cases}\rho_0\left(\frac{3}{4}-\frac{r}{R}\right) & \text { for } r \leq R \\ \text { Zero } & \text { for } r>R\end{cases}$ Where, $r ( r < R )$ is the distance from the centre $O$ (as shown in figure). The electric field at point $P$ will be :Physics Question Image

JEE MainJEE Main 2022Electric Charges and Fields

Solution:

image
$ \oint \overrightarrow{ E } \cdot d \overrightarrow{ s }=\frac{Q_{ in }}{\varepsilon_{ o }} $
$ E \cdot 4 \pi r ^2=\frac{\int\limits_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 r ^2=\frac{\rho_{ o } r ^3}{4 \varepsilon_o}\left\{1-\frac{ r }{ R }\right\} $
$ E =\frac{\rho_{ o } r }{4 \varepsilon_{ o }}\left\{1-\frac{ r }{ R }\right\}$