QJEEM22Phf2a0: JEE Main 2022 Physics - Class 12 JEEM22Phf2a0 Solution: A spherically symmetric charge distribution is considered with…

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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:

JEE Main 2022 (Online) 29th July Morning Shift Physics - Electrostatics Question 87 English

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

Accepted Answer

$\left( {4\pi {r^2}} \right){E_\rho } = {{{Q_{in}}} \over {{\varepsilon _0}}}$

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

$= {{{\rho _0}\pi 4} \over {{\varepsilon _0}}}\left( {{{{r^3}} \over 4} - {{{r^4}} \over {4R}}} \right)$

${E_\rho } = {{{\rho _0}} \over {4{\varepsilon _0}}}\left( {r - {{{r^2}} \over R}} \right)$

$= {{{\rho _0}r} \over {4{\varepsilon _0}}}\left( {1 - {r \over R}} \right)$

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