JEE MAIN - Physics (2022 - 29th July Morning Shift - No. 10)

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

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

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

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

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

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

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

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

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

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