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

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

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

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

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

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Electrostatics

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

ρ (r) = {ρ_{0} (\frac{3}{4} - \frac{r}{R}) Zero for r \leq R for r > R

Where,

r (r < R)

is the distance from the centre O (as shown in figure). The electric field at point P will be :

Answer: (C)

Using Gauss's law for a spherical Gaussian surface of radius

r < R

\oint E \cdot d s = \frac{Q _{in}}{ε _{0}}

. The enclosed charge is

Q_{in} = \int_{0}^{r} ρ_{0} (\frac{3}{4} - \frac{r ^{'}}{R}) 4 π r^{'2} d r^{'} = 4 π ρ_{0} [\frac{3 r ^{3}}{12} - \frac{r ^{4}}{4 R}] = π ρ_{0} r^{3} (1 - \frac{r}{R})

. From Gauss's law:

E \cdot 4 π r^{2} = \frac{π ρ _{0} r ^{3} ( 1 - r / R )}{ε _{0}}

, giving

E = \frac{ρ _{0} r}{4 ε _{0}} (1 - \frac{r}{R})

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option