A spherically symmetric charge distribution is considered with charge density varying as $\rho (r)={\begin{matrix}{\rho }_{0}(\frac{3}{4}-\frac{r}{R}) & \text{ for }r\leq R \\ \mathrm{Zero} & & & \mathrm{for}r>R\end{matrix}$ 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 Physics, Electrostatics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Electrostatics/648b5a6e417cc3fb48d67d90/question_1__q_648b5a6e417cc3fb48d67d90__cdn-question-pool.getmarks.app__bc891dff-30ec-4b12-b920-07c5eeaa772d-image__327b8c199d_final_ppt_sync.png)

![JEE Main 2022 Physics, Electrostatics — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Electrostatics/648b5a6e417cc3fb48d67d90/solution_1__q_648b5a6e417cc3fb48d67d90__cdn-question-pool.getmarks.app__2da854da-3f86-449f-93c8-43d71f49f354-image__bb70a2fbc9_final_ppt_sync.png) According to Gauss's law, the electric flux through a closed surface area is equal to the charge inside the surface divided by ${\epsilon }_{0}$. $\Rightarrow \oint \vec{E\cdot }d\vec{s}=\frac{{Q}_{\mathrm{in}}}{{\epsilon }_{0}}$ $\Rightarrow E4\pi {r}^{2}=\frac{{\int }_{0}^{r}{\rho }_{0}(\frac{3}{4}-\frac{r}{R})4\pi {r}^{2}dr}{{\epsilon }_{0}}$ $\Rightarrow E4\pi {r}^{2}=\frac{{\rho }_{o}4\pi }{{\epsilon }_{o}}(\frac{3}{4}\frac{{r}^{3}}{3}-\frac{{r}^{4}}{4R})$ $\Rightarrow E=\frac{{\rho }_{o}r}{4{\epsilon }_{o}}(1-\frac{r}{R})$

A spherically symmetric charge… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Electromagnetism

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

$\rho (r)={\begin{matrix}{\rho }_{0}(\frac{3}{4}-\frac{r}{R}) & \text{ for }r\leq R \\ \mathrm{Zero} & & & \mathrm{for}r>R\end{matrix}$

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 Physics, Electrostatics — question figure

Official previous-year question

Held on 29 Jul 2022 · Verified 6 Jul 2026.

Options

A
$\frac{{\rho }_{0}r}{4{\epsilon }_{0}}(\frac{3}{4}-\frac{r}{R})$
B
$\frac{{\rho }_{0}r}{3{\epsilon }_{0}}(\frac{3}{4}-\frac{r}{R})$
C
$\frac{{\rho }_{0}r}{4{\epsilon }_{0}}(1-\frac{r}{R})$
D
$\frac{{\rho }_{0}r}{5{\epsilon }_{0}}(1-\frac{r}{R})$

Did you get this right?

Solution

JEE Main 2022 Physics, Electrostatics — solution figure

According to Gauss's law, the electric flux through a closed surface area is equal to the charge inside the surface divided by ${\epsilon }_{0}$ .

$\Rightarrow \oint \vec{E\cdot }d\vec{s}=\frac{{Q}_{\mathrm{in}}}{{\epsilon }_{0}}$

$\Rightarrow E4\pi {r}^{2}=\frac{{\int }_{0}^{r}{\rho }_{0}(\frac{3}{4}-\frac{r}{R})4\pi {r}^{2}dr}{{\epsilon }_{0}}$

$\Rightarrow E4\pi {r}^{2}=\frac{{\rho }_{o}4\pi }{{\epsilon }_{o}}(\frac{3}{4}\frac{{r}^{3}}{3}-\frac{{r}^{4}}{4R})$

$\Rightarrow E=\frac{{\rho }_{o}r}{4{\epsilon }_{o}}(1-\frac{r}{R})$

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Options

$\frac{{\rho }_{0}r}{4{\epsilon }_{0}}(\frac{3}{4}-\frac{r}{R})$

$\frac{{\rho }_{0}r}{3{\epsilon }_{0}}(\frac{3}{4}-\frac{r}{R})$

$\frac{{\rho }_{0}r}{4{\epsilon }_{0}}(1-\frac{r}{R})$

$\frac{{\rho }_{0}r}{5{\epsilon }_{0}}(1-\frac{r}{R})$

Solution

JEE Main 2022 Physics, Electrostatics — solution figure

According to Gauss's law, the electric flux through a closed surface area is equal to the charge inside the surface divided by ${\epsilon }_{0}$ .

$\Rightarrow \oint \vec{E\cdot }d\vec{s}=\frac{{Q}_{\mathrm{in}}}{{\epsilon }_{0}}$

$\Rightarrow E4\pi {r}^{2}=\frac{{\int }_{0}^{r}{\rho }_{0}(\frac{3}{4}-\frac{r}{R})4\pi {r}^{2}dr}{{\epsilon }_{0}}$

$\Rightarrow E4\pi {r}^{2}=\frac{{\rho }_{o}4\pi }{{\epsilon }_{o}}(\frac{3}{4}\frac{{r}^{3}}{3}-\frac{{r}^{4}}{4R})$

$\Rightarrow E=\frac{{\rho }_{o}r}{4{\epsilon }_{o}}(1-\frac{r}{R})$

Options

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Solution

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Solution

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Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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