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 ${\epsilon }_{0}$. $\Rightarrow \oint \vec{E\cdot }d\vec{s}=\frac{{Q}_{\mathrm{in}}}{{\epsilon }_{0}}$ $\Rightarrow E4\pi {r}^{2}=\frac{{\int }_{0}^{r}{\rho }_{0}(\frac{3}{4}-\frac{r}{R})4\pi {r}^{2}dr}{{\epsilon }_{0}}$ $\Rightarrow E4\pi {r}^{2}=\frac{{\rho }_{o}4\pi }{{\epsilon }_{o}}(\frac{3}{4}\frac{{r}^{3}}{3}-\frac{{r}^{4}}{4R})$ $\Rightarrow E=\frac{{\rho }_{o}r}{4{\epsilon }_{o}}(1-\frac{r}{R})$

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 ${\epsilon }_{0}$ .

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

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

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

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

JEE Main 2022 — Physics Electromagnetism

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Solution

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JEE Main 2022 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism