A spherically symmetric charge distribution… - JEE Main 2014 Physics PYQ

JEE Main 2014 — Physics Electromagnetism

2014mcqmedium

A spherically symmetric charge distribution is characterised by a charge density having the following variations: $\rho(r)=\rho_o\left(1-\frac{r}{R}\right)$ for $r < R$ $\rho(\mathrm{r})=0$ for $r \geq \mathrm{R}$ Where $r$ is the distance from the centre of the charge distribution $\rho_{\mathrm{o}}$ is a constant. The electric field at an internal point $(r < R)$ is:

Official previous-year question

Held on 12 Apr 2014 · Verified 6 Jul 2026.

Options

A
$\frac{\rho_{\mathrm{o}}}{4 \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{r}}{3}-\frac{\mathrm{r}^2}{4 \mathrm{R}}\right)$
B
$\frac{\rho_0}{\varepsilon_0}\left(\frac{r}{3}-\frac{r^2}{4 R}\right)$
C
$\frac{\rho_{\mathrm{o}}}{3 \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{r}}{3}-\frac{\mathrm{r}^2}{4 \mathrm{R}}\right)$
D
$\frac{\rho_{\mathrm{o}}}{12 \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{r}}{3}-\frac{\mathrm{r}^2}{4 \mathrm{R}}\right)$

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Solution

Let us consider a spherical shell of radius $x$ and thickness $\mathrm{dx}$ . JEE Main 2014 Physics, Electrostatics — solution figure

Charge on this shell $d q=\rho .4 \pi x^2 d x=\rho_0\left(1-\frac{x}{R}\right) \cdot 4 \pi x^2 d x$ $\therefore$ Total charge in the spherical region from centre to $r(r < R)$ is $\begin{aligned} &q=\int d q=4 \pi \rho_0 \int_0^r\left(1-\frac{x}{R}\right) x^2 d x \\ &=4 \pi \rho_0\left[\frac{x^3}{3}-\frac{x^4}{4 R}\right]_0^r \\ &=4 \pi \rho_0\left[\frac{r^3}{3}-\frac{r^4}{4 R}\right] \\ &=4 \pi \rho_0 r^3\left[\frac{1}{3}-\frac{r}{4 R}\right] \end{aligned}$ $\therefore$ Electric field at $\mathrm{r}, \mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{r}^2}$ $\begin{aligned} &=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 \pi \rho_0 r^3}{r^2}\left[\frac{1}{3}-\frac{r}{4 R}\right] \\ &=\frac{\rho_0}{\varepsilon_0}\left[\frac{r}{3}-\frac{r^2}{4 R}\right] \end{aligned}$

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