Let there be a spherically symmetric charge distribution with charge density varying as \rho ( r ) = \rho _0( 5 \over 4 - r \over R ) upto r=R, and \rho ( r ) = 0 for r>R, where r is the distance from the erigin. The electric field at a distance r( r \lt R ) from the origin is given by

\rho _0r \over 4\varepsilon _0( 5 \over 3 - r \over R )

4\pi \rho _0r \over 3\varepsilon _0( 5 \over 3 - r \over R )

4\rho _0r \over 4\varepsilon _0( 5 \over 4 - r \over R )

\rho _0r \over 3\varepsilon _0( 5 \over 4 - r \over R )

Electrostatics PYQ — JEE Main 2010 · Shift 0 · Q69 with Solution

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Correct answer: A

For a spherically symmetric charge distribution, the electric field at radius $r$ is radial and has magnitude $E(r)$ . By Gauss's law,

$E(r)\cdot 4\pi r^2 = \frac{Q_{\text{enc}}(r)}{\varepsilon_0}$

So,

$E(r)=\frac{Q_{\text{enc}}(r)}{4\pi \varepsilon_0 r^2}$

for $r<R$ .

Given

$\rho(r)=\rho_0\left(\frac{5}{4}-\frac{r}{R}\right), \qquad r\le R$

The charge enclosed within radius $r$ is

$Q_{\text{enc}}(r)=\int_0^r \rho(r')\, dV = \int_0^r \rho(r')\, 4\pi r'^2\,dr'$

Thus,

$Q_{\text{enc}}(r)=4\pi \rho_0\int_0^r \left(\frac{5}{4}-\frac{r'}{R}\right) r'^2\,dr'$

Expand the integrand:

$Q_{\text{enc}}(r)=4\pi \rho_0\left[\frac{5}{4}\int_0^r r'^2\,dr' - \frac{1}{R}\int_0^r r'^3\,dr'\right]$

Now evaluate the integrals:

$\int_0^r r'^2\,dr' = \frac{r^3}{3}, \qquad \int_0^r r'^3\,dr' = \frac{r^4}{4}$

So,

$Q_{\text{enc}}(r)=4\pi \rho_0\left[\frac{5}{4}\cdot \frac{r^3}{3} - \frac{1}{R}\cdot \frac{r^4}{4}\right]$

$Q_{\text{enc}}(r)=4\pi \rho_0\left(\frac{5r^3}{12}-\frac{r^4}{4R}\right)$

$E(r)=\frac{1}{4\pi \varepsilon_0 r^2}\cdot 4\pi \rho_0\left(\frac{5r^3}{12}-\frac{r^4}{4R}\right)$

Simplify:

$E(r)=\frac{\rho_0}{\varepsilon_0 r^2}\left(\frac{5r^3}{12}-\frac{r^4}{4R}\right)$

$E(r)=\frac{\rho_0}{\varepsilon_0}\left(\frac{5r}{12}-\frac{r^2}{4R}\right)$

Factor out $\dfrac{\rho_0 r}{4\varepsilon_0}$ :

$E(r)=\frac{\rho_0 r}{4\varepsilon_0}\left(\frac{5}{3}-\frac{r}{R}\right)$

This matches Option A:

$\boxed{\frac{\rho_0 r}{4\varepsilon_0}\left(\frac{5}{3}-\frac{r}{R}\right)}$

Stored correct answer: A

Derived answer: A

So they agree.