Let there be a spherically symmetric charge distribution with charge density varying as $\rho(r)=\rho_{0}\left(\frac{5}{4}-\frac{r}{R}\right)$ upto $r=R$, and $\rho(r)=0$ for $r>R$, where $r$ is the distance from the origin. The electric field at a distance $r(r<R)$ from the origin is given by

(B) $\frac{\rho_{0} r}{4 \varepsilon_{0}}\left(\frac{5}{3}-\frac{r}{R}\right)$

(A) $\frac{4 \pi \rho_{0} r}{3 \varepsilon_{0}}\left(\frac{5}{3}-\frac{r}{R}\right)$

(C) $\frac{4 \rho_{0} \mathrm{r}}{3 \varepsilon_{0}}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(D) $\frac{\rho_{0} r}{3 \varepsilon_{0}}\left(\frac{5}{4}-\frac{r}{R}\right)$

JEE Main · 2010 · 11 Apr 2010 · Offline · Electrostatics

Let there be a spherically symmetric charge distribution with charge density varying as

ρ (r) = ρ_{0} (\frac{5}{4} - \frac{r}{R})

upto

r = R

, and

ρ (r) = 0

for

r > R

, where

r

is the distance from the origin. The electric field at a distance

r (r < R)

from the origin is given by

Answer: (B)

By symmetry the field at radius

r

is radial, and Gauss's law gives

E (r) = \frac{Q _{enc}}{4 π ε _{0} r ^{2}}

, where

Q_{enc}

is the charge within radius

r

. Integrating the density over thin shells,

Q_{enc} = \int_{0}^{r} ρ_{0} (\frac{5}{4} - \frac{r ^{'}}{R}) 4 π r^{'2} d r^{'} = 4 π ρ_{0} (\frac{5 r ^{3}}{12} - \frac{r ^{4}}{4 R}) = π ρ_{0} r^{3} (\frac{5}{3} - \frac{r}{R})

. Dividing by

4 π ε_{0} r^{2}

, the field inside the distribution is

E (r) = \frac{ρ _{0} r}{4 ε _{0}} (\frac{5}{3} - \frac{r}{R})

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option