A spherically symmetric charge distribution is characterized by a charge density having the following variation : $ρ (r) = ρ_{0} (1 - \frac{r}{R})$ for r < R $ρ (r) = 0$ for r $\geq$ R. Where r is the distance from the centre of the charge distribution and $ρ_{0}$ is a constant. The electric field at an internal point (r<R) is

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$\frac{ρ_{0}}{4 \in_{0}} (\frac{r}{3} - \frac{r^{2}}{4 R})$

$\frac{ρ_{0}}{3 \in_{0}} (\frac{r}{3} - \frac{r^{2}}{4 R})$

$\frac{ρ_{0}}{12 \in_{0}} (\frac{r}{3} - \frac{r^{2}}{4 R})$

$\frac{ρ_{0}}{ϵ_{0}} (\frac{r}{3} - \frac{r^{2}}{4 R})$

answer is A.

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Detailed Solution

When charge density is not constant total charge enclosed upto a distance r can be obtained by

$q = \int 4 {πr}^{2} dr ρ$

$q = \int 4 {πr}^{2} {drρ}_{0} (1 - \frac{r}{R})$

$on integration q = 4 {πρ}_{0} (\frac{r^{3}}{3} - \frac{r^{4}}{4 R})$

$E . 4 {πr}^{2} = \frac{q}{ε_{0}}$

$E = 4 {πρ}_{0} (\frac{r^{3}}{3} - \frac{r^{4}}{4 R}) \frac{1}{4 π ε_{0} r^{2}}$

$E = \frac{ρ_{0}}{ϵ_{0}} (\frac{r}{3} - \frac{r^{2}}{4 R})$

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