A spherically symmetric charge distribution is considered with charge density varying as
$ρ (r) = \{\begin{matrix} ρ_{0} (\frac{3}{4} - \frac{r}{R}) & for r ⩽ R \\ Zero & for r > R \end{matrix}\}$
Where, r(r < R) is the distance from the centre O (as shown in figure). The electric field at point P will be :

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$\frac{ρ_{0} r}{4 ε_{0}} (1 - \frac{r}{R})$

$\frac{ρ_{0} r}{5 ε_{0}} (1 - \frac{r}{R})$

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

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

answer is C.

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Detailed Solution

By Gauss law
Question Image
$\begin{array}{l} ∮ \vec{E} . d \vec{s} = \frac{Q_{in}}{ε_{0}} \\ \Rightarrow E . 4 {πr}^{2} = \frac{\int_{0}^{r} ρ_{0} (\frac{3}{4} - \frac{r}{R}) 4 {πr}^{2} dr}{ε_{0}} \\ \Rightarrow E 4 {πr}^{2} = \frac{ρ_{0} 4 π}{ε_{0}} (\frac{3}{4} \frac{r^{3}}{3} - \frac{r^{4}}{4 R}) \\ \Rightarrow {Er}^{2} = \frac{ρ_{0} r^{3}}{4 ε_{0}} \{1 - \frac{r}{R}\} \\ \Rightarrow E = \frac{ρ_{0} r}{4 ε_{0}} \{1 - \frac{r}{R}\} \end{array}$