Question

Let there be a spherically symmetric charge distribution with charge density varying as $p\left(r\right)=p_0(\frac{5}{4}-\frac{r}{R})$ upto $r=R$, and $p\left(r\right)=0$ 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 :

• A. $\frac{4\pi p_{0}r}{3{ϵ}_0}(\frac{5}{3}-\frac{r}{R})$
• B. $\frac{p_{0}r}{4{ϵ}_0}(\frac{5}{3}-\frac{r}{R})$
• C. $\frac{4p_{0}r}{3{ϵ}_0}(\frac{5}{4}-\frac{r}{R})$
• D. $\frac{p_{0}r}{4{ϵ}_0}(\frac{5}{4}-\frac{r}{R})$

Answer 1

Answer: (B)

$\frac{p_{0}r}{4{ϵ}_0}(\frac{5}{3}-\frac{r}{R})$

Apply shell theorem the total charge upto distance $r$ can be calculated as followed
dq $=4\pi r^2$ .dr.p
$=4\pi r^2.dr.p_0[\frac{5}{4}-\frac{r}{R}]$
$=4\pi p_0[\frac{5}{4}{r}^{2}dr-\frac{r^3}{R}dr]$
$\int dq=q=4\pi p_0{\int }_0^r(\frac{5}{4}{r}^{2}dr-\frac{r^3}{R}dr)$
$=4\pi p_0[\frac{5}{4}\frac{r^3}{3}-\frac{]}{R}\frac{r^4}{4}]$
$E=\frac{kq}{r^2}$
$=\frac{1}{4\pi {ϵ}_0{r}^2}.4\pi p_0\left[\frac{5}{4}\right(\frac{r^3}{3})-\frac{r^4}{4R}]$
$E=\frac{p_{0}r}{4{ϵ}_0}[\frac{5}{3}-\frac{r}{R}]$