Question

Let there be a spherically symmetric charge density varying as $\rho \left(r\right)={\rho }_0(\frac{5}{4}-\frac{r}{R})$ upto $r=R$,and $\rho \left(r\right)=0$ 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. $\frac{{\rho }_{0}r}{{3\epsilon }_0}(\frac{5}{4}-\frac{r}{R})$
• B. $\frac{4\pi {\rho }_{0}r}{{3\epsilon }_0}(\frac{5}{3}-\frac{r}{R})$
• C. $\frac{{\rho }_{0}r}{{4\epsilon }_0}(\frac{5}{3}-\frac{r}{R})$
• D. $\frac{{4\rho }_{0}r}{{3\epsilon }_0}(\frac{5}{4}-\frac{r}{R})$

Answer 1

Answer: (A)

$\frac{{\rho }_{0}r}{{3\epsilon }_0}(\frac{5}{4}-\frac{r}{R})$

$\rho \left(r\right)={\rho }_0(\frac{5}{4}-\frac{r}{R})$

$r=R$ $\rho \left(r\right)=0$

$r>R$

$r+$ the distance from the origin

The electric field at a distance $=r(r<R)$

Apply shell theorem, the total charge up to distance $r$ can be calculated as follows

$d\rho =4\pi r^{2}dr.\rho$

$4\pi r^2.dr.{\rho }_0(\frac{5}{4}-\frac{r}{R})$

($\because$ $dq=\rho dv$)

$=4\pi {\rho }_0(\frac{5}{4}{r}^{2}dr-\frac{r^3}{R}dr)$

$=\int dq-q=4\pi {\rho }_0{\int }_0^t\frac{5}{4}{r}^{2}dr-\frac{r^3}{R}dr$

$=4\pi {\rho }_0[\frac{5}{4}\frac{r^3}{3}-\frac{1}{R}\frac{r^4}{4}]$

As the electric field,

$E=\frac{K_q}{r^2}$

$=\frac{1}{4\pi {ϵ}_0}\frac{1}{r^2}4\pi {\rho }_0[\frac{5}{q}\left(\frac{r^3}{5}\right)-\frac{r^4}{4R}]$

$E=\frac{{\rho }_{0}r}{4{ϵ}_0}(\frac{5}{3}-\frac{r}{R})$