Question

The volume charge density of a sphere carrying charge $Q$ and of radius $R$ is proportional to the square of the distance from the centre. The ratio of the magnitude of the electric field at a distance $2R$ from the centre to that at a distance of $\frac{R}{2}$ from the centre is

Answer 1

Answer: 2

As given,
$\rho =C{r}^2$ [where $\rho =$ volume charge density]
$q(r)=\underset{0}{\overset{r}{\int }}\left(4\pi r^{2}dr\right)\rho$
$=\underset{0}{\overset{r}{\int }}4\pi r^2\cdot C{r}^{2}dr=\frac{4}{5}\pi C{r}^5$
${\left(\text{E}|\right)}_{r=2\text{R}}=\frac{{\left(\text{kq}\right)}_{(2\text{R})}}{(2\text{R}{)}^2}=\frac{\text{k}(\frac{4}{5})\pi {\left(\text{CR}\right)}^5}{4{\left(\text{R}\right)}^2}$
..... $[\because$ sphere has radius $R;$ so $r\le R$ for enclosed charge]
$=\frac{k\pi C{R}^3}{5}$
${\left(\text{E}|\right)}_{r=\text{R}/2}=\frac{{\left(\text{kq}\right)}_{(\text{R}/2)}}{(\frac{\text{R}}{2}{)}^2}=\frac{\text{k}(\frac{4}{5})\pi \text{C}(\frac{\text{R}}{2}{)}^5}{(\frac{\text{R}}{2}{)}^2}=\frac{\text{k}\pi {\left(\text{CR}\right)}^3}{5\times 2}$
$\Rightarrow \frac{E_{(2R)}}{E_{(R/2)}}=2$