Question

A sphere carrying charge $Q$ and of radius $R$ has its volume charge density proportional to the square of the distance from the centre. Determine 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.

Answer 1

Answer: 2

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