Question

A sphere of radius $R$ carries charge such that its volume charge density is proportional to the square of the distance from the centre. What is the ratio of the magnitude of the electric field at a distance $2R$ from the centre to the magnitude of the electric field at a distance of $R/2$ from the centre (i.e., $E_{r=2R}/E_{r=R/2})?$

• A. 2
• B. 1.5
• C. 3
• D. 2.5

Answer 1

Answer: (A)

$\rho =C{r}^2\Rightarrow$ $q=\underset{0}{\overset{r}{\int }}4\pi r^2\cdot dr\rho =\underset{0}{\overset{r}{\int }}4\pi C{r}^{4}dr$
$q=\frac{4}{5}\pi C{r}^5$
$E_{r=2R}=\frac{k{q}_{(r=R)}}{(2R{)}^2}=\frac{k(4/5)\pi C{R}^5}{4R^2}$
$E_{r=R/2}=\frac{k{q}_{(r=R/2)}}{(R/2{)}^2}=\frac{k(4/5)\pi C(R/2{)}^5}{(R/2{)}^2}$
Now, $\frac{E_{r=2R}}{E_{r=R/2}}=2$