Question

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho =k{r}^a$, where $k$ and a are constants and $r$ is the distance from its center. If the electric field at $r=\frac{R}{2}$ is times that at $r=R$, the value of $a$ is

• A. 3
• B. 5
• C. 2
• D. 7

Answer 1

Answer: (C)

$\oint \vec{E}\cdot d\vec{A}=\frac{1}{{\epsilon }_0}\int (\rho dv)$
$=\frac{1}{{\epsilon }_0}\int k{r}^a\times 4\pi r^{2}dr$
or $E\times 4\pi R^2=\left(\frac{4\pi k}{{\epsilon }_0}\right)\frac{R^{(a+3)}}{(a+3)}$
$\therefore E_1=\frac{k{R}^{(a+1)}}{{\epsilon }_0(a+3)}$
$E$ For $r=\frac{R}{2}\cdot E_2=\frac{k{\left(\frac{R}{2}\right)}^{a+1}}{{\epsilon }_0(a+3)}$
Given $E_2=\frac{E_1}{8}$
or $\frac{k{\left(\frac{R}{2}\right)}^{a+1}}{{\epsilon }_0(a+3)}=\frac{1}{8}\frac{k{R}^{(a+1)}}{{\epsilon }_0(a+3)}$
$\therefore \frac{1}{2^{a+3}}=\frac{1}{8}$
or $a=2$.