Question

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho =\kappa r^a$, where $\kappa$ and $a$ are constants and $r$ is the distance from its centre.
If the electric field at $r=\frac{R}{8}$ is $\frac{1}{8}$ times that at $r=R$, find the value of $a$ .

• A. $2.0$
• B. $3.2$
• C. $2.5$
• D. $0.2$

Answer 1

The correct option is A $2.0$

Let us consider a spherical shell of radius $x$ and thickness $dx$. The volume of this shell is $4\pi x^2\left(dx\right)$. The charge enclosed in this sperical shell is
$dq=4\pi x^2\left(dx\right)\times k{x}^a$
$\therefore dq=4\pi k{x}^{2+a}dx$
For $r=R$
The total charge enclosed in the sphere of radius $R$ is
$Q={\int }_0^{R}4\pi k{x}^{2+a}dx=4\pi k\frac{R^{2+a}}{3+a}$
The electric field at $r=R$ is
$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{4\pi k{R}^{2+a}}{(3+a)R^2}=\frac{1}{4\pi {\epsilon }_0}\frac{4\pi k}{3+a}{R}^{1+a}$
For $r=R/2$
The total charge enclosed in the sphere of radius $R/2$ is
$Q^{\prime }={\int }_0^{R/2}4\pi k{x}^{2+a}dx=\frac{4\pi k{(R+2)}^{3+a}}{3+a}$
The electric field at $r=R/2$ is
$E_2=\frac{1}{4\pi {\epsilon }_0}\frac{4\pi k}{3+a}\frac{{(R+2)}^{3+a}}{{\left(R/2\right)}^2}=\frac{1}{4\pi {\epsilon }_0}\frac{4\pi k}{3+a}{\left(\frac{R}{2}\right)}^{1+a}$
Given $E_2=\frac{1}{8}{E}_1$
$\therefore \frac{1}{4\pi {\epsilon }_0}\frac{4\pi k}{3+a}{\left(\frac{R}{2}\right)}^{1+a}=\frac{1}{2^3}\times \frac{1}{4\pi {\epsilon }_0}\frac{4\pi k}{3+a}{R}^{1+a}$
$\Rightarrow 1+a=3\Rightarrow a=2$