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 $r=\frac{R}{2}$ is $\frac{1}{8}$ times that at $r=R$, find the value of $a$

Question

Aicharya · Accepted Answer

Let us consider a spherical shell of radius $x$ and thickness $d x .$ The volume of this shell is $4 \pi x^{2}(d x)$. The charge enclosed in this spherical shell is

$d q=\left(4 \pi x^{2}\right) d x \times k x^{a}$
$\therefore \quad d q=4 \pi k x^{2+a} d x$
For $r=R:$
The total charge enclosed in the sphere of radius $R$ is
$
Q=\int_{0}^{R} 4 \pi k x^{2+a} d x=4 \pi k \frac{R^{3+a}}{3+a} \text { . }
$
$\therefore \quad$ The electric field at $r=R$ is
$
E_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{4 \pi k R^{3+a}}{(3+a) R^{2}}=\frac{1}{4 \pi \varepsilon_{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} d x=\frac{4 \pi k(R / 2)^{3+a}}{3+a}
$
$\therefore \quad$ The electric field at $r=R / 2$ is
$E_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{4 \pi k}{3+a} \frac{(R / 2)^{3+a}}{(R / 2)^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{4 \pi k}{3+a}\left(\frac{R}{2}\right)^{1+a}$
Given, $E_{2}=\frac{1}{8} E_{1}$
$
\begin{array}{l}
\therefore \quad \frac{1}{4 \pi \varepsilon_{0}} \frac{4 \pi k}{(3+a)}\left(\frac{R}{2}\right)^{1+a}=\frac{1}{2^{3}} \times \frac{1}{4 \pi \varepsilon_{0}} \frac{4 \pi k}{3+a} R^{1+a} \\
\Rightarrow \quad 1+a=3 \Rightarrow a=2
\end{array}
$

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho=\kappa…

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