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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 centre. If the electric field at $r=$ $\frac{R}{2}$ is $\frac{1}{8}$ times that at $r=R,$ the value of $a$ is
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The correct answer is:
2
Using Gauss's law, we have
$$
\oint \vec{E} \cdot d \vec{A}=\frac{1}{\epsilon_{0}} \int(\rho d v)=\frac{1}{\epsilon_{0}} \int_{0}^{R} k r^{a} \times 4 \pi r^{2} d r
$$
or $E \times 4 \pi R^{2}=\left(\frac{4 \pi k}{\epsilon_{0}}\right) \frac{R^{(a+3)}}{(a+3)}$
$$
\therefore \quad E_{1}=\frac{k R^{(a+1)}}{\epsilon_{0}(a+3)}
$$
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 $\frac{k\left(\frac{R}{2}\right)^{a+1}}{\epsilon_{0}(a+3)}=\frac{1}{8} \frac{k R^{(a+1)}}{\epsilon_{0}(a+3)}$
$$
\begin{array}{ll}
\therefore & \quad \frac{1}{a+1}=\frac{1}{8} \\
\text { or } & a=2 .
\end{array}
$$
$$
\oint \vec{E} \cdot d \vec{A}=\frac{1}{\epsilon_{0}} \int(\rho d v)=\frac{1}{\epsilon_{0}} \int_{0}^{R} k r^{a} \times 4 \pi r^{2} d r
$$
or $E \times 4 \pi R^{2}=\left(\frac{4 \pi k}{\epsilon_{0}}\right) \frac{R^{(a+3)}}{(a+3)}$
$$
\therefore \quad E_{1}=\frac{k R^{(a+1)}}{\epsilon_{0}(a+3)}
$$
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 $\frac{k\left(\frac{R}{2}\right)^{a+1}}{\epsilon_{0}(a+3)}=\frac{1}{8} \frac{k R^{(a+1)}}{\epsilon_{0}(a+3)}$
$$
\begin{array}{ll}
\therefore & \quad \frac{1}{a+1}=\frac{1}{8} \\
\text { or } & a=2 .
\end{array}
$$
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