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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 $a$.
$1$
$2$
$3$
$4$
Solution
Using Gauss's law, we have
$\oint \vec{E} \cdot d \vec{A}=\frac{1}{\epsilon_0} \int(\rho d v)=\frac{1}{\epsilon_0} \int_0^R k r^{ a } \times 4 \pi r^2 d r$
$\text { or } E \times 4 \pi R^2=\left(\frac{4 \pi k}{\epsilon_0}\right) \frac{R^{(a+3)}}{( a +3)}$
$\text { For } r =\frac{ R }{2} \cdot E _2=\frac{ k \left(\frac{R}{2}\right)^{a+1}}{\epsilon_0(a+3)}$
Given, $E _2=\frac{E_1}{8}$
$\text { or } \frac{k\left(\frac{R}{2}\right)^{a+1}}{\epsilon_0(a+3)}=\frac{1 k R^{(a+1)}}{8 \epsilon_0(a+3)}$
$\therefore 2^{\frac{1}{a+1}}=\frac{1}{8}$
or $a =2$.
