$\oint \vec{E} \cdot d \vec{A}=\frac{1}{\in_0} \int(\rho d v)=\frac{1}{\in_0} \int_0^R k r^{ a } \times 4 \pi r^2 d r$
$\text { or } E \times 4 \pi R^2=\left(\frac{4 \pi k}{\in_0}\right) \frac{R^{(a+3)}}{( a +3)}$
$\text { For } r =\frac{ R }{2} \cdot E _2=\frac{ k \left(\frac{R}{2}\right)^{a+1}}{\in_0(a+3)}$
Given, $E _2=\frac{E_1}{8}$
$\text { or } \frac{k\left(\frac{R}{2}\right)^{a+1}}{\in_0(a+3)}=\frac{1 k R^{(a+1)}}{8 \in_0(a+3)}$
$\therefore 2^{\frac{1}{a+1}}=\frac{1}{8}$
or $a =2$.
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Reason : Mass of the electrons is very small
