QJEEAdv09Ph9be3: JEE Advanced 2009 Physics - Class 12 JEEAdv09Ph9be3 Solution: A solid sphere of radius R has a charge Q distributed in its…

Applying Gauss's theorem, we get

$E(4\pi {r^2}) = {{{q_{encl}}} \over {{t_0}}} = {1 \over t}\int\limits_0^r {k{x^2}(4\pi {x^2})dx}$

Now, $E{r^2} = {k \over {{t_0}}}\int\limits_0^r {{x^{2 + a}}dx = {k \over {{t_0}}}\left( {{{{r^{3 + a}}} \over {3 + a}}} \right)}$

Therefore, $E = {k \over {{t_0}}}{{{r^{1 + a}}} \over {3 + a}}$

That is, $E \propto {r^{1 + a}}$

Now, $E\left( {{R \over 2}} \right) = {1 \over 8}E(R)$

Therefore, ${\left( {{R \over 2}} \right)^{1 + a}} = {1 \over 8}{(R)^{1 + a}} \Rightarrow 8 = {2^{1 + a}}$

where $1 + a = 3$ and hence $a = 2$ .