JEE Advance - Physics (2009 - Paper 2 Offline - No. 17)
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 $$r = R/2$$ is 1/8 times than at $$r = R$$, find the value of $$a$$.
Answer
2
Explanation
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 $$a = 2$$.
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