JEE MAIN - Physics (2016 (Offline) - No. 13)
The region between two concentric spheres of radii $$'a'$$ and $$'b',$$ respectively (see figure), have volume charge density $$\rho = {A \over r},$$ where $$A$$ is a constant and $$r$$ is the distance from the center. A such that the electric field in the region between the spheres will be constant, is :
_en_13_1.png)
$${{2Q} \over {\pi \left( {{a^2} - {b^2}} \right)}}$$
$${{2Q} \over {\pi \,{a^2}}}$$
$${Q \over {2\pi \,{a^2}}}$$
$${Q \over {2\pi \,\left( {{b^2} - {a^2}} \right)}}$$
Explanation
_en_13_2.png)
Applying Gauss's law
$$\oint {s\overrightarrow E .\overrightarrow {ds} } = {Q \over {{ \in _0}}}$$
$$\therefore$$ $$E \times 4\pi {r^2} = {{Q + 4\pi a{r^2} - 4\pi A{a^2}} \over {{ \in _0}}}$$
$$\rho = {{dr} \over {dv}}$$
$$Q = \rho 4\pi {r^2}$$
$$Q = \int\limits_a {{A \over r}} 4{\pi ^2}dr = 4\pi A\,\left[ {{r^2} - {a^2}} \right]$$
$$E = {1 \over {4\pi { \in _0}}}\left[ {{{Q - 4\pi A{a^2}} \over {{r^2}}} + 4\pi A} \right]$$
For $$E$$ to be independent of $$'r'$$
$$Q - 2\pi A{a^2} = 0$$
$$\therefore$$ $$A = {Q \over {2\pi {a^2}}}$$
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