Q. Charge is distributed within a sphere of radius R with a volume charge density $\rho (r) = \frac{A}{r^2} e^{-2r/a}$ where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :

Solution:

$Q = \int \rho dv $
$ = \int^{R}_{0} \frac{A}{r^{2}} e^{-2r/a} \left(4\pi r^{2} dr \right)$
$ = \int^{R}_{0} \frac{A}{r^{2}} e^{-2r/a} \left(4 \pi r^{2} dr\right) $
$ = 4 \pi A \int^{R}_{0} e^{-2r/a} dr $
$ = 4\pi A \left( \frac{e^{-2r/a}}{- \frac{2}{a}}\right) ^{R}_{0} $
$ = 4\pi A \left(- \frac{a}{2}\right) \left(e^{-2 R/a} -1\right) $
$ Q = 2 \pi aA\left(1 - e^{-2R/a} \right) $
$ R = \frac{a}{2} \log \left(\frac{1}{1- \frac{Q}{2\pi aA}}\right) $

Solition Image

You must select option to get answer and solution

Questions from JEE Main 2019

Physics Most Viewed Questions