# 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 :

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Solution:

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

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