Question 41 of 311

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

A$\dfrac{a}{2}\log ⁡\left(\dfrac{1}{1-\dfrac{Q}{2\pi aA}}\right)$B$a\log ⁡\left(\dfrac{1}{1-\dfrac{Q}{2\pi aA}}\right)$C$a\log ⁡\left(1-\dfrac{Q}{2\pi aA}\right)$D$\dfrac{a}{2}\log ⁡\left(1-\dfrac{Q}{2\pi aA}\right)$

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Answer: (A) $\dfrac{a}{2}\log ⁡\left(\dfrac{1}{1-\dfrac{Q}{2\pi aA}}\right)$

$\rho \left(r\right)=\dfrac{A}{{r}^{2}}{e}^{-\dfrac{2r}{a}}$



Charge enclosed between $r$ and $r+dr$ is

$dq=\rho \left(r\right)4\pi {r}^{2}dr$

To get total charge ${Q}^{}$

$Q=\int _{0}^{R}dq=\int _{0}^{R}\dfrac{A}{{r}^{2}}{e}^{-2r/a}4\pi {r}^{2}dr$$=-4\pi A\dfrac{a}{2}{[{e}^{-2r/a}]}_{0}^{R}$

$Q=-4\pi A\dfrac{a}{2}\left[{e}^{-2R/a}-1\right]$

$R=-\ln ⁡\left[1-\dfrac{Q2}{4\pi Aa}\right]\dfrac{a}{2}=\dfrac{a}{2}\ln ⁡\left(\dfrac{1}{1-\dfrac{Q}{2\pi aA}}\right)$