Question

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

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

Answer 1

Answer: (D)

$\frac{a}{2} \log\left( \frac{1}{1- \frac{Q}{2\pi aA}}\right)$

Answer 2

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