Question

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

Answer 1

According to the question

$Q={\int }_0^{R}p\left(r\right)dv$ $[dv=4\pi r^{2}dr]$

$\Rightarrow Q={\int }_0^R\frac{A}{r^2}{e}^{-2r/a}4\pi r^{2}dr$

$Q=4\pi A{\int }_0^R{e}^{-2r/a}dr=4A\pi {[-e^{-2r/a}\times \frac{a}{2}]}_0^R=\frac{4\pi Aa}{2}(-e^{-2R/a}+1)$

$Q=2\pi Aa(1-e^{-2R/a})$

Find radius of the sphere

$1-e^{-2R/a}=\frac{Q}{2\pi Aa}$

$1-\frac{Q}{2\pi Aa}=e^{-2R/a}$

$ln(1-\frac{Q}{2\pi Aa})=\frac{-2R}{a}$

$\frac{-a}{2}ln(1-\frac{Q}{2\pi Aa})=R$

$R=\frac{a}{2}ln\left(\frac{2\pi Aa}{2\pi Aa-Q}\right)$.