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(B) The total charge $Q$ is given by the integral of the volume charge density over the volume of the sphere:$Q = \int_0^R \rho(r) \cdot 4\pi r^2 dr$S

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The total charge $Q$ is given by the integral of the volume charge density over the volume of the sphere:$Q = \int_0^R \rho(r) \cdot 4\pi r^2 dr$Substituting $\rho(r) = \frac{A}{r^2} e^{-2r/a}$:$Q = \int_0^R \left( \frac{A}{r^2} e^{-2r/a} \right) (4\pi r^2) dr = 4\pi A \int_0^R e^{-2r/a} dr$Evaluating the integral:$Q = 4\pi A \left[ \frac{e^{-2r/a}}{-2/a} \right]_0^R = 4\pi A \left( -\frac{a}{2} \right) (e^{-2R/a} - e^0)$$Q = -2\pi aA (e^{-2R/a} - 1) = 2\pi aA (1 - e^{-2R/a})$Rearranging to solve for $R$:$1 - e^{-2R/a} = \frac{Q}{2\pi aA}$$e^{-2R/a} = 1 - \frac{Q}{2\pi aA}$Taking the natural logarithm on both sides:$-\frac{2R}{a} = \log \left( 1 - \frac{Q}{2\pi aA} \right)$$R = -\frac{a}{2} \log \left( 1 - \frac{Q}{2\pi aA} \right) = \frac{a}{2} \log \left( \frac{1}{1 - \frac{Q}{2\pi aA}} \right)$

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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