Question

The potential due to an electrostatic charge distribution is $V(r)=\frac{q{e}^{-\alpha r}}{4\pi {\epsilon }_{0}r}$ where a is positive. The net charge within a sphere centred at the origin and of radius $1/\alpha$ is

• A. $2q/e$
• B. $(1-1/e)q$
• C. $q/e$
• D. $(1+1/e)q$

Answer 1

Answer: (A)

Electric field due to given charge distribution is
$E=-\frac{dV}{dr}=-\frac{d}{dr}\frac{q{e}^{-\alpha r}}{4\pi {\epsilon }_{0}r}$
$=\frac{-q}{4\pi {\epsilon }_0}\frac{d}{dr}\left(\frac{e^{-\alpha r}}{r}\right)$
$=\frac{-q}{4\pi {\epsilon }_0}\left(\frac{-\alpha r{e}^{-\alpha r}-e^{-\alpha r}}{r^2}\right)$
$=\frac{q}{4\pi {\epsilon }_0}\cdot e^{-\alpha r}\cdot \left(\frac{\alpha r+1}{r^2}\right)$
Electric field at $r=\frac{1}{\alpha }$ is
$E\left(r=\frac{1}{\alpha }\right)=\frac{q}{4\pi {\epsilon }_0}\cdot e^{-\alpha \times \frac{1}{\alpha }}\left(\frac{\alpha \times \frac{1}{\alpha }+1}{1/{\alpha }^2}\right)$
$=\frac{(q/e)}{4\pi {\epsilon }_0}\cdot 2{\alpha }^2$
Flux through a sphere of radius $\frac{1}{\alpha }$ is
$\varphi =\int E\cdot dA$
For spherical distribution,
$E\cdot dA=EdA\cos 0^{\circ }=EdA$
and $E=$ uniform.
So, we have
$\varphi =E\int dA$
$=\frac{q/e}{4\pi {\epsilon }_0}\cdot 2{\alpha }^2\cdot 4\pi {\left(\frac{1}{\alpha }\right)}^2$
$\varphi =\frac{2q}{{\epsilon }_{0}e}$
From Gauss' law, we have $\varphi =\frac{q_{\text{enclosed }}}{{\epsilon }_0}$
So, charge enclosed in given sphere is $\frac{2q}{e}$.