Question

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

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