KVPY SX SB Exam 2018 Question Paper

(a) Electric field due to given charge distribution is
$\mathit{E}=-\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{r}}=-\frac{\mathit{d}}{\mathit{d}\mathit{r}}\frac{\mathit{q}{\mathit{e}}^{-\mathit{\alpha }\mathit{r}}}{4\mathit{\pi }{\mathit{\epsilon }}_0\mathit{r}}$
$=\frac{-\mathit{q}}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{\mathit{d}}{\mathit{d}\mathit{r}}\left(\frac{{\mathit{e}}^{-\mathit{\alpha }\mathit{r}}}{\mathit{r}}\right)$
$=\frac{-\mathit{q}}{4\mathit{\pi }{\mathit{\epsilon }}_0}\left(\frac{-\mathit{\alpha }\mathit{r}{\mathit{e}}^{-\mathit{\alpha }\mathit{r}}-{\mathit{e}}^{-\mathit{\alpha }\mathit{r}}}{{\mathit{r}}^2}\right)$
$=\frac{\mathit{q}}{4\mathit{\pi }{\mathit{\epsilon }}_0}\cdot {\mathit{e}}^{-\mathit{\alpha }\mathit{r}}\cdot \left(\frac{\mathit{\alpha }\mathit{r}+1}{{\mathit{r}}^2}\right)$
Electric field at $\mathit{r}=\frac{1}{\mathit{\alpha }}$ is
$\mathit{E}\left(\mathit{r}=\frac{1}{\mathit{\alpha }}\right)$ $=\frac{\mathit{q}}{4\mathit{\pi }{\mathit{\epsilon }}_0}\cdot {\mathit{e}}^{-\mathit{\alpha }\times \frac{1}{\mathit{\alpha }}}\left(\frac{\mathit{\alpha }\times \frac{1}{\mathit{\alpha }}+1}{1∕{\mathit{\alpha }}^2}\right)$
$=\frac{\left(\mathit{q}∕\mathit{e}\right)}{4\mathit{\pi }{\mathit{\epsilon }}_0}\cdot 2{\mathit{\alpha }}^2$
Flux through a sphere of radius $\frac{1}{\mathit{\alpha }}$ is
$\mathit{\phi }=\int \mathit{E}\cdot \mathit{d}\mathit{A}$
For spherical distribution, $\mathit{E}\cdot \mathit{d}\mathit{A}=\mathit{E}\mathit{d}\mathit{A}\cos 0^{\circ }=\mathit{E}\mathit{d}\mathit{A}$ and $\mathit{E}=$ uniform.
So, we have ∴
$\mathit{\phi }=\mathit{E}\int \mathit{d}\mathit{A}$
$=\frac{\mathit{q}∕\mathit{e}}{4\mathit{\pi }{\mathit{\epsilon }}_0}\cdot 2{\mathit{\alpha }}^2\cdot 4\mathit{\pi }{\left(\frac{1}{\mathit{\alpha }}\right)}^2$
or $\mathit{\phi }=\frac{2\mathit{q}}{{\mathit{\epsilon }}_0\mathit{e}}$
From Gauss' law, we have
So, charge enclosed in given sphere is $\frac{2\mathit{q}}{\mathit{e}}$.