(a) Derivation of $\mathit{E}$ along the axial line of dipole
(b) Graph between $\mathit{E}$ vs $\mathit{r}$
(c) (i) Diagrams for stable and unstable equilibrium of dipole
(ii) Torque on the dipole in the two cases

Electric field at $\mathrm{P}$ due to charge
Electric field at $\mathit{P}$ due to charge $-\mathit{q}$ is ${\mathit{E}}_2=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{\mathit{q}}{{\left(\mathit{r}+\mathit{a}\right)}^2}$
Net electric Field at $\mathit{P}$
${\mathit{E}}_1-{\mathit{E}}_2=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_{\mathrm{o}}}\frac{\mathit{q}}{{\left(\mathit{r}-\mathit{a}\right)}^2}$ $-\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_{\mathrm{o}}}\frac{\mathit{q}}{{\left(\mathit{r}-\mathit{a}\right)}^2}$
$=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{2\mathit{p}\mathit{r}}{{\left({\mathit{r}}^2-{\mathit{a}}^2\right)}^2}\left(\mathit{p}=\mathit{q}.2\mathit{a}\right)$
Its direction is parallel to $\overrightarrow{\mathit{p}}$.
(For short Dipole $=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{2\mathit{p}}{{\mathit{r}}^3}$ without drawing the graph)

(i) For stable Equilibrium: $\overrightarrow{\mathit{p}}$ is parallel to $\overrightarrow{\mathit{E}}$.
(ii) For unstable equilibrium: $\overrightarrow{\mathit{p}}$ is antiparallel to $\overrightarrow{\mathit{E}}$
Torque $=0$ for (i) as well as case (ii).
(Also accept, $\overrightarrow{\mathit{\tau }}=\overrightarrow{\mathit{p}}\times \overrightarrow{\mathit{E}}∕\mathit{\tau }=\mathit{p}\mathit{E}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$
OR
(a) Using Gauss's theorem to find $\mathit{E}$ due to an infinite plane sheet of charge
(b) Expression for the work done to bring charge $\mathit{q}$ from infinity to $\mathit{r}$
(a)
$\oint \mathit{E}.\mathit{d}\mathit{s}=\frac{\mathit{q}}{{\mathit{\epsilon }}_0}$
The electric field $\mathit{E}$ points outwards normal to the sheet. The field lines are parallel to the Gaussian surface except for surfaces 1 and 2 .
Hence the net flux $=\oint \mathit{E}.\mathit{d}\mathit{s}=\frac{\mathit{q}}{{\mathit{\epsilon }}_0}=\frac{\mathit{\sigma }\mathit{A}}{{\mathit{\epsilon }}_0}=2\mathit{E}\mathit{A}$
where $\mathit{A}$ is the area of each of the surface 1 and 2.
$\therefore \oint \mathit{E}\cdot \mathit{d}\mathit{s}=\frac{\mathit{q}}{{\mathit{\epsilon }}_0}=\frac{\mathit{\sigma }\mathit{A}}{{\mathit{\epsilon }}_0}=2\mathit{E}\mathit{A}$
$\mathit{E}=\frac{\mathit{\sigma }}{2{\mathit{\epsilon }}_0}$
(b) $\mathit{W}=\mathit{q}\underset{\mathit{\infty }}{\overset{\mathit{r}}{\int }}\overrightarrow{\mathit{E}}\cdot \overrightarrow{\mathit{d}}\mathit{r}$
$=\mathit{q}\underset{\mathit{\infty }}{\overset{\mathit{r}}{\int }}\left(-\mathit{E}\mathit{d}\mathit{r}\right)$
$=-\mathit{q}\underset{\mathit{\infty }}{\overset{\mathit{r}}{\int }}\left(\frac{\mathit{\sigma }}{2{\mathit{\epsilon }}_0}\right)\mathit{d}\mathit{r}$
$=\frac{\mathit{q}\mathit{\sigma }}{2{\mathit{\epsilon }}_0}\left|\mathit{\infty }-\mathit{r}\right|$
$=\left(\mathit{\infty }\right)$