(a) Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction.
(b) Derive an expression for the electric potential at any point along the axial line of an electric dipole. 

(a) 

(b) Electric potential due to dipole along axial line:

The figure shows a dipole consisting of two equal and opposite charges +q and -q  separated by a distance 2 l. Let, P be a point on the end- on position of the dipole. OP is the axial line of dipole.
The potential due to charge +q at point
$\begin{array}{l}\mathrm{P}={\mathrm{V}}_1\\ {\mathrm{V}}_1=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{(\mathrm{r}-\mathrm{l})}\end{array}$
The potential due to charge -q at point P=V₂
${\mathrm{V}}_2=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{(-\mathrm{q})}{(\mathrm{r}+\mathrm{l})}$
Total potential at point $\mathrm{P}=\mathrm{V}={\mathrm{V}}_1+{\mathrm{V}}_2$
$\begin{array}{l}\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{(\mathrm{r}-\mathrm{l})}+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{(-\mathrm{q})}{(\mathrm{r}+\mathrm{l})}\\ =\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{1}{\mathrm{r}-\mathrm{l}}-\frac{1}{\mathrm{r}+\mathrm{l}}\right]\\ =\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{r}+\mathrm{l}-\mathrm{r}+\mathrm{l}}{(\mathrm{r}-\mathrm{l})(\mathrm{r}+\mathrm{l})}\right]\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2\mathrm{ql}}{(\mathrm{r}-\mathrm{l})(\mathrm{r}+\mathrm{l})}\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2\mathrm{ql}}{\left({\mathrm{r}}^2-{\mathrm{l}}^2\right)}\end{array}$
$\mathrm{But} \mathrm{the} \mathrm{dipolemoment},$$\mathrm{P}=2\mathrm{lq}$
 So, $\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{P}}{\left({\mathrm{r}}^2-{\mathrm{l}}^2\right)}$