Deduce an expression for the electric potential due to an electric dipole at any point on its axis. Mention one contrasting feature of electric potential of a dipole at a point as compared to that due to single charge.

Consider, two charges q and -q seperated by distance 2a such that, it's total charge is 0.

 
Potential due to charge q = $\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{q}{r_1}$
Potential due to charge -q = $\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{q}{r_2}$ 
Potential due to dipole, is the sum of potential due to charges q and -q.
Therefore,
                   $\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}\left(\frac{\mathrm{q}}{{\mathrm{r}}_1}-\frac{\mathrm{q}}{{\mathrm{r}}_2}\right)$ 
 where, r₁ and r₂ are distances of charges q and -q from point P.

Now, 
$$\begin{gathered} {{\mathrm{r}}_1}^2 = {\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{ar} \mathrm{cos\theta } \\ {{\mathrm{r}}_2}^2 = {\mathrm{r}}^2+{\mathrm{a}}^2+2\mathrm{ar} \mathrm{cos\theta } \end{gathered}$$ 

Considering r much greater than a, r>>a.

Therefore,
$$\begin{gathered} {{\mathbf{r}}_{\mathbf{1}}}^{\mathbf{2}}\mathbf{=}{\mathbf{r}}^{\mathbf{2}}\mathbf{ }\left(\mathbf{1}\mathbf{-}\frac{\mathbf{2}\mathbf{acos\theta }}{\mathbf{r}}\mathbf{ }\mathbf{+}\frac{{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\cong }{\mathbf{r}}^{\mathbf{2}}\mathbf{ }\left(\mathbf{1}\mathbf{-}\frac{\mathbf{2}\mathbf{acos\theta }}{\mathbf{r}}\right) \\ \mathbf{\parallel }\mathbf{ly}\mathbf{,}\mathbf{ }{{\mathbf{r}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{=}{\mathbf{r}}^{\mathbf{2}}\mathbf{ }\left(\mathbf{1}\mathbf{+}\frac{\mathbf{2}\mathbf{acos\theta }}{\mathbf{r}}\right) \\ \mathbf{Using}\mathbf{ }\mathbf{Binomial}\mathbf{ }\mathbf{theorem}\mathbf{,}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{retaining}\mathbf{ }\mathbf{upto}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{first}\mathbf{ }\mathbf{order}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{a}\mathbf{/}\mathbf{r}\mathbf{;}\mathbf{ }\mathbf{we}\mathbf{ }\mathbf{have}\mathbf{,}\mathbf{ } \\ \frac{\mathbf{1}}{{\mathbf{r}}_{\mathbf{1}}}\mathbf{\cong }\mathbf{ }\frac{\mathbf{1}}{\mathbf{r}\mathbf{ }}{\left(\mathbf{1}\mathbf{-}\frac{\mathbf{2}\mathbf{acos\theta }}{\mathbf{r}}\right)}^{\mathbf{-}\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\mathbf{ }\mathbf{\cong }\frac{\mathbf{1}}{\mathbf{r}}\left(\mathbf{1}\mathbf{+}\frac{\mathbf{a}}{\mathbf{r}}\mathbf{cos\theta }\right) \\ \mathbf{and} \\ \frac{\mathbf{1}}{{\mathbf{r}}_{\mathbf{2}}}\mathbf{\cong }\mathbf{ }\frac{\mathbf{1}}{\mathbf{r}\mathbf{ }}{\left(\mathbf{1}\mathbf{+}\frac{\mathbf{2}\mathbf{acos\theta }}{\mathbf{r}}\right)}^{\mathbf{-}\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\mathbf{ }\mathbf{\cong }\frac{\mathbf{1}}{\mathbf{r}}\left(\mathbf{1}\mathbf{-}\frac{\mathbf{a}}{\mathbf{r}}\mathbf{cos\theta }\right) \end{gathered}$$

Putting these values in the above equation of potential we have,

$$\begin{gathered} \mathbf{V}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\mathbf{.}\frac{\mathbf{2}\mathbf{acos\theta }}{{\mathbf{r}}^{\mathbf{2}}} \\ \mathbf{Dipole}\mathbf{ }\mathbf{moment}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{given}\mathbf{ }\mathbf{by}\mathbf{ }\mathbf{p}\mathbf{=}\mathbf{q}\mathbf{.}\mathbf{2}\mathbf{a} \\ \mathbf{\therefore }\mathbf{V}\mathbf{=}\frac{\mathbf{pcos\theta }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}{\mathbf{r}}^{\mathbf{2}}} \\ \mathbf{p}\mathbf{ }\mathbf{cos\theta }\mathbf{=}\mathbf{p}\mathbf{.}\overbrace{\mathbf{r}} \end{gathered}$$ 

where, $$\begin{gathered} \stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{vector}\mathbf{ }\mathbf{along}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{position}\mathbf{ }\mathbf{vector}\mathbf{ }\mathbf{OP}\mathbf{.} \\ \mathbf{Electric}\mathbf{ }\mathbf{potential}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{dipole}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{given}\mathbf{ }\mathbf{by}\mathbf{,} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{\circ }}}\frac{\mathbf{p}\mathbf{.}\overbrace{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{r}\mathbf{>}\mathbf{>}\mathbf{a}\mathbf{)} \end{gathered}$$

For potential at any point on axis, $\mathbf{\theta }\mathbf{=}\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\pi }\mathbf{]}$ 
V=$\mathbf{\pm }\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\mathbf{ }\frac{\mathbf{p}}{{\mathbf{r}}^{\mathbf{2}}}$ 

Potential is positive when $\mathrm{\theta }=0$ and potential is negative when $\theta =\pi$ .

Electrical potential falls off at large distance, as $\frac{1}{{\mathrm{r}}^2}$ and not as $\frac{1}{\mathrm{r}},$characteristic of the potential due to a single charge.