(a) Derive an expression for the electric field E due to a dipole of length ‘2a’ at a point distant r from the centre of the dipole on the axial line.
(b) Draw a graph of E versus r for r >> a.
(c) If this dipole were kept in a uniform external electric field E0, diagrammatically represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases.

(a) Consider an electric dipole AB. The charges -q and +q of dipole are situated at A and B respectively, as shown in the figure. The separation between the charges is 2a. Electric dipole moment is given by 
$\mathrm{p}=\mathrm{q}\times 2\mathrm{a}$......(1)
Consider a point P on the axis of dipole at a distance r from midpoint O of electric dipole.
The distance of point P from the charge +q at B is, BP = r−a and distance of point P from charge -q at A is AP = r + a

Let E1 and E2 be the electric field strengths at point P due to charges +q and -q respectively. 


The resultant electric field due to electric dipole is given by,

$\begin{array}{l}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\cdot \frac{\mathrm{q}}{(\mathrm{r}-\mathrm{a}{)}^2}-\frac{1}{4{\mathrm{\pi \epsilon }}_0\cdot }\frac{\mathrm{q}}{(\mathrm{r}+\mathrm{a}{)}^2}\\ \mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\mathrm{q}\left[\frac{1}{(\mathrm{r}-\mathrm{a}{)}^2}-\frac{1}{(\mathrm{r}+\mathrm{a}{)}^2}\right]\\ \mathrm{E}=\frac{1}{4{\mathrm{\pi c}}_0}\cdot \mathrm{q}\left(\frac{(\mathrm{r}+\mathrm{a}{)}^2-(\mathrm{r}-\mathrm{a}{)}^2}{(\mathrm{r}-\mathrm{a}{)}^2(\mathrm{r}+\mathrm{a}{)}^2}\right)\\ \mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\mathrm{q}\left[\frac{4\mathrm{ra}}{{\left({\mathrm{r}}^2-{\mathrm{a}}^2\right)}^2}\right]\\ \mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\cdot \frac{2\mathrm{pr}}{{\left({\mathrm{r}}^2-{\mathrm{a}}^2\right)}^2}\left[\text{ from (i) }\right]\end{array}$
 (b) if r$\gg >{\mathrm{a}}_1$
$\begin{array}{r}\mathrm{E}=\frac{1}{4{\mathrm{\pi ϵ}}_0}\cdot \frac{2\mathrm{p}}{{\mathrm{r}}^3}\\ \mathrm{E}={\mathrm{E}}_2-{\mathrm{E}}_1\end{array}$
(b) Graph of E versus r for r >> a

E
(c) Position of dipole in stable equilibrium:

$\begin{array}{l}\overrightarrow{\mathrm{\tau }}=\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{E}}=\mathrm{pEsin}\mathrm{\theta }=\mathrm{pEsin}{0}^{\circ }\\ \overrightarrow{\mathrm{\tau }}=0\end{array}$
In stable equilibrium, 𝜃=0°
Position of dipole in unstable equilibrium
In unstable equilibrium, 𝜃=180°

$$\begin{gathered} \overrightarrow{-}\overrightarrow{\mathrm{\tau }}=\mathrm{pEsin}{180}^{\circ } \\ \overrightarrow{\mathrm{T}}=0 \end{gathered}$$