Q.What is electric dipole? Find an expression for electric potential at any point due to an electric dipole.

An electric dipole is two equal and opposite charges $\pm q$ a distance $2a$ apart, with moment $p=q\,2a$. The potential at a distant point $(r,\theta)$ is $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$.

What is an electric dipole?

An electric dipole consists of two equal and opposite point charges, $+q$ and $-q$, separated by a small distance $2a$. Its strength is measured by the electric dipole moment

$\vec p = q\,(2a),$

a vector directed from the negative charge to the positive charge, with magnitude $p = q\times 2a$.

Potential at any point due to a dipole.

Let $O$ be the centre of the dipole, with $-q$ at $A$ and $+q$ at $B$ ($AB = 2a$). Consider a point $P$ at distance $r$ from the centre $O$, such that $OP$ makes an angle $\theta$ with the dipole axis. Let $r_1$ and $r_2$ be the distances of $P$ from $+q$ and $-q$ respectively.

The potential at $P$ is the sum of the potentials due to the two charges:

$V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r_1} + \frac{1}{4\pi\varepsilon_0}\frac{(-q)}{r_2} = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r_1} - \frac{1}{r_2}\right).$

Approximation for $r \gg a$. Dropping perpendiculars from $A$ and $B$ onto the line $OP$, for a point far from the dipole:

$r_1 \approx r - a\cos\theta, \qquad r_2 \approx r + a\cos\theta.$

Therefore

$\frac{1}{r_1} - \frac{1}{r_2} = \frac{r_2 - r_1}{r_1 r_2} = \frac{(r+a\cos\theta)-(r-a\cos\theta)}{(r-a\cos\theta)(r+a\cos\theta)} = \frac{2a\cos\theta}{r^{2}-a^{2}\cos^{2}\theta}.$

Since $r \gg a$, we neglect $a^2\cos^2\theta$ compared with $r^2$: …