Derive the expression for the electric potential due to an electric dipole at a general point $P$ at a distance $r$ from its center.

(N/A) Consider an electric dipole consisting of charges $+q$ and $-q$ separated by a distance $2a$. Let $O$ be the midpoint of the dipole. We want to find the electric potential at a point $P$ located at a distance $r$ from $O$,such that the line $OP$ makes an angle $\theta$ with the dipole axis.
Let $r_1$ be the distance of point $P$ from charge $+q$ and $r_2$ be the distance of point $P$ from charge $-q$.
The electric potential $V$ at point $P$ due to the dipole is the algebraic sum of the potentials due to individual charges:
$V = V_1 + V_2 = \frac{kq}{r_1} - \frac{kq}{r_2} = kq \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$
Using the law of cosines in the triangles formed by the dipole and point $P$:
$r_1^2 = r^2 + a^2 - 2ar \cos \theta$
$r_2^2 = r^2 + a^2 + 2ar \cos \theta$
For $r \gg a$,we can approximate $r_1$ and $r_2$ using the binomial expansion:
$r_1 \approx r - a \cos \theta$
$r_2 \approx r + a \cos \theta$
Substituting these into the potential expression:
$V = kq \left( \frac{1}{r - a \cos \theta} - \frac{1}{r + a \cos \theta} \right) = kq \left( \frac{r + a \cos \theta - (r - a \cos \theta)}{r^2 - a^2 \cos^2 \theta} \right)$
$V = \frac{kq(2a \cos \theta)}{r^2} = \frac{kp \cos \theta}{r^2}$,where $p = 2aq$ is the dipole moment.