Derive an expression for the potential at a point due to a short dipole. Hence show what will be the potential at an axial and equatorial point:

Hint: We know that the electric potential due to a charge, is defined as the amount of energy needed to move a unit positive charge to infinity. Also the potential at any point is the vector sum of potentials at that point. Also, potential is proportional to the charge and inversely proportional to the square of the distance between the point and the charge.

Complete step by step answer:
Consider an electric dipole \[AB\] of length $2a$ and let $X$ be a point on the axial line such that it is at a distance $x$ from the centre of the dipole \[AB\].

Then the electric potential at the point $X$ is given as,$V_{X}=V_{+q}+V_{-q}$
Or, $V_{X}=K\left[ \dfrac{q}{x-a}+\dfrac{-q}{x+a} \right]=\dfrac{K(2qa)}{x^{2}-a^{2}}=\dfrac{KP}{x^{2}-a^{2}}$, where $P$ is the dipole moment of the electric dipole \[AB\]of length $a$ and $K$ is the coulombs constant which is equal to $9\times 10^{9}Nm^{2}/C$
Consider an electric dipole \[AB\]of length $2a$ and let $X$ be a point on the equatorial line such that it is at a distance $x$ from the centre of the dipole. If $AX=r_{1}$ and $BX=r_{2}$ and $X$ makes angle $\theta$ at $A$ and $B$ t then, potential at $X$ can be resolved as follow:


Now, since the $sin\theta$ components are equal and opposite they cancel each other. The $cos\theta$ components add up, and are also equal to zero, as the sign of the charges are opposite.

Hence the potential due to a dipole at the axial position is given as:$V_{a}=\dfrac{KP}{x^{2}-a^{2}}$
While the potential due to a dipole at the equatorial position is $0$

Note:
The dipole moment $P=2aq$. This is taken for simplification and easy calculations. Also, $K=\dfrac{1}{4\pi \epsilon_{0}}$. It is better to assume the length of the dipole as $2a$ as then, the half length of the dipole will be $a$ which is again easy for calculations.