Electric potential at equatorial point of a small dipole with dipole moment p(At, r distance from the dipole) is
A. zero
B. \[\dfrac{p}{{4\pi {\varepsilon _0}{r^2}}}\]
C. \[\dfrac{p}{{4\pi {\varepsilon _0}{r^3}}}\]
D. \[\dfrac{{2p}}{{4\pi {\varepsilon _0}{r^3}}}\]

Hint:Electric dipole moment measures the separation of the positive and negative electrical charges within a system and is defined as the product of the magnitude of the charge and the distance between them. The electric potential at the equatorial point of a small dipole can be found by using the formula of electric potential at the equatorial point

Formula used:
Magnitude of the intensity of the electric field at a distance r on equatorial line is given as,
\[{E_e} = k \times \dfrac{p}{{{r^3}}}\]
Electric potential at the equatorial point is given as,
\[{V_e} = k \times \dfrac{p}{{{r^2}}}\]
Where p is dipole moment, \[{V_e}\] is the electric potential on the equatorial line, r is distance from dipole and k is coulomb’s constant,\[k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}\].

Complete step by step solution:
As we know that magnitude of the intensity of electric field at a distance r on equatorial line is,
\[{E_e} = k \times \dfrac{p}{{{r^3}}}\]
As electric potential at equatorial point is,
\[{V_e} = k \times \dfrac{p}{{{r^2}}} = 0{\rm{(cos9}}{{\rm{0}}^0} = 0)\]
For the equatorial axis of the dipole,
\[\theta = {90^0}\]
The angle \[\theta \] between the line joining the point with the centre of the dipole and the axial line of the dipole. Therefore, when the given point is on the equatorial axis, the electric potential is zero

Hence option A is the correct answer.

Note: An electric dipole consists of a pair of two charges equal in magnitude (q) but opposite in nature (i.e. one is a positive charge and the other is a negative charge). These two charges are separated by a distance of length d. An electric dipole moment can be calculated as the product of charge and the distance between them.