Question

The electric potential in equatorial position of an electric dipole is

Answer 1

we know,
electric field at axial point or End on is given by, $\vec{E_a}=\frac{1}{4\pi\epsilon_0}\frac{2\vec{p}}{r^3}$

and electric field at equatorial position or broad on is given by, $\vec{E_e}=\frac{1}{4\pi\epsilon_0}\frac{-\vec{p}}{r^3}$

here, p denotes dipole moment, r is the separation of observation point to midpoint of dipole.

you did mistake in typing potential is scalar quantity. you doesn't say potential in equatorial position or potential in axial position .

I think you want to find electric field in equatorial position, and that is $\vec{E_e}=\frac{1}{4\pi\epsilon_0}\frac{-\vec{p}}{r^3}$

Answer 2

Answer:

The electric potential in equatorial position of an electric dipole is zero.

Explanation:

Let the dipole be consists of the charges $\pm q$, separated by distance $d$.

Consider a point P, on the equatorial line passing through the center of the dipole, at distance $a$ from the center of the dipole.

The electric potential due to a charge $q$ at a point $r$ distance away from it is given by

$V = \dfrac{kq}{r}.$

where, $k$ is the Coulomb's constant.

The distance of point P from both the charges is given as

$r = \sqrt{\left ( \dfrac{d}{2} \right )^2+a^2}.$

The electric potential at the point P due to the positive charge is given by

$V_+=\dfrac{kq}{\sqrt{\left ( \dfrac{d}{2} \right )^2+a^2}}.$

The electric potential at the point P due to the negative charge is given by

$V_-=\dfrac{k(-q)}{\sqrt{\left ( \dfrac{d}{2} \right )^2+a^2}}.$

Since, the electric potential is a scalar quantity, the net electric potential at point P due to both the charges is equal to the algebraic sum of electric potentials at point P due to both the charges.

Therefore, the net electric potential at point P,

$V = V_++V_-=\dfrac{kq}{\sqrt{\left ( \dfrac{d}{2} \right )^2+a^2}}+\dfrac{k\ (-q)}{\sqrt{\left ( \dfrac{d}{2} \right )^2+a^2}}=0.$