What is an electric dipole moment and how is it related to the potential?

An electric dipole moment (denoted by $\vec{p}$) is a vector quantity that characterizes an electric dipole. It is defined as the product of the magnitude of either charge ($q$) and the distance ($2a$) separating them, directed from the negative charge to the positive charge. So, $p = q(2a)$. The electric potential due to a dipole is directly proportional to the magnitude of the dipole moment. A larger dipole moment means a stronger dipole, leading to a higher potential at a given point (assuming other factors like distance and angle are constant).

Why is the electric potential zero on the equatorial line of an electric dipole?

On the equatorial line, any point is equidistant from the positive charge ($+q$) and the negative charge ($-q$) of the dipole. Since potential is a scalar quantity and depends on the sign of the charge, the potential due to $+q$ will be positive and the potential due to $-q$ will be negative, but of equal magnitude at any point on the equatorial line. According to the principle of superposition, the total potential is the algebraic sum of these two potentials, which results in zero ($V = V_+ + V_- = V_{positive} + (-V_{positive}) = 0$). This holds true for all points on the equatorial plane.

How does the potential due to an electric dipole vary with distance, compared to a point charge?

For a single point charge, the electric potential $V$ varies inversely with the distance $r$ from the charge, i.e., $V \propto 1/r$. However, for an electric dipole, at distances much larger than the dipole's length, the electric potential $V$ varies inversely with the square of the distance $r$, i.e., $V \propto 1/r^2$. This faster decay ($1/r^2$ vs $1/r$) is because the effects of the positive and negative charges of the dipole tend to cancel each other out more effectively as the distance from the dipole increases.

Is the electric field also zero on the equatorial line where the potential is zero?

No, not necessarily. While the electric potential is zero on the equatorial line of an electric dipole, the electric field is generally not zero there. Electric field is a vector quantity, and its components from the positive and negative charges do not completely cancel out on the equatorial line. In fact, the electric field on the equatorial line of a dipole is directed opposite to the dipole moment vector. This is a common point of confusion for students and highlights the difference between scalar potential and vector field.

What is the significance of the angle $\theta$ in the dipole potential formula?

The angle $\theta$ represents the angle between the dipole moment vector (from $-q$ to $+q$) and the position vector of the point where potential is being calculated (from the center of the dipole to the point). This angle is crucial because it accounts for the spatial orientation of the point relative to the dipole. The $\cos\theta$ term in the formula $V = \frac{p \cos\theta}{4\pi\epsilon_0 r^2}$ dictates how the potential varies with direction. It ensures that potential is maximum on the axial line ($\theta = 0^circ$ or $180^circ$) and zero on the equatorial line ($\theta = 90^circ$), reflecting the anisotropic nature of the dipole's field.

Potential due to Electric Dipole

Physics

NEET UG

Version 1Updated 22 Mar 2026

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The electric potential due to an electric dipole at a point in space is defined as the scalar sum of the potentials due to the two individual point charges constituting the dipole. For a point $P$ located at a distance $r$ from the center of the dipole, making an angle $heta$ with the dipole axis, and assuming the distance $r$ is much larger than the separation between the charges ( $r gg a$ ), the…

Quick Summary

An electric dipole consists of two equal and opposite point charges, $+q$ and $-q$ , separated by a small distance $2a$ . The electric dipole moment, $\vec{p}$ , is a vector from $-q$ to $+q$ with magnitude $p = q(2a)$ .

The electric potential at a point due to a dipole is the scalar sum of potentials from its constituent charges. For points far from the dipole ( $r \gg a$ ), the potential $V$ at a distance $r$ from the center and at an angle $\theta$ with the dipole axis is given by $V = \frac{p \cos\theta}{4\pi\epsilon_0 r^2}$ .

This formula shows a characteristic $1/r^2$ dependence, which is faster than the $1/r$ dependence for a single point charge. On the axial line ( $\theta = 0^circ$ or $180^circ$ ), the potential is $V = \pm \frac{p}{4\pi\epsilon_0 r^2}$ .

Crucially, on the equatorial line ( $\theta = 90^circ$ ), the potential is always zero, although the electric field is not. This angular dependence is a key feature distinguishing dipole potential from point charge potential.

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Electric Dipole — Two equal and opposite charges (

+q, -q

) separated by

2a

Dipole Moment —

\vec{p} = q(2\vec{a})

, direction from

-q

+q

General Potential Formula (for $r \gg a$) —

V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}

Axial Line Potential —

V_{axial} = \pm \frac{p}{4\pi\epsilon_0 r^2}

(for

\theta = 0^circ, 180^circ

)

Equatorial Line Potential —

V_{equatorial} = 0

(for

\theta = 90^circ

)

Distance Dependence —

V_{dipole} \propto 1/r^2

(vs.

V_{point charge} \propto 1/r

)

Nature — Potential is a scalar quantity.

Potential Energy of Dipole in External Field —

U = -\vec{p} \cdot \vec{E} = -pE \cos\theta

To remember the dipole potential formula: 'P Cosey on R Squared'.

P — Dipole moment (

p

)

Cosey —

\cos\theta

R Squared —

r^2

in the denominator

So, $V = \frac{p \cos\theta}{4\pi\epsilon_0 r^2}$ .

For equatorial line: 'Equator is Zero' (potential is zero).