What an Electric Dipole Is
An electric dipole is a pair of equal and opposite point charges, $+q$ and $-q$, separated by a fixed distance $2a$. The line joining the two charges defines a direction in space; by convention this direction runs from $-q$ to $+q$. The mid-point of the two charges is the centre of the dipole.
The total charge of a dipole is exactly zero. That zero, however, does not make the field zero. Because $+q$ and $-q$ sit at slightly different places, the fields they produce do not cancel everywhere — they cancel only in the limit of very large distance, and even there a residue survives. Understanding that residue is the whole point of the topic.
The dipole moment vector $\mathbf{p}$ points from the negative charge to the positive charge, with magnitude $p = q\times 2a$.
The Dipole Moment Vector
The explicit field calculation (below) shows that the dipole's far-field never depends on $q$ and $a$ separately — only on their product $qa$. NCERT promotes this product to a defining quantity, the electric dipole moment:
$$\mathbf{p} = q \times 2a\,\hat{\mathbf{p}} \qquad p = q(2a)$$
It is a vector. Its magnitude is the charge $q$ times the full separation $2a$, and its direction $\hat{\mathbf{p}}$ runs from $-q$ to $+q$. The SI unit is the coulomb-metre (C m). Every later formula in this topic is written compactly in terms of $\mathbf{p}$.
Field on the Axis vs the Equator
The field at a general point is the vector sum of the fields of $+q$ and $-q$ by the parallelogram law. The algebra closes neatly for two special locations: a point on the axial line (the extended dipole axis) and a point on the equatorial plane (perpendicular to the axis through the centre). For a short dipole — meaning $r \gg a$ — NCERT gives the two results verbatim.
On the axis the field points along $\mathbf{p}$; on the equator it points opposite to $\mathbf{p}$, and is half as strong at equal distance.
$$\text{Axial: }\;\; E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\,\frac{2p}{r^{3}} \qquad (r \gg a)$$
$$\text{Equatorial: }\;\; E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^{3}} \qquad (r \gg a)$$
Two facts are encoded here, and NEET tests both. First, the magnitude on the axis is exactly twice that on the equator at the same distance — the famous $2:1$ ratio. Second, the directions differ: the axial field is parallel to $\mathbf{p}$, while the equatorial field is anti-parallel to $\mathbf{p}$.
| Location | Field magnitude (r ≫ a) | Direction | Relative strength |
|---|---|---|---|
| Axial line | (1/4πε₀)·2p/r³ | Along $\mathbf{p}$ (−q → +q) | 2 units |
| Equatorial plane | (1/4πε₀)·p/r³ | Opposite to $\mathbf{p}$ | 1 unit |
| Ratio axial : equatorial | 2 : 1 | Anti-parallel to each other | — |
Axial is double the equatorial — and points the other way
The single most common slip is mixing up the factor of 2 or forgetting the direction. The axial field carries the factor $2p$; the equatorial field carries plain $p$. They are not just different in size — the axial field is along $\mathbf{p}$ and the equatorial field is opposite to $\mathbf{p}$.
Axial $= 2\times$ equatorial, same $r$. Axial $\parallel \mathbf{p}$; equatorial anti-$\parallel \mathbf{p}$.
The 1/r³ Falloff
A lone point charge produces a field that decays as $1/r^2$. A dipole's field decays faster, as $1/r^3$. The reason is structural: the net charge of a dipole is zero, so the leading $1/r^2$ contribution from the two charges cancels at large $r$, leaving the next-order $1/r^3$ term as the survivor. NCERT states it directly — the dipole field at large distances "falls off not as $1/r^2$ but as $1/r^3$."
This faster decay is exactly what NEET 2022 probed: a $-q$, $+q$ pair viewed from $R \gg L$ behaves as a dipole, so its field varies as $1/R^3$. The dependence on the angle between $\mathbf{r}$ and $\mathbf{p}$ is also part of the picture — the field magnitude is not the same in every direction at a fixed $r$, which is why axial and equatorial values differ.
A dipole's field is just superposition of two point-charge fields. Revisit Electric Field & Field Lines to see how a single charge's $1/r^2$ field is defined and drawn.
Dipole in a Uniform Field
Now place a permanent dipole in a uniform external field $\mathbf{E}$, oriented at angle $\theta$ to $\mathbf{p}$. The charge $+q$ feels a force $q\mathbf{E}$; the charge $-q$ feels $-q\mathbf{E}$. Because the field is uniform, these forces are equal and opposite, so the net force is zero — the dipole does not translate.
But the two forces act at different points, $2a$ apart. Equal, opposite, non-collinear forces form a couple, and a couple produces a torque. NCERT computes its magnitude as the force times the perpendicular arm of the couple:
$$\tau = qE \times 2a\sin\theta = pE\sin\theta \qquad \boldsymbol{\tau} = \mathbf{p}\times\mathbf{E}$$
Equal, opposite forces on $+q$ and $-q$ give zero net force but a couple of torque $pE\sin\theta$ that turns $\mathbf{p}$ toward $\mathbf{E}$.
The torque always tends to align $\mathbf{p}$ with $\mathbf{E}$; once aligned ($\theta = 0$), the torque vanishes. Associated with this restoring action is a potential energy, defined relative to the $\theta = 90^\circ$ position:
$$U = -pE\cos\theta = -\mathbf{p}\cdot\mathbf{E}$$
Energy is lowest when $\mathbf{p}$ lines up with $\mathbf{E}$, which is why that orientation is the resting state of a free dipole.
Zero force does not mean zero torque
In a uniform field the net force on a dipole is zero, but the torque $pE\sin\theta$ is generally non-zero. Students often write "force $=0$, therefore equilibrium" and miss the rotation. Equilibrium needs both zero force and zero torque — that happens only at $\theta = 0^\circ$ or $\theta = 180^\circ$. Separately, in a non-uniform field even the net force is non-zero.
Uniform field: $F_{\text{net}}=0$, $\tau = pE\sin\theta$. Non-uniform field: $F_{\text{net}}\neq 0$.
Torque and Energy at Key Angles
The three orientations $\theta = 0^\circ,\,90^\circ,\,180^\circ$ are the ones NEET returns to. Reading torque and potential energy off a single table prevents sign errors under pressure.
| Angle θ | Torque τ = pE sinθ | Energy U = −pE cosθ | State |
|---|---|---|---|
| 0° (p ∥ E) | 0 | −pE (minimum) | Stable equilibrium |
| 90° (p ⟂ E) | pE (maximum) | 0 | Maximum torque |
| 180° (p anti-∥ E) | 0 | +pE (maximum) | Unstable equilibrium |
U = −pE cosθ is a minimum at θ = 0, not a maximum
The minus sign matters. At $\theta = 0$, $U = -pE$ — the most negative, hence the lowest energy and stable equilibrium. At $\theta = 180^\circ$, $U = +pE$ — the highest energy and unstable equilibrium. Work done to rotate from $\theta_1$ to $\theta_2$ is $W = U(\theta_2) - U(\theta_1) = pE(\cos\theta_1 - \cos\theta_2)$.
Lowest energy at alignment ($\theta=0$); highest at anti-alignment ($\theta=180^\circ$).
Physical Significance
The dipole is not an abstraction — it is the everyday language of molecules. In molecules such as CO₂ and CH₄ the centres of positive and negative charge coincide, so the permanent dipole moment is zero; a field can still induce one. But in molecules such as water, H₂O, the two centres do not coincide, giving a permanent dipole moment even with no field applied. These are the polar molecules, and their dipole moments govern how matter responds to electric fields.
The non-uniform-field result explains a classroom staple: a comb run through dry hair attracts neutral bits of paper. The comb's field polarises the paper, inducing a dipole, and because that field is non-uniform the induced dipole feels a net force toward the comb.
Electric Dipole in one screen
- A dipole is $+q$ and $-q$ separated by $2a$; moment $\mathbf{p} = q(2a)$ points from $-q$ to $+q$.
- Axial field $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$ (along $\mathbf{p}$); equatorial field $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}$ (opposite to $\mathbf{p}$).
- Axial : equatorial $= 2:1$, in opposite senses. The dipole field falls as $1/r^3$, faster than a point charge's $1/r^2$.
- Uniform field: net force $= 0$, but torque $\boldsymbol{\tau} = \mathbf{p}\times\mathbf{E}$, magnitude $pE\sin\theta$.
- Potential energy $U = -\mathbf{p}\cdot\mathbf{E} = -pE\cos\theta$: minimum ($-pE$) at $\theta=0$ (stable), maximum ($+pE$) at $\theta=180^\circ$ (unstable).
- Non-uniform field gives a net force; polar molecules (e.g. H₂O) carry permanent dipole moments.