Electric dipole and its properties

Rub a balloon on your hair and it sticks to a wall — even though the wall is electrically neutral. Something in the neutral wall responds to the balloon's charge. That "something" is a swarm of tiny electric dipoles: pairs of positive and negative charge, glued together and separated by a whisker of distance, that line up in the balloon's field and pull it in. A water molecule is one such dipole, permanently and forever, because its oxygen atom hoards electrons more greedily than its two hydrogens. A single point charge cannot explain this — a lone charge either has a field or it does not — but a pair of equal and opposite charges, held a fixed tiny distance apart, is a genuinely new object with its own physics. This is the electric dipole: two charges $+q$ and $-q$ , always equal in magnitude, separated by a small distance $2a$ . On its own it is electrically neutral — add up the charges and you get zero — yet it is not "nothing." It has a direction, captured by a single vector called the dipole moment $\vec p = q\,(2\vec a)$ , pointing from the negative charge to the positive one. That one vector is enough to predict how the dipole's field fades with distance, how strongly it responds when you place it in an external field, and why polar molecules like water behave so differently from non-polar ones like carbon dioxide. This article builds the idea from the ground up: what a dipole is and how its moment is defined and computed, the short-dipole approximation that makes all the later formulas usable, and a full worked derivation of the dipole moment for a realistic charge configuration. It sets up — but does not yet compute — the axial and equatorial field expressions and the torque in a uniform field, which belong to the next linked node. By the end you should be able to define a dipole precisely, calculate its moment for any given charge-separation pair, and explain in words why a neutral object can still produce a field that never quite vanishes.

Key takeaways

An electric dipole is a pair of equal and opposite charges $+q$ , $-q$ separated by a small fixed distance $2a$ ; the pair is electrically neutral overall.
The dipole moment $\vec p = q\,(2\vec a)$ is a vector pointing from $-q$ to $+q$ , with SI unit coulomb-metre (C·m).
Even though net charge is zero, $\vec p \neq 0$ whenever the charges are separated — this is what lets a neutral object still produce and feel electric effects at a distance.
The short-dipole (point-dipole) approximation treats $2a \ll r$ , the separation as negligible compared with the distance at which the dipole's effect is measured; this is what makes $p$ alone (not $q$ and $a$ separately) the only quantity that matters for far-field formulas.
Real molecules such as water are permanent dipoles because of unequal charge sharing between atoms (electronegativity difference); symmetric molecules like $\text{CO}_2$ have zero net dipole moment despite polar bonds, because the two bond dipoles cancel by symmetry.
A dipole's field falls off faster with distance ( $\propto 1/r^3$ ) than a single point charge's field ( $\propto 1/r^2$ ) — the opposite charges' fields partially cancel at large distance, leaving only the "leftover" effect of the separation.
Key idea: a dipole is a neutral pair of equal and opposite charges whose separation, captured by the single vector p = q(2a), is the only thing that survives at a distance, making the dipole moment (not the individual charges) the natural language for describing its far-field behaviour.

The core insight

An electric dipole's entire far-field identity is packed into one vector, the dipole moment $\vec p = q(2\vec a)$ pointing from $-q$ to $+q$ , whose magnitude sets how strongly a neutral but separated pair of charges is felt at a distance and whose survival despite zero net charge is why polar molecules and charge pairs matter in electrostatics.

Where it fits in the syllabus

This topic sits inside CBSE Class 12 Physics, Unit I "Electrostatics," chapter "Electric Charges and Fields," which explicitly lists "electric dipole; dipole moment $\vec p = q\cdot 2\vec a$ " as prescribed content, alongside the field due to a dipole (axial/equatorial) and the torque on a dipole in a uniform field. This section covers the definition, the dipole moment vector, its calculation, and the short-dipole approximation — the foundation every later dipole formula rests on. The axial/equatorial field expressions and the torque formula $\vec\tau = \vec p \times \vec E$ are deliberately left to the next linked node so each piece gets full treatment; the dipole's field at a general (non-axial, non-equatorial) point and its behaviour in a non-uniform field are both explicitly out of the CBSE 2025-26 syllabus and are not attempted here.

The physical picture

Picture two point charges of equal magnitude $q$ but opposite sign, held rigidly a fixed small distance apart. Call the vector running from the negative charge to the positive charge $2\vec a$ — the factor of $2$ is a deliberate convention, because $\vec a$ is then the vector from the centre of the dipole (the midpoint of the line joining the charges) out to each charge. The straight line through both charges is the dipole axis.

\vec p = q\,(2\vec a)

where: $q$ is the magnitude of each charge (in coulombs), $2\vec a$ is the displacement vector from the negative charge to the positive charge (in metres), and $\vec p$ is the electric dipole moment (in coulomb-metres, C·m), a vector that always points from $-q$ towards $+q$ . The magnitude alone is $p = q(2a)$ , where $2a$ is the straight-line separation between the two charges. Reading this physically: dipole moment is not a new kind of charge — it is a bookkeeping device for "how much charge, separated by how much distance, and in which direction." Doubling either the charge or the separation doubles $p$ ; a dipole with the charges pushed together to $a \to 0$ has $p \to 0$ and behaves, from any distance, exactly like no charge at all.

Figure

A pair of charges labelled plus q and minus q separated along a horizontal axis with the centre marked and the dipole moment vector drawn from the negative charge toward the positive charge

Neutral overall, but not "nothing" at a distance. Add the two charges: $q + (-q) = 0$ , so the dipole carries no net charge and produces no field that behaves like a single point charge's $1/r^2$ field at large distance — the two charges' fields largely cancel. But they do not cancel exactly, because they sit at slightly different distances from any external point. That leftover mismatch is what the dipole moment $\vec p$ captures, and it is why the dipole's field dies off faster, as $1/r^3$ , rather than vanishing outright (the precise axial and equatorial expressions are built from this idea in the next node).

The short-dipole approximation. All standard dipole formulas — the field along the axis, the field on the equatorial line, and the torque in a uniform field — are derived, and only valid, in the limit that the separation is small compared with the distance of interest: $2a \ll r$ . This is called the short-dipole or point-dipole approximation. It is what allows every later formula to be written purely in terms of $p$ , without needing $q$ and $a$ separately. Physically it means: from far enough away, all that matters about the pair of charges is the single number and direction packed into $\vec p$ — the detailed geometry of exactly how the charges are arranged washes out.

Real dipoles: permanent and induced. A permanent dipole exists even with no external field, because of the molecule's own geometry and electronegativity difference — water ( $\text{H}_2\text{O}$ ) is the standard example: oxygen pulls the shared electrons closer to itself than the two hydrogens do, leaving oxygen slightly negative and the hydrogens slightly positive, and the bent (not straight) shape of the molecule means these effects do not cancel, giving water a nonzero permanent dipole moment. By contrast, carbon dioxide ( $\text{CO}_2$ ) has polar carbon-oxygen bonds, but because the molecule is linear and symmetric, the two bond dipole moments point in exactly opposite directions and cancel — $\text{CO}_2$ has zero net dipole moment despite having polar bonds. (How a neutral dielectric develops an induced dipole moment when placed in an external field is a qualitative idea that belongs to the next chapter, "Electrostatic Potential and Capacitance," and is only mentioned here for context.)

Derivation AFinding the dipole moment of a two-charge configuration

The goal is to compute the dipole moment vector — magnitude and direction — for a concrete, physically realistic pair of charges placed at known positions, using only the definition $\vec p = q(2\vec a)$ and elementary vector subtraction; no calculus is needed, only correctly identifying which charge is which and measuring the separation between them along the line joining them.

Step 1. Identify the two charges and their positions. Suppose a charge $-q$ sits at position $x = -a$ on an axis and $+q$ sits at $x = +a$ , so the two charges are placed symmetrically about the origin, which is therefore the centre of the dipole. The full separation between the charges is the distance from $-q$ to $+q$ , namely $(+a) - (-a) = 2a$ , measured along the axis.

Step 2. Form the displacement vector from the negative charge to the positive charge. By definition this vector is $2\vec a$ , and it points along the positive $x$ -direction here (from $x=-a$ to $x=+a$ ), with magnitude $2a$ . Getting the direction right is the entire content of this step: the vector must run from $-q$ to $+q$ , never the reverse.

Step 3. Apply the definition of dipole moment, $\vec p = q\,(2\vec a)$ . Substituting the magnitude of the charge and the displacement vector from Step 2 gives a vector of magnitude $p = q(2a)$ , pointing in the same direction as $2\vec a$ — that is, along the positive $x$ -axis, from the negative charge towards the positive charge.

Step 4. Check the physical sense of the result. If either charge is doubled, $p$ doubles; if the charges are dragged together so $a \to 0$ , $p \to 0$ and the pair becomes indistinguishable, from a distance, from having no charge at all — consistent with the pair's net charge being permanently zero. This confirms that $\vec p$ correctly measures the "separation effect" and nothing else: it does not care about net charge (which is always zero for a dipole) but entirely about how far apart, and in which sense, the two charges sit.

Worked examples

Board level - dipole moment of a simple charge pair

Question: Two charges $+4\ \text{nC}$ and $-4\ \text{nC}$ are separated by a distance of $2\ \text{cm}$ . Find the magnitude and direction of the dipole moment.

Step 1. List the knowns and match them to the definition. Here the charge magnitude is $q = 4\times10^{-9}\ \text{C}$ , and the full separation between the two charges is $2a = 2\ \text{cm} = 0.02\ \text{m}$ (this is already the full separation, not the half-distance $a$ ).

Step 2. Apply $p = q(2a)$ directly. $p = (4\times10^{-9}\ \text{C})(0.02\ \text{m}) = 8\times10^{-11}\ \text{C}\,\cdot\,\text{m}$ .

Step 3. State the direction using the definition of $\vec p$ : it points from the negative charge towards the positive charge, i.e. along the line joining the two charges, from $-4\ \text{nC}$ to $+4\ \text{nC}$ .

Answer: $p = 8\times10^{-11}\ \text{C}\,\cdot\,\text{m}$ , directed from the negative charge to the positive charge.

Competitive level - effective charge of a polar molecule

Question: A molecule has a dipole moment of $p = 6.2\times10^{-30}\ \text{C}\,\cdot\,\text{m}$ . If the effective charge separation is $1.0\times10^{-10}\ \text{m}$ (a typical atomic-scale distance), estimate the effective charge $q$ at each end, and express it as a fraction of the elementary charge $e$ .

Step 1. Rearrange the definition to solve for the unknown. From $p = q(2a)$ , isolate the charge: $q = \dfrac{p}{2a}$ .

Step 2. Substitute the given values, being careful that $2a$ (not $a$ ) is the full separation already stated in the problem: $q = \dfrac{6.2\times10^{-30}\ \text{C}\,\cdot\,\text{m}}{1.0\times10^{-10}\ \text{m}} = 6.2\times10^{-20}\ \text{C}$ .

Step 3. Express this as a fraction of one elementary charge, $e = 1.6\times10^{-19}\ \text{C}$ , to see how it compares with a single electron's charge: $\dfrac{q}{e} = \dfrac{6.2\times10^{-20}}{1.6\times10^{-19}} \approx 0.39$ .

Step 4. Interpret the fraction physically. A value of about $0.39\,e$ is not literally an electron transferred from one atom to the other — real molecular bonding involves partial, shared electron distributions, not a whole discrete charge hopping across. The dipole moment formula gives an effective charge that reproduces the same $p$ , useful for comparison, without claiming that exactly $0.39$ of an electron physically sits at each end.

Answer: The effective charge is about $6.2\times10^{-20}\ \text{C} \approx 0.39\,e$ , reflecting a partial, shared charge separation typical of a polar molecule rather than a whole electron transferred. Timing note: this rearrange-and-substitute style question should take under ninety seconds in a competitive exam — the only trap is forgetting that the "effective separation" given in the problem is already the full $2a$ , not the half-distance $a$ ; plugging it in as $a$ instead of $2a$ produces an answer exactly double the correct one.

Common traps

Thinking a dipole has a net electric charge because it has a nonzero dipole moment — a dipole is electrically neutral overall ( $+q$ and $-q$ sum to zero); the dipole moment measures charge separation, not net charge.
Confusing the separation used in $p = q(2a)$ : the formula uses the full distance between the two charges, written as $2a$ , not the half-separation $a$ alone — a common source of a factor-of-two error in both directions (using $a$ where $2a$ is meant, or vice versa).
Forgetting that $\vec p$ is a vector with a specific direction (from $-q$ to $+q$ ), not just a magnitude — direction matters when a dipole is later placed in an external field.
Assuming dipole formulas hold exactly at any distance from the charges — the short-dipole approximation requires $2a \ll r$ ; it is not exact at points close to the dipole, only in the far-field limit that this syllabus works with.
Assuming any molecule with polar bonds must have a nonzero dipole moment — molecular symmetry can cancel individual bond dipoles entirely, as in linear $\text{CO}_2$ , even though each carbon-oxygen bond is individually polar.

You should be able to…

Define an electric dipole and state clearly why it is electrically neutral overall.
Write the definition $\vec p = q(2\vec a)$ from memory, including the direction convention (from $-q$ to $+q$ ).
Calculate the dipole moment, or the effective charge, or the separation, given any two of the three quantities $p$ , $q$ , $2a$ .
Explain, in words, why a neutral dipole still produces a nonzero field at a distance, and why that field falls off faster than a single point charge's field.
State the short-dipole approximation ( $2a \ll r$ ) and explain why it is needed for the far-field formulas.
Decide, given a molecule's shape and bond polarities, whether it should have a nonzero permanent dipole moment (e.g. water yes, carbon dioxide no).

Ask the tutor: want to see why a dipole's field falls off as $1/r^3$ instead of $1/r^2$ , or work through the axial and equatorial field formulas and the torque on a dipole in a uniform field? Ask the AI tutor to walk through it.

If a dipole has zero net charge, how can it produce any electric field at all?

Net charge and dipole moment measure two different things. Net charge asks "how much charge is left over if you add everything up" — for a dipole that is always zero. Dipole moment instead measures how that (zero) charge is arranged in space. Because the positive and negative charges sit at slightly different distances from any outside point, their individual $1/r^2$ fields do not cancel perfectly — a small leftover survives, and that leftover is the dipole's field. It fades faster with distance than a single charge's field, but it is not zero.

Why is the separation written as 2a instead of just a?

It is a convention chosen so that $\vec a$ itself is the vector from the dipole's centre to each charge — a genuinely useful quantity, since the centre is the natural reference point for later formulas (like the axial and equatorial fields in the next topic, which are measured from the centre). Writing the full separation as $2a$ keeps that "half-distance" $a$ available as its own symbol. The cost is that you must always remember $p = q(2a)$ uses the full separation, not $a$ alone — this is the single most common numerical slip on this topic.

Does every neutral molecule have a dipole moment?

No. Whether a molecule has a nonzero permanent dipole moment depends on both bond polarity and molecular shape. Water is bent, so its two O–H bond dipoles add up to a nonzero net vector. Carbon dioxide is linear and symmetric, so its two C=O bond dipoles point in exactly opposite directions and cancel exactly, leaving zero net dipole moment even though each individual bond is polar. A fully symmetric, non-polar molecule (like $\text{N}_2$ or $\text{CH}_4$ ) has essentially no permanent dipole moment at all.

Is the field of a dipole at any point around it covered in this syllabus?

Not in general. The CBSE 2025-26 syllabus explicitly restricts the electric field of a dipole to two special lines: along the dipole's own axis, and along the equatorial line (the perpendicular bisector of the line joining the charges) — both covered in the next linked node. The field at a completely general point off these two lines, and a dipole's behaviour in a non-uniform external field, are both explicitly outside the prescribed syllabus, so you will not be asked to derive or quote them.