University Physics II · Electric Fields · 2.6

Electric Dipoles: Fields, Torque & Energy

Equal and opposite charges a short distance apart leave one vector behind: p⃗ = q d⃗. It sets a field that falls as 1/r³, and it fixes the torque and the energy the pair feels in a field of its own.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A neutral pair of charges is not electrically silent. Separate +q and −q by d⃗ and one vector survives at a distance: p⃗ = q d⃗, pointing from the negative charge to the positive one. Because the total charge is zero the 1/r² terms cancel, so the leading field falls as 1/r³ — twice as strong on the axis as on the perpendicular bisector, and reversed in direction between them.

Far away p⃗ is the whole description; q and d stop being separately visible. The same moment settles what an external field does to the dipole. Equal and opposite forces mean a uniform field cannot push it anywhere, but it can turn it: torque p⃗ × E⃗, orientation energy −p⃗·E⃗, lowest when p⃗ lies along E⃗ and highest when it lies against it.

Everything in between is a pendulum — and for molecules at room temperature, a pendulum that thermal collisions keep knocking over.

Simple definition: An electric dipole is a pair of equal and opposite charges a short distance apart, described by the moment p⃗ = q d⃗ — the one vector that fixes both the field it produces far away and the torque it feels in an applied field.
Example: A water molecule has p = 6.2 × 10⁻³⁰ C m: the oxygen end sits slightly negative and the hydrogen end slightly positive. The molecule is neutral, yet it turns in an applied field.

Electric dipole momentp⃗ = q d⃗

d⃗ runs from −q to +q, so p⃗ points from the negative charge to the positive one.

C m; 1 D = 3.336 × 10⁻³⁰ C m, and water is 1.85 D

Axial field, exact then farE = 2kpr/(r² − d²/4)² → 2kp/r³

On the axis E⃗ is parallel to p⃗; the 1/r³ form is 0.5% low at r = 10d.

N C⁻¹; k = 1/4πε₀ = 8.99 × 10⁹ N m² C⁻²

Equatorial field, exact then farE = kp/(r² + d²/4)³⁄² → kp/r³

Antiparallel to p⃗, and far away exactly half the axial magnitude.

N C⁻¹; measured on the perpendicular bisector

Far field at any angleEᵣ = 2kp cos θ/r³, E_θ = kp sin θ/r³

Valid for r ≫ d. The magnitude never reaches zero at any angle.

θ from p⃗; |E⃗| = (kp/r³)√(1 + 3cos²θ)

Torque in a uniform fieldτ⃗ = p⃗ × E⃗, |τ⃗| = pE sin θ

Largest at θ = 90°, zero at 0° and 180°, always turning p⃗ towards E⃗.

N m; the net force is zero, so τ⃗ is the same about any point

Orientation energy and small swingsU = −p⃗·E⃗ = −pE cos θ, ω = √(pE/I)

Flipping an aligned dipole costs 2pE; near θ = 0 it swings like a pendulum.

J and rad s⁻¹; U = 0 is chosen at θ = 90°

One vector replaces the pair

Take charges +q and −q held a distance d apart. Their total charge is zero, so the pair has no monopole field at all; what it has is a moment. Define p⃗ = q d⃗, with q the magnitude of either charge and d⃗ the displacement from −q to +q. So p⃗ points from the negative charge towards the positive one — the opposite sense to the field in the gap between them, which is a standard place to lose a sign. The unit is the coulomb metre; molecules are quoted in debyes, with 1 D = 3.336 × 10⁻³⁰ C m, so water's 1.85 D is 6.2 × 10⁻³⁰ C m. Because the total charge is zero, p⃗ does not depend on where you put the origin: shifting every position by a⃗ adds (Σq)a⃗ = 0. That independence is what lets one vector stand in for the pair. An ideal point dipole is the limit d → 0 and q → ∞ with p held fixed, and most of what follows works in that limit.

Axial and equatorial fields, exactly

Put the dipole on the x axis with its centre at the origin. On the axis at distance r > d/2 the two contributions are antiparallel and subtract: E = kq/(r − d/2)² − kq/(r + d/2)², which tidies to Eₐₓᵢₐₗ = 2kpr/(r² − d²/4)², directed along p⃗. On the perpendicular bisector both charges lie the same distance √(r² + d²/4) away, so their components across the axis cancel and their components along it add, each carrying a factor (d/2)/√(r² + d²/4). That gives E_eq = kp/(r² + d²/4)³⁄², directed antiparallel to p⃗. Two features survive: the axial field runs with p⃗ while the equatorial field runs against it, and at equal r the axial field is the larger — even though on the axis the two contributions subtract. With q = 2.0 nC and d = 1.0 mm, p = 2.0 × 10⁻¹² C m, and at r = 10 cm the axial field is 36 N C⁻¹ against 18 N C⁻¹ on the equator.

Why the far field falls as 1/r³

Let r ≫ d. Then r² ± d²/4 → r², and both exact results collapse onto one power: Eₐₓᵢₐₗ → 2kp/r³ and E_eq → kp/r³, a clean factor of two apart. The steeper falloff is the signature of a neutral source. Each charge alone gives 1/r², those terms cancel because the charges are equal and opposite, and what survives is the first-order difference between two inverse squares — one power of r smaller. At a general angle θ measured from p⃗, Eᵣ = 2kp cos θ/r³ and E_θ = kp sin θ/r³, with magnitude (kp/r³)√(1 + 3cos²θ): 2kp/r³ at θ = 0, kp/r³ at θ = 90°, and never zero anywhere. How far is far enough? For the pair above, at r = 5d the 1/r³ axial form is 2.0% low, and at r = 10d it is 0.5% low. The equatorial form errs the other way, 1.5% high at r = 5d and 0.4% high at r = 10d. Ten separations out, either version is good to better than a percent.

A uniform field turns a dipole but cannot move it

In a uniform E⃗ the forces are +qE⃗ and −qE⃗ — equal, opposite, and applied at different points. The net force is zero, so the centre of mass does not accelerate. The couple is not zero. Taking moments about the centre, each charge contributes (d/2)qE sin θ, giving τ⃗ = p⃗ × E⃗ with magnitude pE sin θ, where θ is the angle from E⃗ to p⃗. Because the net force vanishes, that torque is the same about every point, so you may take moments wherever it is convenient. It peaks at θ = 90°, vanishes at θ = 0 and θ = 180°, and always swings p⃗ towards E⃗. For numbers, mount two 0.50 g beads carrying ±50 nC at the ends of a light 4.0 cm rod, pivoted at its centre so gravity exerts no net torque. Then p = 2.0 × 10⁻⁹ C m, and in E = 5.0 × 10⁴ N C⁻¹ the maximum torque is pE = 1.0 × 10⁻⁴ N m.

Energy, alignment, and the temperature that undoes it

Turning a dipole against that torque stores energy, and integrating gives U(θ) = −p⃗·E⃗ = −pE cos θ, with the zero placed at θ = 90°. This is orientation energy in the external field, not the dipole's internal energy. The minimum is −pE at θ = 0, stable; the maximum is +pE at θ = 180°, unstable; flipping an aligned dipole costs 2pE, which for the rod above is 2.0 × 10⁻⁴ J. Near θ = 0 the torque is nearly linear, I d²θ/dt² = −pE sin θ ≈ −pE θ, so the dipole oscillates at ω = √(pE/I) exactly as a pendulum does. That rod has I = 2m(d/2)² = 4.0 × 10⁻⁷ kg m², so ω = 15.8 rad s⁻¹ and the period is 0.40 s. Molecules do not settle like that. Water in a strong laboratory field of 1.0 × 10⁷ N C⁻¹ has pE = 6.2 × 10⁻²³ J, while k_B T at 300 K is 4.1 × 10⁻²¹ J, 67 times larger. Collisions win, and the mean alignment is only ⟨cos θ⟩ ≈ pE/(3k_B T) = 0.005.

Where the model runs out

Three assumptions carry all of this. The 1/r³ field needs r ≫ d; nearer in, use the exact expressions, and between the charges the field looks nothing like a dipole field. The torque and energy results need E⃗ to be uniform across the dipole — that is exactly what makes the two forces cancel. Relax it and they do not cancel: a dipole aligned along x in a field varying along x feels Fₓ = p dEₓ/dx, a net pull towards the stronger field. That is why a charged rod picks up scraps of paper. The rod's own non-uniform field induces a dipole in the paper, then pulls on it, and the pull is attractive whatever the rod's sign, because the induced p⃗ follows E⃗. Finally, p is treated as fixed. A permanent dipole such as water carries its moment into any field; an induced dipole has p⃗ = αE⃗ and orientation energy −½αE², the half being the work spent pulling the charge apart.

Change one variable at a time

Make the relationship visible.

Interactive model

Angle θ of p⃗ from E⃗55 °

Charge separation d4.0 cm

Field E3.0 × 10⁴ N C⁻¹

Sweep the angle and watch the two force arrows: they never change length or direction, so nothing ever pushes the dipole anywhere — it can only turn. The torque peaks at 90°, which is exactly where the energy readout passes through zero, and the only two angles with no torque are the ones where the rod lies along the field: 0° is the stable one, 180° the unstable one. Then lengthen d, or raise E, and see both readouts scale together.

MOMENT p = qd0.80 nC m

TORQUE pE sin θ19.66 μN m

ENERGY −pE cos θ-13.77 μJ

FLIP COST 2pE48.00 μJ

Live interpretationMOMENT p = qd: 0.80 nC m. TORQUE pE sin θ: 19.66 μN m. ENERGY −pE cos θ: −13.77 μJ. FLIP COST 2pE: 48.00 μJ

Catch the common trap

Explain before calculating.

A small dipole of moment p⃗ sits at the origin pointing along +x, with charge separation d. Point A lies on the +x axis at distance r ≫ d; point B lies on the +y axis at the same distance r. How do the fields at A and B compare?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA pair of charges ±4.0 nC is held 3.0 mm apart. Find the dipole moment, and the largest torque the pair can feel in a uniform field of 2.5 × 10⁵ N C⁻¹.

p = qd, with q the magnitude of either charge: p = 4.0 × 10⁻⁹ C × 3.0 × 10⁻³ m = 1.2 × 10⁻¹¹ C m, pointing from −q to +q.
τ = pE sin θ, and sin θ is at most 1, so the maximum is τ = pE at θ = 90° — the dipole broadside to the field.
τ(max) = 1.2 × 10⁻¹¹ C m × 2.5 × 10⁵ N C⁻¹ = 3.0 × 10⁻⁶ N m. The net force stays zero at every angle; only the couple changes.

Answerp = 1.2 × 10⁻¹¹ C m, from −q to +q; τ(max) = 3.0 × 10⁻⁶ N m at θ = 90°

MediumA dipole is built from ±5.0 nC held 1.0 mm apart. Find the field 4.0 cm from its centre on the axis, and 4.0 cm from its centre on the perpendicular bisector, giving the direction of each.

p = qd = 5.0 × 10⁻⁹ C × 1.0 × 10⁻³ m = 5.0 × 10⁻¹² C m.
r = 4.0 cm is forty separations out, so the r ≫ d forms are safe — the axial one runs about 0.03% low here. r³ = (0.040 m)³ = 6.4 × 10⁻⁵ m³.
Axial: E = 2kp/r³ = 2 × 8.99 × 10⁹ × 5.0 × 10⁻¹² ÷ 6.4 × 10⁻⁵ = 8.99 × 10⁻² ÷ 6.4 × 10⁻⁵ = 1.40 × 10³ N C⁻¹, parallel to p⃗.
Equatorial: E = kp/r³ is exactly half the axial value, 7.02 × 10² N C⁻¹ — and antiparallel to p⃗, not along it.

Answerp = 5.0 × 10⁻¹² C m; E(axial) = 1.4 × 10³ N C⁻¹ along p⃗, E(equatorial) = 7.0 × 10² N C⁻¹ against p⃗

HardTwo 0.80 g beads carrying ±40 nC sit at the ends of a light 5.0 cm rod pivoted at its centre, in a uniform field E = 6.0 × 10⁴ N C⁻¹. The rod is held at θ = 60° to the field and released. Find the torque at release, the kinetic energy it has on reaching θ = 0, and the frequency of small oscillations about θ = 0.

p = qd = 40 × 10⁻⁹ C × 0.050 m = 2.0 × 10⁻⁹ C m, so pE = 2.0 × 10⁻⁹ × 6.0 × 10⁴ = 1.2 × 10⁻⁴ N m (the same number in joules for energy).
Torque at release: τ = pE sin 60° = 1.2 × 10⁻⁴ × 0.866 = 1.04 × 10⁻⁴ N m, turning p⃗ towards E⃗.
Energy: U = −pE cos θ, so ΔU = −pE(cos 0° − cos 60°) = −1.2 × 10⁻⁴ × (1 − 0.500) = −6.0 × 10⁻⁵ J, and that loss appears as kinetic energy at θ = 0.
Moment of inertia about the centre, two point masses at d/2: I = 2m(d/2)² = 2 × 0.80 × 10⁻³ kg × (0.025 m)² = 1.0 × 10⁻⁶ kg m².
Near θ = 0 the torque is linear in θ, so ω = √(pE/I) = √(1.2 × 10⁻⁴ ÷ 1.0 × 10⁻⁶) = √120 = 11.0 rad s⁻¹.
f = ω/2π = 10.95 ÷ 6.283 = 1.7 Hz. Note the rod passes θ = 0 rather than stopping — it swings on to θ = −60°.

Answerτ = 1.04 × 10⁻⁴ N m at release; KE = 6.0 × 10⁻⁵ J at θ = 0; f = 1.7 Hz