Electric dipoles: fields, torque and energy

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.

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.

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.

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.

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.

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.