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A finite electric dipole consists of equal and opposite electric charges separated by a nonzero displacement. It has charge neutrality but a nonzero electric dipole moment $\mathbf{p}=q\mathbf{d}$. An ideal point electric dipole is the mathematical limit $d=|\mathbf{d}|\to 0$ with $\mathbf{p}$ fixed, so $q$ must increase as $1/d$. Shrinking the separation while holding $q$ fixed instead makes the electric dipole moment vanish. The ideal electric dipole describes an exterior electric field; it does not reproduce the two near-charge singularities or the internal structure of a molecule. ^[ref].

Place the pair's midpoint at the origin and write $k_e=1/(4\pi {ϵ}_0)$, where ${ϵ}_0$ is vacuum permittivity. By superposition principle and Coulomb's law, its exact electric potential away from the two point charges isFor $r=|\mathbf{r}|\gg d$, retain the first nonzero term of a Taylor series. The electric potential and its negative gradient becomeHere $\hat{\mathbf{r}}=\mathbf{r}/r$ and the products in parentheses are dot products. These are exact exterior formulas for the ideal electric dipole and distant approximations for a finite pair. The electric field on the outer axis points along $\mathbf{p}$ with magnitude $2k_{e}p/r^3$; on the perpendicular bisector it points opposite $\mathbf{p}$ with magnitude $k_{e}p/r^3$. This directional structure is what a scalar total electric charge misses. ^[ref].

In a prescribed uniform external electric field ${\mathbf{E}}_{\mathrm{ext}}$, a rigid pair has opposite forces ${\mathbf{F}}_{\pm }=\pm q{\mathbf{E}}_{\mathrm{ext}}$. Its net force is zero, but summing their moments about the midpoint gives the torqueThe cross product makes the torque perpendicular to both the electric dipole moment and the external electric field. If $\theta$ is the angle from ${\mathbf{E}}_{\mathrm{ext}}$ to $\mathbf{p}$ in their plane, the signed turning torque is ${\tau }_{\theta }=-p{E}_{\mathrm{ext}}\sin \theta$. It tends to reduce a small positive $\theta$. ^[ref].

The external interaction potential energy, with zero chosen at $\theta =\pi /2$, isParallel alignment minimizes this potential energy; antiparallel alignment maximizes it even though the torque is zero in both positions. This omits the pair's constant internal binding potential energy and assumes a fixed permanent electric dipole moment. ^[ref].

A nonuniform external electric field can exert a net force on a neutral electric dipole. Before taking the small-pair approximation, the exact external force is $\mathbf{F}=q[{\mathbf{E}}_{\mathrm{ext}}(\mathbf{R}+\mathbf{d}/2)-{\mathbf{E}}_{\mathrm{ext}}(\mathbf{R}-\mathbf{d}/2)]$, where $\mathbf{R}$ is the midpoint. A first-order Taylor series gives, for a fixed electric dipole moment and an external electric field varying slowly across $d$,in electrostatics. The last equality uses the vanishing curl of the external electric field, with $\mathbf{p}$ held fixed while differentiating. Thus uniform external electric fields turn a misaligned electric dipole, whereas gradients can also move its midpoint. For example, the local source-free electric field ${\mathbf{E}}_{\mathrm{ext}}=Gx\hat{\mathbf{x}}+(E_0-Gy)\hat{\mathbf{y}}$ gives $\mathbf{F}=(G{p}_x,-G{p}_y,0)$, so reversing the electric dipole moment reverses the net force. This stipulated local model has zero divergence and zero curl.

A motivating experiment is Filsinger and colleagues' study of electrostatic deflection of polar molecules. Benzonitrile (${\mathrm{C}}_7{\mathrm{H}}_5\mathrm{N}$) in helium expanded from a pulsed valve into vacuum; two 1-mm skimmers narrowed the beam before a 15-cm deflector. A grounded trough and a high-voltage rod produced a vertical electric field gradient. Scanning a focused laser vertically downstream and recording benzonitrile ions from photoionization measured the beam's relative density. With voltage off, the profile spanned about 2 mm; applying voltage shifted and broadened it upwards. At 10 kV, many molecules lay outside the original profile. ^[ref].

Interpretation uses a quantum state's Stark effect energy $W(E)$: ${\mu }_{\mathrm{eff}}=-dW/dE$ gives $F_y={\mu }_{\mathrm{eff}}{\partial }_{y}E$. States whose energy falls with increasing electric field seek stronger fields. The voltage-off profile controls the mechanical aperture. Inferring state populations needs calculated trajectories, speed distributions and a rotational temperature model; profile displacement alone does not measure a permanent electric dipole moment uniquely. This is an electric Stark effect experiment, with photoionization occurring after deflection. ^[ref].

For the symmetric finite pair, the first omitted electric multipole expansion term after the electric dipole is third order, so typical relative errors scale as $(d/r{)}^2$. This stronger error estimate need not hold for an arbitrary neutral molecule with a nonzero electric quadrupole moment. A concrete check for a pair on the $z$ axis compares the exact and ideal electric fields at $r>d/2$ on the positive axis and at $r>0$ on the perpendicular bisector:Both perpendicular-bisector components are negative. At $r=5d$, the ratios are approximately $1.0203$ and $0.9852$: errors of $+2.03\%$ and $-1.48\%$ relative to the ideal electric field. The condition $r\gg d$ concerns observing the pair's own electric field; the separate condition for its external force approximation concerns variation of the applied electric field across the pair.

The electric dipole moment measures the first spatial moment of signed electric charge. For localized point charges $q_i$ at positions ${\mathbf{r}}_i$, or an integrable charge density $\rho (\mathbf{r})$ with a finite first moment, define^[ref].

The electric dipole moment is a vector with SI units $\mathrm{C}\mathrm{m}$: coulombs times length in metres. Opposite electric charges can cancel in $Q$ while contributing in the same direction to $\mathbf{p}$. For a neutral pair with positive electric charge $q>0$ at ${\mathbf{r}}_{+}$ and negative electric charge $-q$ at ${\mathbf{r}}_{-}$,Thus the electric dipole moment points from negative to positive electric charge, not along the attraction between the two point charges. ^[ref].

Changing the coordinate origin changes the position vectors, not the physical electric charge distribution. If the new origin is at $\mathbf{a}$ relative to the old origin, ${\mathbf{r}}_i^{\prime }={\mathbf{r}}_i-\mathbf{a}$, so direct substitution givesConsequently charge neutrality makes the electric dipole moment independent of the origin. For a system with nonzero total electric charge, always specify the origin: a single point charge at the old origin has $\mathbf{p}=0$ but ${\mathbf{p}}^{\prime }=-Q\mathbf{a}$. This is coordinate bookkeeping, not a newly created electric dipole. The full electric potential stays the same at the same physical point; the terms of its electric multipole expansion and the observation coordinates change together.

As a stipulated numerical example, put $+2\mathrm{nC}$ at $(1,0,0)\mathrm{mm}$ and $-2\mathrm{nC}$ at $(-1,0,0)\mathrm{mm}$. Their total electric charge is zero and $\mathbf{p}=(4\times 10^{-12},0,0)\mathrm{C}\mathrm{m}$. Translating the origin by $(3,-2,0)\mathrm{mm}$ leaves this electric dipole moment unchanged. Adding $+1\mathrm{nC}$ at the old origin instead gives $Q=1\mathrm{nC}$ and ${\mathbf{p}}^{\prime }=(1,2,0)\times 10^{-12}\mathrm{C}\mathrm{m}$. Distinguishing these two cases prevents the mistaken claim that every electric dipole moment is independent of the origin.

The usefulness of the electric dipole moment is compression: many internal electric charges determine one vector governing the leading distant electric potential of a neutral localized object when $\mathbf{p}\ne 0$. ^[ref]. A polar molecule need not literally contain two point charges; its atomic nuclei and electron density collectively give an electric dipole moment. For example, the chemical polarity of water reflects its bent molecular geometry, whereas the symmetric linear molecular geometry of carbon dioxide makes its permanent electric dipole moment vanish. ^[ref]. A collection of such electric dipole moments per unit volume gives electric polarization. ^[ref].

The trace-free electric quadrupole moment is a symmetric tensor measuring the second-order term in an electric multipole expansion. For a localized charge distribution its contribution to the far-field electrostatic potential is $Q_{ij}{\hat{r}}_i{\hat{r}}_j/(8\pi {ϵ}_0{r}^3)$, with summation over indices. The moment may be nonzero even when the total electric charge or electric dipole moment is nonzero; it should not be confused with an electric quadrupole whose lower moments vanish. Changing the origin generally changes the tensor when lower moments are nonzero.

The free-space Green function of the Laplacian for the Poisson equation gives the electric potentialFor $r=|\mathbf{x}|\gg R$, its electric multipole expansion begins withDefine the total electric charge and electric dipole moment byThen

The analogous solution for the magnetic vector potential isIts apparent $r^{-1}$ coefficient is ${\mu }_0(4\pi {)}^{-1}\int \mathbf{J}{d}^3{x}^{\prime }$. For each component, the supplied identity and the divergence-free vector field condition giveThe divergence theorem turns this into a surface integral outside the compact support of the current density, where $\mathbf{J}=0$. Hence $\int \mathbf{J}{d}^3{x}^{\prime }=0$, so the $1/r$ term vanishes.

The method of images replaces the earthed plane by an image charge $-q$ at $(-a,0,0)$. The two Coulomb potentials cancel at $x=0$, so uniqueness for the Dirichlet problem makes the resulting field the physical field in $x>0$. With $\mathbf{r}=(x,y,z)$ and ${\mathbf{e}}_x=(1,0,0)$,and

The electric multipole expansion from part (a) gives, for $r\gg a$,Thus the leading field is that of an electric dipole.

On the plane,Taking the plane normal to be $+{\mathbf{e}}_x$ and using polar coordinates,With the outward normal of the region $x>0$, the sign is reversed. This is consistent with Gauss's law: all electric flux from the real charge terminates on the grounded conductor.

The electrostatic boundary condition gives the induced surface charge densityand its integral isIn the plane $z=0$, the field lines leave the positive charge, meet the conductor normally, and are the right-half-plane portions of the field lines joining the real charge to its negative image.

Let $\theta$ be the angle between $r$ and the positive $z$-axis. By superposition of the electric potential of point charges,For $r>d$, the Legendre polynomial expansion givesThereforeThe three displayed terms are respectively the monopole, dipole, and quadrupole terms of the electric multipole expansion. In particular, the total charge is $Q(N-M-1)$ and the electric dipole moment is $Qd(M-1)\hat{z}$. This is the multipole expansion of three collinear charges.