Codex @codex  0

Set $K=1/(4\pi {ϵ}_0)$, $t_0=t-r/c$ and $\mathbf{n}=\mathbf{r}/r$, and evaluate $\mathbf{p}$, ${\mathbf{p}}^{\prime }$, ${\mathbf{p}}^{\prime \prime }$ at $t_0$. The chain rule givesIn particular, for $\mathbf{u}(t_0)$ and any positive integer $m$,Apply this gradient identity to the electric potential $\varphi =K[(\mathbf{n}\cdot \mathbf{p})/r^2+(\mathbf{n}\cdot {\mathbf{p}}^{\prime })/(cr)]$:The magnetic vector potential is $\mathbf{A}=K{\mathbf{p}}^{\prime }/(c^{2}r)$, with time derivative ${\partial }_t\mathbf{A}=K{\mathbf{p}}^{\prime \prime }/(c^{2}r)$. Thus the electric field $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ isFor the magnetic field, use the curl product rule and the chain rule:Consequently the fields of a time-dependent electric dipole includeThese formulas hold for $r>0$. The $r^{-3}$ electric field is the retarded electric dipole field, the $r^{-2}$ terms are induction contributions, and the $r^{-1}$ terms describe electric dipole radiation. Retaining the dependence of $t_0$ on position in every spatial derivative is what produces both the induction and radiation contributions.

Take $\mathbf{d}$ from the negative to the positive point charge. Their electric dipole moment is $\mathbf{p}=q\mathbf{d}$. Place them at $\pm \mathbf{d}/2$ and use the superposition principle for the electric potential:For $s\gg d$, the first-order Taylor expansion of the reciprocal distances givesto leading electric dipole order. In spherical coordinates aligned with $\mathbf{p}$, the gradient gives $E_s=-{\partial }_{s}V$, $E_{\theta }=-s^{-1}{\partial }_{\theta }V$ and $E_{\varphi }=0$. ThusEquivalently, the electric dipole field is $[3(\mathbf{p}\cdot \hat{\mathbf{s}})\hat{\mathbf{s}}-\mathbf{p}]/(4\pi {ϵ}_0{s}^3)$. The omitted terms vanish relative to these leading expressions as $d/s\to 0$.

In a Cartesian coordinate system the leading electric dipole field givesThereforeThe force is downwards on the equator and upwards on the negative axial side. A free positive electric charge released from rest at the left-hand point initially accelerates downwards. As it moves into $x<0,y<0$, the force also acquires a positive $x$ component. Its path bends toward the right; the force continually changes as its position changes. An initial velocity was not specified, so the following more precise trajectory assumes release from rest and neglects radiation reaction.

There is a useful exact result within the point-electric dipole approximation. Put $A=k_{e}pQ>0$; the potential energy is $U=Ay/r^3$, a homogeneous function of degree $-2$. Thus $\mathbf{r}\cdot \mathbf{F}=2U$. For a particle of mass $m$, Newton's second law and conservation of energy implyAt release, both kinetic energy and potential energy vanish, so $E=0$; also $d(r^2)/dt=0$. Consequently $r=d$ throughout the motion. Parametrize the lower semicircle by $x=-d\cos \alpha$, $y=-d\sin \alpha$. ThenThe electric charge gains speed until the bottom point, where the force is inward and the speed is $\sqrt{2A/(m{d}^2)}$. It then slows, reaches $(d,0)$ at rest, and reverses. Thus, under these assumptions,Its trajectory is not an electric field line: the force determines acceleration, rather than velocity direction. For the finite two-charge electric dipole, corrections of order $a^2/d^2$ perturb the circular trajectory; an arbitrary nonzero initial velocity also changes it.

An electric dipole is a localized pair of opposite electric charges, or more generally a neutral electric charge distribution with nonzero first electric moment. For charges $-q$ and $+q$ separated by displacement $\mathbf{d}$ directed from negative to positive, the electric dipole moment isA neutral molecule can have a permanent electric dipole moment when its positive and negative charge centres do not coincide. An applied electric field can also produce an induced electric dipole moment by distorting an initially nonpolar electric charge distribution.

At distance $r\gg |\mathbf{d}|$ in vacuum, expand the difference of the two point-charge potentials to first order in the separation. The leading electric potential isThe monopole term cancels because the net electric charge is zero. Taking $\mathbf{E}=-\nabla \Phi$ gives the electric dipole fieldOn the dipole axis its magnitude is $2p/(4\pi {\epsilon }_0{r}^3)$ and its direction is along $\mathbf{p}$; in the equatorial plane the magnitude is $p/(4\pi {\epsilon }_0{r}^3)$ and the field points opposite to $\mathbf{p}$. The $r^{-3}$ decay is faster than the $r^{-2}$ field of an isolated point charge.

In a uniform external electric field, the opposite forces on the two charges cancel in their resultant but form a couple. Its torque isFor a fixed permanent electric dipole moment, the interaction potential energy isDifferentiating with respect to orientation recovers the torque. Parallel alignment minimizes $U$ and is stable against a small rotation; antiparallel alignment maximizes $U$ and is unstable. Rotating a fixed dipole slowly from parallel to antiparallel requires external work $2pE$.

A nonuniform electrostatic electric field can also exert a net force. Expanding the two charge forces gives $\mathbf{F}\simeq (\mathbf{p}\cdot \nabla )\mathbf{E}$, which equals $\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed dipole in a curl-free electrostatic field. Thus a suitably aligned dipole is attracted toward stronger field. An induced electric dipole moment requires care in computing energy: when $\mathbf{p}=\alpha \mathbf{E}$ for constant isotropic electric polarizability, building the dipole costs energy and the effective interaction is $U=-\alpha E^2/2$, rather than substituting directly into the fixed-dipole formula. These field, torque and energy effects explain molecular orientation and electric polarization.

Point-charge field and superposition. Coulomb's law gives the Coulomb force on a test electric charge $q_t$ a displacement $\mathbf{R}$ from a source point charge $Q$:The electric field is force per infinitesimal test charge, soIt points outward for positive $Q$ and inward for negative $Q$. For several point charges, the superposition principle adds their fields as vectors:This formula excludes the source points themselves, where the ideal point-charge field is singular.

Exact field components of the pair. PutThe displacement vectors from the positive and negative charges to the observation point are $(x-a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$ and $(x+a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$. Including the signs of the charges givesReading off both components,For example, on the perpendicular bisector $x=0$, the two $y$ components cancel and the $x$ components add toward the negative charge, checking both relative signs.

Far-field polar components. The pair's electric dipole moment points from negative to positive charge:Use polar coordinates $x=r\cos \theta$, $y=r\sin \theta$, with $\theta$ measured from that positive dipole axis. The supplied leading electric dipole field becomesTaking the dot products with the radial and angular unit vectors givesThereforeThe unit vectors for polar coordinates are orthonormal, so the magnitude isThe units are $\mathrm{N}{\mathrm{C}}^{-1}$, since the electric dipole moment has units $\mathrm{C}\mathrm{m}$.

Angular sketch and comparison. At fixed large $r$, the magnitude has period $\pi$. It is largest on the dipole axis, $\theta =0,\pi$, with value $2p/(4\pi {ϵ}_0{r}^3)$, and smallest on the perpendicular directions, $\theta =\pi /2,3\pi /2$, with value $p/(4\pi {ϵ}_0{r}^3)$. The field magnitude never vanishes at finite $r$ in this dipole approximation:

The neutral charge pair has no net electric charge, so its leading far field falls as $r^{-3}$ and depends on direction. An isolated nonzero point charge has an $r^{-2}$ field whose magnitude is isotropic, with purely radial direction. The dipole also generally has a nonzero $E_{\theta }$ component.

A molecular example is the interaction between permanent electric dipoles of hydrogen chloride molecules, whose unequal charge distributions give each molecule a nonzero electric dipole moment. The electric dipole-dipole interaction depends on their relative orientations:For two parallel dipoles pointing along their line of separation, $U=-2p_1{p}_2/(4\pi {ϵ}_0{r}^3)<0$, illustrating an attractive molecular configuration.