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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 $d/s\to 0$.

Use the electric dipole above a grounded conducting plane approximation, valid for $d\ll a$. Let $\hat{\mathbf{z}}$ point upward, and choose $\mathbf{p}=p(\sin \theta \hat{\mathbf{x}}+\cos \theta \hat{\mathbf{z}})$. Reflecting the two constituent point charges and reversing their signs produces an image electric dipole momentat height $-a$. Thus the parallel electric dipole moment reverses and the normal one is unchanged. At the real electric dipole, the separation from its image is $2a\hat{\mathbf{z}}$, giving the image electric fieldThe physical torque is $\mathbf{\tau }=\mathbf{p}\times {\mathbf{E}}_{\mathrm{im}}$, soFor $0<\theta <\pi /2$, its direction reduces $\theta$, tending to align the electric dipole with the upward normal. In general it favours either normal orientation; a perfectly parallel orientation has zero torque but is unstable to tilting.

The conductor-response potential energy includes one half of the real-image interaction:This factor accounts for the induced surface charge changing as the physical source is assembled. Differentiation with respect to height at fixed orientation givesThere is no horizontal force, since translating the entire arrangement parallel to the plane leaves $U$ unchanged. For example, the normal and parallel orientations give respective downward force magnitudes $3p^2/(32\pi {ϵ}_0{a}^4)$ and $3p^2/(64\pi {ϵ}_0{a}^4)$.

A single physical point charge induces total plane electric charge $-q$. One can verify this by integrating its induced surface charge density, $-qh/[2\pi ({\varrho }^2+h^2{)}^{3/2}]$, over the plane. The two physical electric charges therefore induce opposite totals, givingThe local surface charge density is generally nonzero. The zero total holds for the finite neutral pair as well as its ideal electric dipole limit.

The electric dipole moment is $\mathbf{p}=qa\hat{\mathbf{y}}$. Write $\theta$ for the angle measured from $+y$, with $\hat{\mathbf{\theta }}$ pointing toward increasing $\theta$. The exact electric potential isFor $r\gg a$, expand the reciprocals to first order in $a/r$. Their common $1/r$ terms cancel, giving the electric dipole approximationThe radial electric field follows from $E_r=-\partial \varphi /\partial r$, and the tangential electric field from $E_{\theta }=-(1/r)\partial \varphi /\partial \theta$. The omitted terms are of relative order $a^2/r^2$ away from zeros of a component. Multiplying by the test electric charge gives the force componentsThe electric field lines emerge from the positive electric charge and terminate on the negative electric charge. They are mirror-symmetric about the electric dipole axis and bow out across its equatorial plane. In the point-electric dipole limit their meridional curves obey $dr/(rd\theta )=E_r/E_{\theta }=2\cot \theta$, hence $r=C{\sin }^2\theta$.

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.

Apply the superposition principle to the result of part (a). If $F(a,h)=-qh/[2\pi {ϵ}_0(a^2+h^2{)}^{3/2}]$, then the exact normal electric field from the two real point charges and their image charges isFor $d\ll h$, Taylor's theorem gives $E_z=-d{\partial }_{h}F+O(d^2)$. Sincethe leading electric dipole result isThe electric dipole moment is $\mathbf{p}=-qd\hat{\mathbf{z}}$. The circle $a=\sqrt{2}h$ is the zero of the leading normal electric field and hence of the induced surface charge density. Inside it, induced surface charge density is negative; outside it, positive. The tangential electric field is zero everywhere on the grounded electrical conductor, so the leading full electric field vanishes on that circle. The corresponding dipole-induced charge on a grounded plane integrates to zero over the whole plane: the negative central region is balanced by the positive exterior.

For finite $d$, $a=\sqrt{2}h$ is only the leading-order zero. With $u=h^{2/3}$ and $v=(h+d{)}^{2/3}$, the exact zero satisfies $a^2=uv(u+v)$, giving $a=\sqrt{2}(h+d/2)+O(d^2/h)$. For example, inserting any $d>0$ in the exact expression at $a=\sqrt{2}h$ does not give zero. This is a controlled electric dipole approximation, not an exact finite-separation identity.

The electric field within a conductor in electrostatic equilibrium vanishes. The normal electric field boundary condition, with outward normal vector $+\hat{\mathbf{z}}$, therefore gives $\sigma ={ϵ}_0{E}_z(0^{+})$. Directly below the monopole, $a=0$:For the electric dipole,The exact finite-separation answer iswhose Taylor series agrees with the boxed leading term. Although the real point charges have zero total electric charge, their unequal distances leave negative induced surface charge density immediately beneath them.

An electric dipole has equal and opposite charges separated by a small vector from negative to positive, with electric dipole moment $\mathbf{p}=q\mathbf{d}$. More generally, a localized neutral distribution has $\mathbf{p}=\int {\mathbf{r}}^{\prime }\rho ({\mathbf{r}}^{\prime })d^3{r}^{\prime }$, independent of origin. At distances large compared with its size, the leading nonzero term in the multipole expansion isThe field points along $\mathbf{p}$ on either axial side and opposite $\mathbf{p}$ in the equatorial plane; its axial magnitude is twice the equatorial magnitude at equal radius. Near the charges the finite-charge expression must replace the far-field approximation.

A permanent dipole in an externally imposed slowly varying electrostatic field has interaction energy $-\mathbf{p}\cdot \mathbf{E}$, torque $\mathbf{\tau }=\mathbf{p}\times \mathbf{E}$ and force $\nabla (\mathbf{p}\cdot \mathbf{E})$ for fixed $\mathbf{p}$. A uniform field aligns it but gives no net force. For a linearly induced dipole $\mathbf{p}=\alpha \mathbf{E}$, the formation-inclusive energy is $-\alpha E^2/2$, not $-\alpha E^2$: the moment itself grows while the field is applied. Dipole density gives electric polarization, whose bound charges are ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\sigma }_b=\mathbf{P}\cdot \hat{\mathbf{n}}$.

An electric dipole consists of equal and opposite electric charges separated by a displacement $\mathbf{d}$. Its electric dipole moment is $\mathbf{p}=q\mathbf{d}$, directed from negative to positive electric charge. At distances large compared with the separation, its leading electric potential and electric field areThere is no monopole term because the net electric charge is zero. The axial electric field points along $\mathbf{p}$ with magnitude $2p/(4\pi {ϵ}_0{r}^3)$, while the equatorial electric field points oppositely with half that magnitude.

A uniform applied electric field exerts zero net force on an ideal fixed electric dipole, but it exerts torque $\mathbf{\tau }=\mathbf{p}\times \mathbf{E}$. Its orientation-dependent potential energy is $U=-\mathbf{p}\cdot \mathbf{E}$, so alignment with the electric field is stable. A nonuniform applied electric field can also exert a net force, $\mathbf{F}=\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed permanent electric dipole moment.

A permanent electric dipole moment exists without an applied electric field. An induced electric dipole moment instead develops in response to the local electric field; in an isotropic linear model $\mathbf{p}=\alpha \mathbf{E}$. Creating that induced moment requires accounting for the work of polarization, so its mechanical potential energy in a specified applied electric field is $-\alpha E^2/2$, rather than $-\alpha E^2$. In matter, the dipole moment per unit volume is the polarization density $\mathbf{P}$.

From molecular dipoles to macroscopic fields. A dielectric contains charges which remain bound to its atoms or molecules. An applied electric field displaces positive and negative charge centres and can orient permanent electric dipoles. The electric polarization $\mathbf{P}$ is electric dipole moment per unit volume, in $\mathrm{C}{\mathrm{m}}^{-2}$. Its associated bound charge isat the boundary of the polarized material, where $\hat{\mathbf{n}}$ points outward. A uniform electric polarization has no volume bound charge, but generally has surface bound charge. The electric field remains the force per unit test charge; it includes the fields of both free and bound charges.

Introducing the electric displacement field $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ packages the free-charge source separately:For a homogeneous linear isotropic dielectric,The electric susceptibility ${\chi }_e$ is dimensionless; the permittivity $ϵ$ has units $\mathrm{F}{\mathrm{m}}^{-1}$. This is a constitutive approximation. An anisotropic dielectric requires a permittivity tensor, while a nonlinear dielectric need not have electric polarization proportional to electric field. The field in these equations is the macroscopic field, not necessarily the local field at an individual molecule.

Interfaces. Let $\hat{\mathbf{n}}$ point from medium 1 into medium 2. Integrating Gauss's law over a thin pillbox, and the electrostatic curl equation around a thin loop, gives the electrostatic interface boundary conditionsThus normal electric displacement is continuous when there is no free surface charge, whereas tangential electric field is continuous even if the permittivity changes. Normal electric field generally jumps because the interface carries bound charge. The electric potential is continuous for an ordinary interface without an ideal dipole sheet. These conditions, together with Laplace's equation in each homogeneous charge-free region, determine the field.

Energy and electrical sources. Reversibly assembling free charge requires incremental work $\delta W=\int \mathbf{E}\cdot \delta \mathbf{D}dV$. For a linear dielectric whose response can be reversibly integrated from zero field, the electrostatic energy isThis includes the material polarization response, rather than simply using ${ϵ}_0{E}^2/2$ everywhere. In a capacitor, $W=Q^2/(2C)=C{V}^2/2$. Inserting a dielectric increases capacitance. At fixed charge the stored energy decreases. At fixed voltage the stored field energy increases, but the voltage source supplies $VdQ$; the appropriate mechanical energy including that source changes by $-V^{2}dC/2$. Both cases predict attraction of dielectric into the capacitor. A force calculation must state which electrical quantity is fixed. A lossy or hysteretic dielectric also dissipates energy, so the reversible quadratic expression does not describe its full history.

Shape and depolarization. Suppose the dielectric is surrounded by vacuum and the distant imposed electric field is ${\mathbf{E}}_0$. Bound surface charge produces a depolarization field. For a uniformly polarized ellipsoid, along a principal axis,Here $N_d$ is the depolarization factor of a dielectric ellipsoid. This self-consistent field explains why the same material responds differently in different shapes.

An infinite slab with electric field normal to its faces has $N_d=1$, giving $E_{\mathrm{in}}=E_0/{ϵ}_r$; its opposite surface bound charges reduce the internal field. For electric field parallel to the faces, $N_d=0$, giving $E_{\mathrm{in}}=E_0$. End effects are neglected.

A very long circular rod in an axial electric field has negligible depolarization away from its ends, so $E_{\mathrm{in}}\simeq E_0$. In a transverse electric field the infinite-cylinder solution has $N_d=1/2$, giving $E_{\mathrm{in}}=2E_0/({ϵ}_r+1)$. To see the latter directly, take the applied field along $x$. The interior and exterior electric potentials can be written ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +A\cos \theta /r$. Continuity of potential and normal electric displacement at $r=a$ gives $B=2E_0/({ϵ}_r+1)$.

For a sphere of radius $a$, $N_d=1/3$ andFor example, write ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +C\cos \theta /r^2$. Continuity of potential gives $B=E_0-C/a^3$, and continuity of normal electric displacement gives $ϵB={ϵ}_0(E_0+2C/a^3)$. Solving yields $B=3E_0/({ϵ}_r+2)$ and $C=a^3{E}_0({ϵ}_r-1)/({ϵ}_r+2)$. The $1/r^2$ potential is that of the electric dipole moment above. For ${ϵ}_r=1$ all these induced effects vanish; as ${ϵ}_r$ becomes very large the normal-field sphere and slab approach the corresponding conductor field exclusion. The ideal infinitely long rod in an exactly axial imposed field has a different, singular end-free limit.

The electric field is approximately radial in the long interior, but near an electrode end its lines spread into a fringing field of a capacitor. The electric field magnitude falls on moving outward through that end region. The dielectric's induced electric dipoles experience a net force toward the stronger field, pulling the material into the capacitor. For a weakly perturbing polarizable element this is expressed by $\mathbf{F}=\frac{1}{2}\alpha \nabla E^2$, with positive electric polarizability $\alpha$.

In the sketch the field lines run from the positive electrode to the negative one, curving outward near their ends. The material is attracted toward the interior where the field is stronger. A perfectly uniform interior field by itself would not give a longitudinal dipole-gradient force; the end region is essential to the local explanation. Neglecting its small contribution to the total capacitance does not justify omitting the finite insertion force that it transmits. The energy argument of part (d) gives that force without needing the detailed fringe solution.

An electrical insulator has no freely conducting bulk charge in the ideal static description, but an applied electric field can displace bound positive and negative charge relative to one another. The resulting electric dipole moments per unit volume define the polarization density $\mathbf{P}$. Distortion of electronic charge clouds, relative displacement of ions, and partial alignment of permanent molecular electric dipoles can all contribute. Thermal agitation opposes the last mechanism.

For a weak field in a linear isotropic dielectric, the constitutive law isHere ${\chi }_e$ is the electric susceptibility, $ϵ$ is the permittivity, and $\mathbf{D}$ is the electric displacement field. In an anisotropic material, ${\chi }_e$ is a tensor and $\mathbf{P}$ need not be parallel to $\mathbf{E}$. These response laws describe the macroscopic field, not necessarily the microscopic field acting on one molecule.

Spatial variation and surface termination of the polarization density give bound charge and bound current through the bound charge densitieswhere $\hat{\mathbf{n}}$ is outward from the material. Thus uniform polarization density has no bulk bound charge, but generally has opposite bound charges on opposing surfaces. Gauss's law can be rewritten asWith no free surface charge, the normal component of $\mathbf{D}$ is continuous across an interface; the tangential electric field is continuous in electrostatics.

For example, fill a parallel-plate capacitor of plate area $A$ and gap $d$ with a uniform dielectric, neglecting fringing. At fixed free plate charge $Q$, $D=Q/A$, soThe induced bound surface charges oppose the original field, reducing the electric field and potential difference at fixed $Q$. At fixed applied potential difference, the field remains $V/d$ and the battery instead supplies more free charge. The electric displacement field separates these free-charge and bound-charge roles.

An electric dipole consists of opposite charges separated by a small displacement, with electric dipole moment $\mathbf{p}=q\mathbf{d}$ directed from negative to positive. At distances large compared with its size,The net charge term vanishes; the field decays faster than a monopole field. In an imposed slowly varying electric field,for a permanent fixed dipole. A uniform field produces a torque but no net force. For a linear induced dipole $\mathbf{p}=\alpha \mathbf{E}$, the interaction energy is instead $-\alpha E^2/2$, because building the dipole also costs energy.

A small planar current loop has magnetic dipole moment $\mathbf{\mu }=IA\hat{\mathbf{n}}$, with the normal fixed by the right-hand rule. Outside the source,Magnetic field lines form closed loops; a current-loop dipole does not require magnetic monopoles. Dipole moments describe the leading directional field of a neutral localized source, and their torques explain alignment in external fields.

An electric dipole consists of opposite electric charges separated by a short displacement. Its electric dipole moment is $\mathbf{p}=q\mathbf{d}$, directed from negative to positive electric charge. At distances large compared with the separation, its leading electric potential and electric field in vacuum areThe absence of net electric charge removes the monopole term; the dipole electric potential therefore falls as $r^{-2}$ rather than $r^{-1}$. On its axis the electric field is parallel to $\mathbf{p}$ with magnitude $2p/(4\pi {ϵ}_0{r}^3)$; on the equatorial plane it points oppositely with half that magnitude.

In a uniform applied electric field, the two forces cancel as a net translation but form a couple. The torque and potential energy areAlignment parallel to the electric field is stable. A nonuniform electric field can also exert a net force, $\mathbf{F}=\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed permanent electric dipole moment. An induced electric dipole instead depends on the applied electric field, so its energy must account for the work needed to create 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.

Normalization and angular pattern. A scattering phase function describes the distribution of single-scattered power over outgoing solid angle, for a specified incident direction, wavelength and polarization convention. Use the common normalizationThe scattering asymmetry parameter is the mean cosine of the scattering angle: forward scattering has positive $g$, backward scattering negative $g$, and a forward–backward symmetric pattern has $g=0$.

Let $C=E_0(2\pi /\lambda {)}^4[(n^2-1)/(n^2+2){]}^2{r}^6$. The given small-sphere angular intensity is $I=C(1+{\mu }^2)/2$, where $\mu =\cos \Theta$. Its total scattered power isNormalizing givesDimensionally the supplied $I$ is radiant intensity, power per solid angle. If interpreted as detector irradiance at a distance, the appropriate inverse-square distance factor is also needed; it cancels out of the normalized angular pattern.

The polar plot has equal forward and backward lobes, each with radius $3/2$, and lateral values $3/4$. A small nonabsorbing particle behaves as an induced electric dipole. Radiation transverse to the dipole and the average over two orthogonal incident polarizations produce the $1+{\cos }^2\Theta$ form. Its odd angular moment vanishes:As the size parameter $2\pi r/\lambda$ grows, Rayleigh scattering in the electric-dipole approximation no longer suffices. Higher multipoles and interference produce structure in the angular pattern, and large particles generally develop a strong narrow forward diffraction peak. Their $g$ is normally positive, although detailed values and angular features depend on refractive index, size distribution and shape; they are not fixed by radius alone. Large spheres may show rainbows and backscattering structure, while nonspherical ice introduces further angular features.

Optically thick transmission. In a nonabsorbing two-stream or transport approximation, it is the transport optical depth $(1-g)\tau$ that controls the loss of forward directional memory. Using the same idealized albedo closure as in option 1 gives total forward-hemisphere transmittanceFor the same large but finite extinction optical depth,The respective transport-depth factors are $0.15$, $0.30$ and $1$. The water cloud scatters most strongly forward and returns the least light backward. All transmittances tend to zero in the infinite-depth limit. The strictly unscattered direct-beam transmittance is instead $e^{-\tau }$, identical in the three cases: the ranking concerns total transmitted radiation, including forward-scattered light, not that direct component alone. Full radiative transfer can also depend on higher angular moments beyond $g$.

For the polybutadiene chain in 1,4-dioxane, define $R_x$ from the negative end charge toward the positive one. The electric dipole energy is $-eE{R}_x$, so the force conjugate to end separation is $eE$, even though the two endpoint forces are opposite. Balancing it against the entropic spring force givesNumerically,The zero-field rms of just the $x$ component is $60/\sqrt{3}=34.6\mathrm{nm}$, a different comparison. The mean extension is still much smaller than the contour length $Na=6\mu \mathrm{m}$, consistent with the Gaussian ideal chain force law. This is an end-charged ideal polymer in an electric field; the direct endpoint Coulomb interaction is omitted as stipulated.

Permittivity describes how an electric field and a material's electric polarization combine into the electric displacement field. In a static, local response with linearity and isotropy, and no electric polarization at zero electric field, the constitutive equation isHere ${ϵ}_0$ is vacuum permittivity, while ${ϵ}_r$ is the relative permittivity, a dimensionless quantity. The SI unit of permittivity is farad per metre, $F/m$: multiplying it by an electric field in $V/m$ gives an electric displacement field in $C/{\mathrm{m}}^2$. A larger positive permittivity lets the same parallel-plate capacitor hold more plate electric charge at the same electric potential difference.

The material contribution is electric polarization $\mathbf{P}$, the electric dipole moment per unit volume. The electric displacement field and electric susceptibility ${\chi }_e$ satisfySubstitution gives the first constitutive equation. The electric field here is the averaged, total field inside the material, not just the field of the external plate electric charges. The continuum approximation averages over many chemical atoms or molecules; locality means that the electric polarization at a position depends on the electric field there. Static equilibrium, fixed material conditions and sufficiently small electric fields are needed for this scalar proportionality. ^[ref].

For an ideal parallel-plate capacitor completely filled by a uniform dielectric, hold the free plate electric charge magnitude $Q$ fixed and neglect the fringing field of a capacitor. Writing $A$ for the plate area, Gauss's law gives $D=Q/A$. ThusThe electric dipoles point along the electric field. Their surface electric charges, described by polarization density, oppose the field from the free plate electric charges. For a hypothetical dielectric with ${ϵ}_r=4$, the internal electric field is one quarter of its vacuum value, while $P=3D/4$. At fixed electric potential difference instead, the internal electric field stays approximately $V/d$ and the plate electric charge grows fourfold. These are different experimental constraints. ^[ref].

Permittivity is a useful material summary, not a universal constant of the material under every condition. Temperature, frequency and electric field strength can change the response. In a linear anisotropic dielectric, permittivity is a tensor, $D_i={\sum }_j{ϵ}_{ij}{E}_j$, so the electric displacement field need not be parallel to the electric field. At nonzero frequency, delayed electric polarization requires complex permittivity and may produce dielectric loss. ^[ref].

For the field-free, spin-free linear rigid rotor in one fixed internal quantum state, the electric-dipole interaction is $-\mathbf{d}\cdot \mathbf{E}(t)$. A nonzero permanent electric dipole moment along the molecular axis allows pure rotational photon absorption with $\Delta J=+1$ and photon emission with $\Delta J=-1$. The matrix element between initial and final quantum states must be nonzero: the existence of an energy gap alone is insufficient. Relative to a chosen $z$ axis, polarization of an electromagnetic wave along $z$ selects $\Delta M=0$; transverse polarization of an electromagnetic wave permits $\Delta M=\pm 1$.^[ref]

Hydrogen chloride and carbon monoxide are examples with a permanent electric dipole moment. By symmetry, molecular hydrogen and carbon dioxide have no permanent electric dipole moment in the ideal symmetric internal state. They still rotate and have rotational energy eigenvalues, but the simple permanent-electric dipole spectrum described here is absent. This restriction concerns this interaction and these assumptions, not every possible way of detecting molecular rotation.

The geometrical optics account of an optical trap adds the momentum flux changes of refracted and reflected light rays. It applies when the bead is large compared with the surrounding medium's wavelength. When the bead is much smaller, an induced electric dipole gives a different approximation. For a homogeneous dielectric sphere of radius $a$, define $\mathbf{p}(t)=\mathrm{Re}[\alpha {\mathbf{E}}_0{e}^{-i\omega t}]$, with ${\mathbf{E}}_0$ the peak complex electric field and $\alpha$ its SI polarizability. In a transparent, nonmagnetic, nondispersive host,where $n_p,n_m$ are particle and host refractive indices, $I$ is cycle-averaged light intensity, ${ϵ}_0$ is the vacuum permittivity and $c$ is the speed of light. The quasistatic electric dipole approximation requires $2\pi n_{m}a/{\lambda }_0\ll 1$, with ${\lambda }_0$ the vacuum wavelength, and a field nearly uniform across the bead.

For real quasistatic $\alpha$, building an induced electric dipole gives instantaneous potential energy $-\alpha E^2/2$. Averaging over one optical cycle givesThus $n_p>n_m$ gives positive polarizability and attraction toward larger light intensity; $n_p<n_m$ reverses this contribution. The factor $1/4$ uses a peak field: substituting a root mean square field without changing that factor would halve the force incorrectly. Dimensionally $\alpha |E_0{|}^2$ is energy, so its gradient is force.

This leading conservative term does not remove the optical scattering force. Even a nonabsorbing induced electric dipole reradiates light: at fixed wavelength and index contrast, its Rayleigh scattering cross-section scales as $a^6$, while $\alpha$ scales as $a^3$. A smaller bead can therefore improve the ratio of gradient to scattering forces while weakening the well relative to $k_{B}T$, where $k_B$ is the Boltzmann constant and $T$ the temperature. Optical force balance and resistance to thermal escape are separate questions. Intermediate bead sizes require a wave-scattering calculation; neither a ray calculation nor this small-dipole expression is uniformly valid there.