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In electrostatics, Gauss's law is ${\oint }_{\partial \Omega }\mathbf{E}\cdot \mathbf{n}dS=Q_{\mathrm{enclosed}}/{ϵ}_0$, or locally $\nabla \cdot \mathbf{E}={\rho }_e/{ϵ}_0$, where ${\rho }_e$ is charge density and ${ϵ}_0$ is vacuum permittivity. Spherical symmetry makes the electric field radial, and Gauss's law givesChoose the additive constant of the electrostatic potential by $\varphi (\infty )=0$. Using $\mathbf{E}=-\nabla \varphi$ and continuity of potential across each charged shell givesThe field jumps at the shell surfaces, so the displayed formulas are for the requested open regions. The potential difference of the spherical capacitor is $\Delta \varphi =Q(1/a-1/b)/(4\pi {ϵ}_0)$. Thus its capacitance and electrostatic energy areThe energy follows either by charging work ${\int }_0^{Q}q/Cdq$, or directly from $W=({ϵ}_0/2)\int |\mathbf{E}{|}^2{d}^{3}x=Q^2(1/a-1/b)/(8\pi {ϵ}_0)$.

For any bounded volume $V$ with piecewise smooth closed boundary and outward unit normal vector $\mathbf{n}$, Gauss's law in integral form isThus the outward electric flux equals the enclosed electric charge divided by the vacuum permittivity. To prove it, apply the divergence theorem to the electric field and use its given divergence:The classical proof assumes the requisite differentiability of the electric field. For piecewise smooth charge densities, one may apply the divergence theorem piece by piece and cancel the internal fluxes.

Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:In electrostatics, the electric field is the negative gradient of the electrostatic potential, $\mathbf{E}=-\nabla V$. Substitution into the differential equation form of Gauss's law gives Poisson's equation ${\nabla }^{2}V=-\rho /{ϵ}_0$. Consequently, in a region containing no charge density,which is Laplace's equation. Charges on the bounding conductors enter through the boundary conditions; they need not vanish for Laplace's equation to hold inside the empty gap.

In a source-free vacuum, the Maxwell equations in SI arewhere ${ϵ}_0$ and ${\mu }_0$ are the vacuum permittivity and vacuum permeability. Take the curl of the last equation, substitute Faraday's law of induction, and use the curl of the curl identity:The zero divergence therefore gives the wave equationFor a propagating plane wave, write $\mathbf{B}=\mathrm{Re}[{\mathbf{B}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$. The wave equation gives ${\omega }^2=c^2|\mathbf{k}{|}^2$, and Gauss's law for magnetism gives $\mathbf{k}\cdot {\mathbf{B}}_0=0$. Thus the oscillating magnetic field is a transverse wave: it is perpendicular to the propagation wavevector. A static background magnetic field, if present, is separate from this propagating component.

Interpret the coordinates as a plane in ordinary three-dimensional electrostatics. Set $k_e=1/(4\pi {ϵ}_0)$, where ${ϵ}_0$ is the vacuum permittivity. The work required to bring the second point charge in quasistatically is its electric charge times the first charge's electric potential:The individual point-charge self-energies are excluded; this is the electrostatic potential energy of point charges relative to infinite separation.

By the superposition principle, for $\mathbf{r}=(x,y)$ and $\mathbf{d}=(d,0)$ the electric field isAt the first position both contributions point toward positive $x$, whereas at the second they oppose one another. ThusAt $(d/2,d)$ the two distances are $\sqrt{5}d/2$. The vertical electric field components cancel and the horizontal components add:Dimensional convention. The source's description of space as two-dimensional cannot literally mean two-dimensional electrostatics with the same inverse-square formula. In a genuinely two-dimensional Gauss's law, the circumference grows as $r$, giving an electric field proportional to $1/r$ and a logarithmic electric potential. For example, integrating $1/r$ from $d$ to infinity diverges, so the same finite reference potential energy at infinite separation does not exist. The displayed results use the conventional three-dimensional Coulomb's law restricted to the stated plane.

For this space-charge-filled parallel-plate capacitor, place the first plate at $x=0$ and the second at $x=d=x_o$, ignoring fringing field of a capacitor. With vacuum permittivity between the plates and fixed charge density $\rho$, Poisson's equation and the prescribed boundary electric potentials giveIntegrating twice and imposing both values yieldsThe electric field is minus the gradient of the electric potential, henceThus the electric potential is a parabola, concave downward for $\rho >0$ and upward for $\rho <0$. The signed electric field is a straight line. Its magnitude has a V-shaped corner if its zerolies inside the gap; otherwise its magnitude is a single straight segment. An interior zero requires $|V_1-V_2|<|\rho |d^2/(2{ϵ}_0)$. At that point the electric potential has its interior extremum. For $\rho =0$, these reduce to a linear electric potential and a uniform electric field.

In vacuum, Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:Take opposite electric charges $\pm Q$ on the two electrical conductors, with surface charge densities $\pm \sigma =\pm Q/A$. A Gaussian pillbox crossing one isolated large sheet gives $2ES=\sigma S/{ϵ}_0$, so each sheet contributes $\sigma /(2{ϵ}_0)$ on either side. By the superposition principle, the electric fields add between the sheets and cancel outside. Ignoring the fringing field of a capacitor, the gap electric field is $E=\sigma /{ϵ}_0$. The electric potential difference and capacitance of this parallel-plate capacitor are consequentlyThe approximation requires $d$ much smaller than both transverse sheet dimensions. A uniform filling with permittivity $ϵ$ replaces ${ϵ}_0$ by $ϵ$.

Choose zero electrostatic potential at infinity. By Coulomb's law and the superposition principle, the two point charges giveHere ${ϵ}_0$ is the vacuum permittivity. For a nondegenerate zero-equipotential curve, the electric charges must have opposite signs. Let $d_1,d_2$ be the two distances. The equation $\Phi =0$ becomes $q_1{d}_2=-q_2{d}_1$, and henceFor a circle $r=\rho >0$ centred at the origin, its $\cos \theta$ coefficient must vanish: $q_1^2{r}_2=q_2^2{r}_1$. If $r_1,r_2>0$ and $r_1\ne r_2$, this givesSubstitute $q_2^2/q_1^2=r_2/r_1$ in the constant term:ThereforeThis is an Apollonius circle: the distance ratio $d_2/d_1$ is constant. The distinct-position assumption matters. For example, $r_1=r_2=1$, $q_2=-q_1$ gives zero electrostatic potential everywhere away from the coincident point charges, so there is no distinguished zero circle. If only one position is at the origin and the other is not, no nontrivial positive-radius circle centred at the origin is possible.

For a smooth isolated conducting sphere, Gauss's law gives the surface electric field $E_s=Q/(4\pi {ϵ}_0{a}^2)$ and electric potential $V=Q/(4\pi {ϵ}_{0}a)=a{E}_s$. At the specified dielectric strength, $V_{\max }=3.0\times 10^3\mathrm{V}$. The capacitance of an isolated conducting sphere is $C=4\pi {ϵ}_{0}a$, so its capacitor energy isThis is the ideal breakdown-limited energy of a conducting sphere, taking air's permittivity approximately equal to vacuum permittivity and neglecting nearby objects and surface irregularities. The vacuum permittivity used numerically is $8.8542\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$.^[ref]

Put the charged sheet at $z=0$ and the grounded plates at $z=\pm d/2$. By symmetry the electric field points away from the positive sheet with equal magnitude in the two gaps. A Gaussian pillbox crossing the sheet givesThe plate boundary conditions fix their electric potentials to zero. For the charged sheet between grounded parallel plates, the central potential isWith vacuum permittivity ${\epsilon }_0=8.85419\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, $V_s\simeq 28.24\mathrm{V}$. An electron has electric charge $-e$, so moving from a grounded plate to this positive potential decreases its potential energy by $e{V}_s$. Conservation of energy then givesThe unrounded value is $3.15\times 10^6\mathrm{m}{\mathrm{s}}^{-1}$, about $0.0105c$. Either plate gives the same speed. The ideal calculation neglects the electron's disturbance of the prescribed field and image-force effects at its release point, consistently with treating it as a test particle.

Write $a=0.020\mathrm{m}$ for the side length. In an equilateral triangle, the distance from centre to a vertex is $a/\sqrt{3}$; it is two thirds of the altitude $\sqrt{3}a/2$. By the superposition principle, the electric potential at the centre due to the three fixed electrons isHere $e>0$ is the elementary charge and ${\epsilon }_0$ the vacuum permittivity. The approaching electron has electric charge $-e$, so its potential energy, measured relative to infinity, is positive:The fixed sources do not move, and radiation is excluded. Hence conservation of mechanical energy givesUsing $e=1.60218\times 10^{-19}\mathrm{C}$, $m_e=9.10938\times 10^{-31}\mathrm{kg}$ for the electron mass, and $(4\pi {\epsilon }_0{)}^{-1}=8.98755\times 10^9\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2}$ givesThis speed is small enough to justify classical kinetic energy. There is an idealization in the word “stops”: along the line normal to the triangle through its centre, this threshold energy makes the electron approach the central equilibrium asymptotically. A smooth, conservative trajectory cannot reach an equilibrium with exactly zero velocity in finite time unless it was already there. The energy threshold, and therefore the requested initial speed, is nevertheless unambiguous.

In a nematic liquid crystal, elongated molecules have orientational order around a nematic director $\mathbf{n}$, but lack the long-range positional order of a crystal. Head-tail symmetry identifies $\mathbf{n}$ and $-\mathbf{n}$. Molecular anisotropy and sufficient density favor collective alignment: for long hard rods this can increase translational entropy by reducing excluded-volume constraints, while attractive interactions can also favor alignment. Thermal orientational disorder competes with these effects.

The scalar nematic order parameter convention here isThus $S=0.44$ means $⟨{\cos }^2\theta ⟩=0.6267$. The effective orientation angle from nematic order isThis is not enough information to determine the literal mean angle $⟨\theta ⟩$: different orientation distributions can have the same second moment. Isotropic orientations have $S=0$, and perfect axial alignment has $S=1$.

In the supplied threshold expression, $d$ is the cell thickness; $K_1,K_2,K_3$ are the splay, twist and bend elastic constants of the Frank elastic energy. The dimensionless relative dielectric anisotropy of a nematic is $\Delta ϵ={ϵ}_{\parallel }-{ϵ}_{\perp }$, and ${ϵ}_0$ is the vacuum permittivity. Positive anisotropy favors a director along the applied electric field. Strong surface anchoring fixes orthogonal in-plane directions on the two plates. The low-field director therefore twists by a quarter-turn through the cell. Between crossed polarizers, with the entrance polarization matched to the first surface director, adiabatic optical following in a twisted nematic rotates the polarization and allows transmission. A sufficiently large normal field tilts the interior director towards the field, suppressing that rotation and making the crossed-polarizer cell dark; the anchored boundary layers remain distorted. Reversing the polarizer arrangement can reverse the optical contrast.

The threshold is a Fréedericksz transition: the electric energy gain overcomes the restoring elastic energy of an allowed distortion. Using the expression explicitly supplied by the paper givesThe thickness cancels for this strong-anchoring model. This is the onset of director reorientation, not an assertion that every molecule instantaneously becomes perfectly aligned. The quoted elastic combination is taken as given for this numerical task; a threshold derived from a specified full director-energy model must use that model's own elastic combination consistently. The companion illustration in A2(b) shows the plate anchoring, polarizers and bulk field-aligned state.

The Bohr model assumes Coulomb attraction and classical circular motion between transitions; selected stationary orbits do not radiate; their angular momentum is $mvr=n\hslash$; and a transition exchanges a photon of energy $h\nu =E_i-E_f$. The first two dynamical equations giveHere $n=1,2,\dots$, $m$ is the electron mass under the fixed-proton approximation, $e$ its charge magnitude, and ${ϵ}_0$ the vacuum permittivity. The stationary-orbit postulate deliberately suspends the classical radiation prediction; this is a historical model, not a consistent deduction from classical electrodynamics.

Two stabilization mechanisms. Steric stabilization coats each colloid with adsorbed or grafted polymer layers. Bringing two particles close overlaps or compresses those layers, increasing their free energy through loss of chain conformations and, in a suitable solvent, increased local osmotic pressure. The resulting repulsion can keep the cores apart. A well-solvated nonionic coating can remain effective when added salt would screen electrostatic forces. Coatings can fail through desorption, poor solvent conditions or bridging, so simply adding arbitrary free polymer does not guarantee stabilization.

Charge stabilization of colloids instead uses similarly charged surfaces and their diffuse electric double layers. Their approach costs electrostatic and ionic free energy. In DLVO theory, that repulsion competes with attractive Van der Waals forces. It is particularly sensitive to salt concentration, counterion valence and surface ionization.

Bjerrum length. The standard Bjerrum length is the separation at which the Coulomb interaction energy of two elementary charges equals thermal energy:At $T\simeq 298\mathrm{K}$ and ${\epsilon }_r=78$, this isIt compares ionic electrostatic coupling with thermal agitation. For ions of signed valences $z_1,z_2$, $U/(k_{B}T)=z_1{z}_2{\ell }_B/r$. The standard reference length uses $e$, not the whole colloid charge: assigning a colloid charge $Ze$ does not change the solvent's tabulated ${\ell }_B$.

Screening and the printed normalization. In a symmetric $z:z$ electrolyte, $z$ is the ionic valence magnitude, $e$ the elementary electric charge, $n_{\infty }$ the bulk number density of each ionic species, ${\epsilon }_0$ the vacuum permittivity, ${\epsilon }_R$ the solvent's relative permittivity, $k_B$ the Boltzmann constant and $T$ absolute temperature. Mobile ions rearrange around a charged surface; the Debye–Hückel screening length ${\kappa }^{-1}$ sets the range of the resulting weak potential.

There is an SI normalization error in the original PDF. Combining the Boltzmann distribution $n_{\pm }=n_{\infty }{e}^{\mp ze\psi /(k_{B}T)}$ with the SI Poisson equation and linearizing at $|ze\psi |\ll k_{B}T$ givesThe extra $4\pi$ in the PDF denominator is inconsistent with those SI charge, permittivity and density conventions. Keeping the printed algebra would make its screening length $\sqrt{4\pi }$ times the SI value at the same ion density. No numerical screening length is required here, and the qualitative dependence ${\kappa }^{-1}\propto n_{\infty }^{-1/2}$ is the same in either normalization. Added salt shortens the range of repulsion, allowing the attractive potential to capture particles more readily.

Interaction-potential sketches. Let $h$ be the surface separation between two spheres of radius $a$. A useful schematic DLVO theory potential has the formaway from the microscopic contact region. Here $A_H>0$ is a Hamaker constant; excluded-core repulsion and molecular details regularize contact. The exponential is the double-layer repulsion and the inverse-gap term the short-gap Van der Waals force contribution. A fixed $B$ illustrates the changing screening range; actual charge regulation can change the repulsive amplitude as well.

At low salt there is a high positive barrier between a deep primary minimum near contact and the dispersed state at $U(\infty )=0$. A shallow secondary minimum can occur outside the barrier and support weak reversible association. Increasing salt narrows the repulsive range and lowers the barrier. Thermally activated primary aggregation becomes appreciable when the barrier is only a few $k_{B}T$; the idealized barrier-free condition defines the fast-aggregation threshold associated with the critical coagulation concentration. Above that threshold, collisions can result in rapid capture in the primary minimum.

Aggregate structures. Well above the critical coagulation concentration, the sticking probability is close to one. In diffusion-limited cluster aggregation, the first encounter attaches particles or clusters at accessible outer branches before they can explore many alternative configurations. The result is an open, tenuous structure. Typical ideal three-dimensional cluster models have mass fractal dimension $d_f\simeq 1.8$.

Near the threshold on its lower-salt, slow-aggregation side, a remaining barrier makes most encounters unsuccessful. Reaction-limited cluster aggregation then permits repeated approaches and orientations before attachment, commonly giving denser clusters with $d_f\simeq 2.1$. Since particle count scales as $N\propto (R_g/a{)}^{d_f}$, the larger exponent means more particles at a given large cluster size. This explanation does not require already formed bonds to rearrange: unsuccessful encounters themselves alter which configurations eventually bind.

The side of the threshold matters. At the ideal barrier-free threshold itself, or just above it, aggregation can already be diffusion limited; proximity to the critical coagulation concentration alone does not universally imply a denser reaction-limited structure. The quoted dimensions are typical of irreversible approximately isotropic cluster models, not exact values for every colloid. Strong bond rearrangement or anisotropic attractions can change this comparison.

Stabilization and screening. Charge stabilization of colloids means suppressing aggregation through repulsion between similarly charged particles and their electric double layers. In DLVO theory, this repulsion competes with attractive Van der Waals forces. A substantial repulsive barrier compared with thermal energy $k_{B}T$ can make aggregation slow, even when the dispersed state is only metastable.

The Debye–Hückel screening length ${\lambda }_D={\kappa }^{-1}$ is the characteristic range over which a weak electric potential is attenuated by rearrangement of mobile ions. It measures the thickness of a diffuse screening cloud, not a molecular diameter. For a plane, a weak potential decays as $e^{-\kappa x}$ into the bulk.

Derivation and the source convention. Put a planar interface at $x=0$, with electrolyte in $x>0$, bulk potential zero, and absolute permittivity $\epsilon ={\epsilon }_0{\epsilon }_r$. For a symmetric $z:z$ electrolyte, let $n_{\infty }$ be the bulk number density of each ionic species. The Boltzmann distribution givesso the mobile charge density isSubstitution into the SI Poisson equation, ${\nabla }^2\psi =-{\rho }_e/\epsilon$, gives the Poisson-Boltzmann equationFor $|ze\psi |\ll k_{B}T$, linearize the hyperbolic sine. The Debye–Hückel approximation is thenHere $z$ is the ionic valence magnitude, $e$ the elementary electric charge, $k_B$ the Boltzmann constant, $T$ absolute temperature, ${\epsilon }_0$ the vacuum permittivity, and ${\epsilon }_r$ the solvent's relative permittivity. For several ionic species, the numerator $2z^2{e}^2{n}_{\infty }$ is replaced by $e^2{\sum }_i{z}_i^2{n}_{i,\infty }$.

The original PDF includes an additional $4\pi$ in the denominator. That factor does not follow from SI Poisson equation with number densities and the stated permittivities. The Bjerrum length provides a useful check: ${\ell }_B=e^2/(4\pi \epsilon k_{B}T)$ and the consistent result is ${\kappa }^2=8\pi {\ell }_B{z}^2{n}_{\infty }$. A Gaussian-unit convention moves factors of $4\pi$ together with the charge and permittivity conventions; it does not justify inserting this extra factor while retaining all the given SI quantities. Both numerical interpretations are reported below rather than silently changing the source.

Conditions favoring stability. Increasing the surface-potential or surface-charge magnitude usually strengthens the electrostatic barrier. Lower ionic strength increases ${\lambda }_D$ and generally makes repulsion longer ranged; added salt and especially multivalent counterions shorten it and can promote aggregation. Suitable pH can keep an ionizable surface charged, while conditions near an isoelectric state weaken charge repulsion. The solvent-dependent Van der Waals forces, charge regulation and particle geometry also affect the barrier, so a long screening length is not by itself sufficient. At fixed density and valence, the linear formula gives ${\lambda }_D\propto \sqrt{\epsilon T}$, but actual temperature changes can also alter the permittivity and surface chemistry.

Surface charge at small potential. Take the usual one-sided planar-interface model, with negligible electric displacement on the nonaqueous side. The solution decaying into the solution isThe electric field is $E_x=-{\psi }^{\prime }=\kappa {\psi }_0{e}^{-\kappa x}$. The electrostatic interface boundary conditions then give the surface charge densityOne can also derive it by integrating the diffuse charge density:Thus the diffuse layer neutralizes the surface charge. This is the planar Debye-Hückel double layer. If instead the object is an isolated sheet wetted by identical electrolyte on both sides, each side screens half its total charge and ${\sigma }_{\mathrm{sheet}}=2\epsilon \kappa {\psi }_0$. The one-sided assumption is the conventional local model for the exterior of a large charged colloid; it must be stated because the question does not specify the material on the other side.

Numerical ion density. For $T=298.15\mathrm{K}$, $z=1$, $\kappa =10^9{\mathrm{m}}^{-1}$ and $\epsilon =80{\epsilon }_0=7.0834\times 10^{-10}\mathrm{F}{\mathrm{m}}^{-1}$, the consistent SI result isThis is $\boxed{0.0943\mathrm{mol}{\mathrm{L}}^{-1}}$ for each ion species, and hence also the salt concentration for a monovalent $1:1$ salt. The combined number density of both signs is $2n_{\infty }$.

Literal substitution into the formula printed in the PDF instead givesThese values are distinct consequences of the source's extra factor; they cannot both describe the same physical screening length in SI with the same definition of $n_{\infty }$.

Surface potential and validity. The requested linear estimate in the one-sided model isHowever, the thermal voltage is $k_{B}T/e=25.7\mathrm{mV}$, so $e{\psi }_{0,\mathrm{linear}}/(k_{B}T)\simeq 13.7$. Thus the small-potential approximation is not self-consistent for the supplied surface charge. The boxed value is its formal extrapolation, not a reliable prediction of the actual surface potential.

For a valid nonlinear comparison within the ideal one-sided planar Poisson-Boltzmann equation, multiply the differential equation by ${\psi }^{\prime }$ and use zero potential and field in the bulk. This givesUsing the surface boundary condition yields the Grahame equation and its inverse:With the SI screening length, the latter gives $\boxed{{\psi }_{0,\mathrm{PB}}\simeq 0.135\mathrm{V}}$. This is a nonlinear mean-field estimate under explicit assumptions, not a unique experimentally determined potential: compact interfacial layers, finite ion size and surface chemistry were not supplied. The source's normalization problem and its use of a linear formula far outside the small-potential regime are separate issues.

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].

A radio wave traversing a dilute cold plasma has a group velocity that depends on its frequency. In the ideal unmagnetized limit, the dispersion relation is ${\omega }^2=c^2{k}^2+{\omega }_p^2$, where ${\omega }_p$ is the angular plasma frequency. For $\omega \gg {\omega }_p$, differentiate to obtain $v_g=c\sqrt{1-{\omega }_p^2/{\omega }^2}$, then expand $1/v_g$ to first order. The excess travel time over a vacuum path isHere $f=\omega /(2\pi )$ is ordinary frequency, $e$ is the magnitude of the elementary charge, $m_e$ is the electron mass, ${ϵ}_0$ is the vacuum permittivity, $n_e$ is electron number density, and $c$ is the speed of light. This approximation neglects magnetic corrections, collisions and multipath scattering; its positive sign means lower frequencies arrive later.

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.

Consider a fixed, homogeneous, nonabsorbing, nonmagnetic sphere of radius $a$ in vacuum. Let $m$ be its relative refractive index, a fixed real number; in vacuum this equals the sphere's refractive index. Treat $m$ as independent of wavelength over the range being compared, neglecting optical dispersion. Air can be approximated by vacuum here because its refractive index is close to one. These assumptions describe a toy particle, rather than a literal model of an individual air molecule.

The Rayleigh scattering cross-section $\sigma$ is an effective area: the total scattered power is $P_{\mathrm{sca}}=I_{\mathrm{inc}}\sigma$, where $I_{\mathrm{inc}}$ is the incident light intensity. For $ka\ll 1$ its leading approximation isThe units are area, since $k^4{a}^6$ has units ${\mathrm{m}}^{-4}{\mathrm{m}}^6={\mathrm{m}}^2$. This Rayleigh scattering cross-section measures scattering, not conversion of the incident electromagnetic radiation into heat. The definition follows ^[ref].

To see the mechanism, the nearly uniform applied electric field produces an electric dipole moment with peak magnitudewhere $E_0$ is the incident electric field peak wave amplitude and ${ϵ}_0$ is vacuum permittivity. This is the sphere's response in electrostatics, with its permittivity divided by vacuum permittivity equal to $m^2$ under the nonmagnetic assumption. ^[ref]. The averaged electric dipole radiation power is $P_{\mathrm{sca}}={\omega }^4{p}_0^2/(12\pi {ϵ}_0{c}^3)$, with angular frequency $\omega =ck$ and speed of light $c$. Dividing by $I_{\mathrm{inc}}={ϵ}_{0}c{E}_0^2/2$ gives the displayed Rayleigh scattering cross-section. Thus doubling radius multiplies the electric dipole moment by eight and the scattered power by $64$, provided the larger sphere still satisfies $ka\ll 1$. ^[ref].

The dimensionless ratio of a medium's permittivity to vacuum permittivity. The constant scalar value used in a Wannier-Mott exciton is an approximation to the relevant material response.