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The leading electric dipole radiation fields areThe radiation electric field is transverse, $\mathbf{n}\cdot {\mathbf{E}}_{\mathrm{rad}}=0$. The vector triple product therefore gives the leading Poynting vectorIf $\theta$ is the angle between ${\mathbf{p}}^{\prime \prime }(t-r/c)$ and $\mathbf{n}$, thenUsing ${\mu }_0{ϵ}_0{c}^2=1$, the final result isThe leading Poynting flux is outward, vanishes along the instantaneous retarded ${\mathbf{p}}^{\prime \prime }$ axis, and is maximal in its perpendicular plane. If ${\mathbf{p}}^{\prime \prime }=0$, the radiation term is zero and no angle needs to be assigned. The large-$r$ expansion assumes bounded retarded electric dipole moment and time derivatives; it may also be read at fixed retarded time.

In SI units, the sourced Maxwell equations areTake the divergence of the last equation. The divergence of a curl vanishes; using the first equation gives the necessary local conservation lawFor sources supported in a fixed region, integrate over a containing surface where the current vanishes. The divergence theorem givesThus the total charge is constant. For each component of the electric dipole moment, integration by parts givesThe surface term again vanishes, soSmooth compact support suffices for these manipulations; equivalently one may integrate the charge continuity equation over all space and use the fixed support to identify the integrals with those over $V$.

Take the electric dipole moment along the polar axis, $\mathbf{p}=p\hat{\mathbf{z}}$. The electric field is minus the gradient of the electric potential; in spherical polar coordinates,Since $\mathbf{p}=p(\cos \theta {\hat{\mathbf{e}}}_r-\sin \theta {\hat{\mathbf{e}}}_{\theta })$ and $\mathbf{p}\cdot {\hat{\mathbf{e}}}_r=p\cos \theta$, the radial and angular components combine asThe origin is excluded because the point-dipole potential is singular there.

For a localized charge distribution and a field point $r$ outside it, expand the Coulomb kernel:Integrating against the charge density gives the electrostatic multipole expansionwhere $Q=\int \rho d^{3}x$, $p_i=\int \rho x_i{d}^{3}x$ is the electric dipole moment, and the symmetric trace-free electric quadrupole moment is $Q_{ij}=\int \rho (3x_i{x}_j-|x{|}^2{\delta }_{ij})d^{3}x$. This states the tensor normalization explicitly.

For the tetrahedral point charges, the total charge is zero and the three other vertex vectors sum to minus $(1,1,1)$. HenceEvery vertex has $|x{|}^2=3$ and each $x_i^2=1$, so each diagonal quadrupole entry vanishes. For each distinct coordinate pair the charge-weighted sum of $x_i{x}_j$ is $4q$, yieldingThe tensor is trace-free as required. If the alternative raw second-moment convention is used, its off-diagonal entries are $4q$ and the kernel contraction has the corresponding factor of three; the physical potential is unchanged.

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 electric potential of a point charge $Q$ at the origin isFor the two charges forming the dipole,The first-order Taylor expansion at large $|\mathbf{x}|$ isand therefore, with the electric dipole moment $\mathbf{p}=Q\mathbf{d}$,to leading order. Taking the interaction of a second dipole with the corresponding electric field gives the stated electric dipole-dipole interaction

Let the lattice spacing be $d$ and write the central dipole asIts two horizontal neighbours have moment $p(\cos \theta ,-\sin \theta )$. For either one, the expression in parentheses, after extracting $d^{-3}$, isIts two vertical neighbours have moment $p(-\cos \theta ,\sin \theta )$, and each contributes insteadAdding all four nearest-neighbour interactions and using the Pythagorean trigonometric identity givesThe angle has cancelled, so the energy is independent of $\theta$.

Successive reflections in the two grounded planes require four charges:Their total charge and electric dipole moment vanish. Applying the second-order expansion from part (a), the terms proportional to $a^2{x}^2$, $b^2{y}^2$, and $a^2+b^2$ also cancel, while the mixed terms add. ThusThis is an electric quadrupole potential.

An anomalous chiral redefinition moves phase between the bare QCD angle and the quark mass matrix, so the measurable parameter isThe neutron electric dipole moment requires $|\bar{\theta }|$ to be extremely small, although no generic Standard Model symmetry enforces this. This is the Strong CP problem.

In Peccei–Quinn theory, an anomalous spontaneously broken global symmetry replaces $\bar{\theta }$ by a dynamical axion field. Nonperturbative QCD generates an axion potential whose minimum lies at the CP-conserving value, dynamically relaxing the effective angle to zero.

The far field of a dipole is $[3(p\cdot \hat{x})\hat{x}-p]/(4\pi {\epsilon }_0{r}^3)$. Comparison yields the induced electric dipole moment

The hydrogen ground-state charge cloud is spherically symmetric and has even parity, so it has no permanent electric dipole moment. A uniform electric field couples through a perturbation proportional to $\mathbf{E}\cdot \mathbf{r}$, which is odd under parity. Its expectation in the ground state therefore vanishes, giving no first-order energy correction.

The field can nevertheless polarize the atom: it admixes odd-parity excited states and creates an induced dipole. Such mixing contributes from second order onward, producing the quadratic Stark effect of the hydrogen ground state.

The electric dipole moment of a localized charge density isDifferentiate and use the continuity equation $\dot{\rho }+\nabla \cdot \mathbf{J}=0$:Localization makes the boundary integral vanish. Component by component,This is the time derivative of the electric dipole moment.

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.

Write $C_4=\alpha Q^2/[2(4\pi {ϵ}_0{ϵ}_r{)}^2]$. Differentiating the stipulated potential givesFor positive $\alpha$, the force points toward smaller separation for either sign of $Q$.

Its physical origin is charge-induced dipole attraction, rather than London dispersion forces. A point-charge field induces $\mathbf{p}=\alpha \mathbf{E}$ and the energy cost of inducing that moment gives $u=-\alpha E^2/2$. The charge changes the electric dipole moment; it does not, within this linear model, change the intrinsic coefficient $\alpha$. For a conducting sphere in a uniform medium, $\alpha =4\pi {ϵ}_0{ϵ}_r{a}^3$; other dielectric contrasts give another electric polarizability.

There is a distance inconsistency in the stated model: the point-charge field uses center-to-center separation $R$, whereas the question calls $r$ a surface gap. For a point ion and sphere, $R=a+r$. A controlled far-field dipole model therefore uses $u\simeq -C_4/(a+r{)}^4$, with $R\gg a$. The boxed derivative is the literal result for the given mathematical $u(r)$; it is not a justification for extending that dipole formula to contact.

For a Hertzian dipole of length $\ell$ driven at peak electric current $I_0$, the electric dipole moment amplitude is $p_0=I_0\ell /\omega$. ConsequentlyThe Hertzian dipole fixed-current power scaling is thus $\overline{P}\propto I_0^2{\ell }^2/{\lambda }^2$, givingThe inverse-fourth-power wavelength scaling would hold at fixed electric dipole moment, not at the fixed electric current specified here.

Molecules in clean air are much smaller than visible wavelengths. Their induced electric dipole moments undergo Rayleigh scattering. Away from molecular resonances, the induced dipole amplitude is approximately $p=\alpha E$, with only weak wavelength variation in the polarizability $\alpha$. The radiation-zone electric field is proportional to $\ddot{p}$, so the scattered intensity satisfiesThis is the Rayleigh scattering wavelength dependence. For equal incident intensities, $450\mathrm{nm}$ light scatters about $(650/450{)}^4=4.35$ times as strongly as $650\mathrm{nm}$ light. Thus light scattered out of the direct solar beam and into our line of sight is enriched in blue wavelengths.

The dipole mechanism also determines the Rayleigh scattering angular pattern. For unpolarized light, its angular factor is $1+{\cos }^2\vartheta$; near $90^{\circ }$ scattering, the scattered light has strong linear polarization. The sky's perceived colour depends additionally on the solar spectrum, atmospheric transmission and the eye's spectral response, so the ${\lambda }^{-4}$ law alone does not predict a violet sky. Over a long near-horizontal path the direct beam loses more short-wavelength light, explaining red sunsets. Large aerosol particles instead produce Mie scattering, which is less strongly wavelength-selective and can make haze look whitish.

Put ${\omega }_0^2=k/m$, where $k$ is the binding spring constant rather than an optical wavenumber. The steady complex amplitude solution of the forced Lorentz oscillator model isThe acceleration complex amplitude is $-{\omega }^2{\mathbf{x}}_0$. HenceThis is the scattering cross-section of a harmonically bound charge. The displacement and induced electric dipole moment are parallel to the incident linear polarization, and integrating their dipole radiation over all directions gives the total cross section above.

For $\omega \ll {\omega }_0$, binding dominates andrecovering Rayleigh scattering with static electric polarizability $q^2/k$. For $\omega \gg {\omega }_0$, inertia dominates: ${\mathbf{x}}_0\simeq -q{\mathbf{E}}_0/(m{\omega }^2)$, so its second time derivative becomes frequency independent andThis is a classical Thomson scattering regime only while $\hslash \omega \ll m{c}^2$. Exactly at $\omega ={\omega }_0$, an undamped driven oscillator has no finite steady-state amplitude; the divergent formula must not be used. Finite damping and radiation reaction regulate the physical resonant response. Coherent contributions from many charges in an extended medium can also require a structure calculation rather than multiplying this single-charge result by a particle count.

In Rayleigh scattering, a small bound system develops an induced electric dipole moment $\mathbf{p}(t)=\mathrm{Re}[\alpha (\omega ){\mathbf{E}}_0{e}^{-i\omega t}]$. Well below its binding resonances, the electric polarizability is approximately its static value $\alpha (0)$, independent of frequency. The mean radiated power and incident light intensity areDividing gives the total scattering cross sectionThis Rayleigh scattering wavelength dependence comes from the two time derivatives of an oscillating electric dipole moment, whose radiated field is proportional to ${\omega }^2{p}_0$. It requires the same scatterer with approximately constant low-frequency electric polarizability; resonances and strong optical dispersion invalidate the simple fourth-power law.

Work in the dilute, single-scattering limit and add the diffuse scattered light intensities of randomly positioned particles. The incident unpolarized light has equal, mutually uncorrelated $x$ and $y$ electric components. Each particle therefore has an induced electric dipole moment along its permitted axis, with the same mean squared magnitude proportional to $|\alpha {|}^2$.

For an observer in direction $\hat{\mathbf{n}}=(\sin \psi ,0,\cos \psi )$, choose transverse unit directions ${\mathbf{e}}_{\parallel }=(\cos \psi ,0,-\sin \psi )$ in the scattering plane and ${\mathbf{e}}_{\perp }=\hat{\mathbf{y}}$ perpendicular to it. The electric dipole radiation field is proportional to the transverse projection of the electric dipole. Hence an $x$ electric dipole contributes a wave amplitude factor $\cos \psi$ along ${\mathbf{e}}_{\parallel }$, while a $y$ electric dipole contributes its full wave amplitude along ${\mathbf{e}}_{\perp }$.

Let $C>0$ collect the common incident light intensity, electric polarizability, distance and frequency factors. ThenThe two components are uncorrelated, so the coherency matrix is diagonal in this basis. The degree of polarization isA signed degree of linear polarization, defined positive perpendicular to the scattering plane, has the same expression without the absolute value. The dominant direction is $\hat{\mathbf{y}}$ when $N_2>N_1{\cos }^2\psi$, and ${\mathbf{e}}_{\parallel }$ when the inequality reverses. If the denominator vanishes there is no scattered signal and its degree of polarization is undefined. A coherent exactly forward component, including the transmitted incident beam, is not part of this diffuse-intensity calculation; random particle positions do not by themselves destroy forward optical coherence.

The prescribed surface charge density is that of a uniformly polarized sphere and has zero total charge because $\int \cos \theta d\Omega =0$. Its electric dipole moment at $t=0$ isThe transverse components cancel by axisymmetry, while the $z$ component isThusThe angular surface distribution is a pure degree-one spherical harmonic. Hence its static external electric potential is exactly that of this dipole, with no higher electric multipole moments; it is not an assertion that the potential inside the finite sphere equals a singular point-dipole potential.

For right-handed uniform rotation about $+\hat{\mathbf{x}}$, the rigidly attached dipole isConsequently $|\ddot{\mathbf{p}}{|}^2=p_0^2{\omega }^4$ is constant. The electric dipole radiation formula givesThere is no extra factor of one-half: the squared sine and cosine contributions of the two perpendicular oscillating components sum to one. This is the leading electric-dipole approximation for radiation, requiring $\omega R/c\ll 1$ and observation in the radiation zone. The stated permanent surface distribution is interpreted as the actual charge producing the given external dipole; adding another dielectric screening factor would contradict the dipole moment the question asks to establish. The relative permittivity enters the induced-polarization problem in part (d), not this prescribed total dipole moment.

At $(x,0)$ the distances from the two centres are equal:By the electrostatic shell theorem, a uniformly charged spherical shell gives zero field inside and its point-charge field outside. For $x<x_1$ the point is inside both shells. For $x_1<x<x_2$ only the small, negatively charged shell contributes. For $x>x_2$ both contribute, soThe requested magnitude isIt jumps from zero to $k_{e}q/a^2$ at $x_1$, decreases to $k_{e}q/(4a^2)$ just before $x_2$, jumps down to $k_{e}q/(8a^2)$ just after $x_2$, and then falls as $x^{-3}$. The ideal surface charge makes the field discontinuous at each crossing, so the one-sided limits, rather than a unique value on the sheet, specify the graph. The far field is the equatorial field of an electric dipole with electric dipole moment $\mathbf{p}=-qa\hat{\mathbf{y}}$.