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Normalize a point source by $\int \rho d^{3}xdz=q_1$ on the compact spatial direction. The periodic delta has the Fourier seriesWriting the potential as $\varphi (r,z)={\sum }_n{\varphi }_n(r)e^{inz/R}$ reduces the Poisson equation toAway from the source, the decaying radial solution is proportional to $e^{-|n|r/R}/r$. Its distributional normalization follows from integrating the Laplacian over a small sphere: the radial flux of $1/r$ is $-4\pi$, soThis is the contribution of the $n$th Kaluza-Klein mode, with inverse range $|n|/R$. At equal compact coordinate, the sum is a geometric series,Therefore the pair interaction energy, without the individual self-energies, isIts normalization is fixed by the stated $-4\pi R\rho$ source, rather than by importing another convention for Coulomb's law. More generally summing the phases gives the periodic electrostatic Green function with one compact dimension,An independent check uses the method of images in four spatial dimensions. Since ${\nabla }_4^2{s}^{-2}=-4{\pi }^2{\delta }^4$, its periodic image sum is $R{q}_1{\sum }_n[r^2+(z+2\pi nR{)}^2{]}^{-1}/\pi$, with the same normalization and short-distance singularity.

For $r\gg R$, only the zero Kaluza-Klein mode survives:For $r\ll R$, $\coth x=x^{-1}+x/3+O(x^3)$ givesThe leading potential thus changes from inverse distance to inverse squared distance. Differentiating gives the signed outward forceIt scales as $r^{-2}$ at large distance and as $2q_1{q}_{2}R/(\pi r^3)$ at small distance. A precision inverse-square-law test over separations approaching $R$ would reveal the extra dimension through this change of force law or its exponential corrections. The model assumes localized sources at equal compact coordinate and an electromagnetic field able to explore the circle; a field confined to ordinary spacetime would not show this effect. Finite source size and ordinary screening must be controlled when comparing a real force measurement with this point-source prediction.

With the proton treated as fixed and electron mass $m$, the classical energy and angular momentum areFor a circular orbit, $L=mrv$ and Coulomb's law determines its centripetal acceleration:Eliminating $v$ gives $r=4\pi {ϵ}_0{n}^2{\hslash }^2/(m{e}^2)$. Thus the Bohr radius and orbit radii areThe kinetic energy is half the magnitude of the attractive potential energy. HenceThis is the circular-orbit Bohr model; including proton recoil replaces $m$ by the reduced mass.

The time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\Psi =E\Psi$. For the hydrogen atom, separate variables as $\Psi =g(r)Y_{\ell m}(\theta ,\phi )$. The spherical-coordinate Laplacian has radial part $r^{-2}{\partial }_r(r^2{\partial }_r)$ and angular part $r^{-2}{\Delta }_{S^2}$. A spherical harmonic obeys ${\Delta }_{S^2}{Y}_{\ell m}=-\ell (\ell +1)Y_{\ell m}$. The first displayed term is therefore radial kinetic energy; the second is the attractive Coulomb's law potential energy $-e^2/(4\pi {\epsilon }_{0}r)$ of electron and proton; the third is the angular kinetic energy, also called the centrifugal potential. Regular single-valued angular states have $\boxed{\ell =0,1,2,\dots }$, with orbital angular momentum squared ${\hslash }^2\ell (\ell +1)$.

For $g=r^{\alpha }{e}^{-\beta r}$, direct differentiation givesSubstitution into the Radial Schrodinger equation for the hydrogen atom and equality of the coefficients of $r^{-2}$, $r^{-1}$ and the constant term giveThe regular radial solution selects $\alpha =\ell$, rather than the singular branch $-\ell -1$. At $\ell =0$, $r^{-1}$ is locally square integrable but is still inadmissible as a regular Coulomb eigenfunction: its Laplacian has a point-source contribution at the origin. Bound-state decay requires $\beta >0$. Thus, with the Bohr radius $a_0=4\pi {\epsilon }_0{\hslash }^2/(m{e}^2)$,These are the nodeless states with principal quantum number $n=\ell +1$. Increasing $\ell$ raises the energy towards zero, so emission is from the state labelled $\ell +1$ to the one labelled $\ell$. Using the Planck constant $h=2\pi \hslash$ and photon energy $h\nu$, the frequency is

Place the four fixed charges at $(sa,ta)$ with $s,t\in \{-1,1\}$, and write the moving particle's position as $(x,y)$. Introduce its mass $m>0$, which is needed for an oscillation frequency but is not named in the question. Let $C=Qq/(4\pi {ϵ}_0)>0$. By Coulomb's law and superposition, the moving particle's potential energy isThe electrostatic potential itself is $V/q$. Put $r^2=x^2+y^2$ and ${\delta }_{st}=-(sx+ty)/a+r^2/(2a^2)$. Near the centre, a Taylor expansion givesOn summing over the four corners, linear terms cancel. To quadratic order, $\sum {\delta }_{st}=2r^2/a^2$ and $\sum {\delta }_{st}^2=4r^2/a^2+O(r^4/a^4)$, since $\sum st=0$. Reflection symmetry removes all odd-degree terms. HencewhereThus the centre has zero planar force and a positive-definite planar Hessian matrix $\kappa I_2$. It is a strict local minimum of the constrained potential energy, so it is a stable equilibrium. More explicitly, conservation of the total energy confines sufficiently small-energy motions within a small neighbourhood of this isolated local minimum.

Keeping only the quadratic potential, the two coordinates obey independent small oscillations:Both planar modes have the same angular frequency, and the ordinary frequency isThe planar constraint is essential to this constrained electrostatic equilibrium. Along the perpendicular axis, $V(0,0,z)=4C/(2a^2+z^2{)}^{1/2}=V(0,0)-C{z}^2/(\sqrt{2}{a}^3)+O(z^4)$, so the same stationary point is unstable to unconstrained normal displacements. That also checks the cancellation of the three-dimensional Hessian trace required by Laplace's equation away from charges; it does not contradict the positive-definite restriction to the given plane.

A stellar equation of state supplies pressure and internal energy as functions of density, temperature and composition, together with thermodynamic derivatives needed for stability and transport. Hydrostatic equilibrium fixes the pressure gradient, but does not determine which microscopic components provide the pressure. In ordinary dense interiors local thermodynamic equilibrium is a useful starting point. A consistent mixture iswhere ions, electrons, radiation and interaction corrections are distinguished. The classical electron pressure and electron degeneracy pressure are two limits of the same electron contribution, and must not be added as if they belonged to different particles. The finite-temperature electron equation of state interpolates between them.

In a fully ionized, nondegenerate, nonrelativistic gas, $P_g=\rho k_{B}T/(\mu m_u)$ and specific thermal energy is $u_g=3k_{B}T/(2\mu m_u)$. With nuclear mass fractions $X_A$, charges $Z_A$ and mass numbers $A_A$, the mean molecular weight satisfies ${\mu }^{-1}={\sum }_A{X}_A(1+Z_A)/A_A$, and ${\mu }_e^{-1}={\sum }_A{X}_A{Z}_A/A_A$ is the mean molecular weight per electron. This regime describes much of an ordinary main sequence interior. Toward cooler layers, ionization and molecular dissociation change particle numbers and consume heat. The Saha equation relates ionization to both temperature and electron density: there is no universal horizontal ionization boundary. These regions have larger heat capacity and can have a reduced stellar adiabatic exponent. The simple fully ionized formula is then insufficient.

Equilibrium photons give radiation pressure $P_{\gamma }=a_r{T}^4/3$ and energy per volume $a_r{T}^4$, or specific energy $u_{\gamma }=a_r{T}^4/\rho$. In the nondegenerate gas regime, equality with gas pressure gives the radiation-to-gas pressure boundarya line of slope $1/3$ on a $(\log \rho ,\log T)$ plot. Higher temperatures at fixed mass density favor photon support. A monatomic gas has stellar adiabatic exponent $5/3$, while radiation alone has $4/3$; their adiabatic exponents of a monatomic gas-radiation mixture are not obtained by assuming a fixed pressure fraction during compression. Radiation support is particularly important in massive stars.

For electrons, Pauli exclusion principle and the Fermi-Dirac distribution determine occupation numbers. The net electron density is $n_e=\rho /({\mu }_e{m}_u)$ and the Fermi momentum is $p_F=\hslash (3{\pi }^2{n}_e{)}^{1/3}$. Define the kinetic electron Fermi temperatureFor $T\gg T_F$ the electrons are nearly classical; for $T\ll T_F$ they are strongly degenerate and their pressure depends primarily on density. The intermediate region requires finite-temperature electron equation of state integrals, not a discontinuous switch of formulas.

The equation of state of a cold electron gas gives, in its two limits,These are respectively the $n=3/2$ and $n=3$ pressure-density powers, explaining the approximate white dwarf polytropic mass-radius relation sequence and the Chandrasekhar limit. The kinetic energy per volume is $3P_e/2$ in the nonrelativistic limit and $3P_e$ in the ultrarelativistic limit. Ions can still supply much of the heat capacity even when the electron pressure supplies the mechanical support.

The electron degeneracy crossover $T\sim T_F$ has slope $2/3$ at low mass density and $1/3$ at high mass density. The electron relativistic density threshold isa vertical marker where $p_F=m_{e}c$. It is different from the thermal electron relativistic threshold $k_{B}T\sim m_e{c}^2$, near $5.93\times 10^9\mathrm{K}$, a horizontal temperature scale. Hot dilute matter can have relativistic thermal electrons without degeneracy; cold dense matter can have relativistic degenerate electrons without reaching that temperature.

Electron degeneracy also does not automatically imply that electrons dominate the total pressure. Comparing the cold electron limit with radiation gives the radiation-to-degeneracy pressure boundaryIts logarithmic slopes are $5/12$ for nonrelativistic electrons and $1/3$ for ultrarelativistic electrons. One must compare the pressures separately from the degeneracy criterion; extrapolating the classical gas-radiation line into a degenerate region is incorrect.

The stellar equation-of-state regime diagram uses an illustrative fully ionized carbon composition, ${\mu }_e=2$, $\mu =12/7$. It displays the electron-degeneracy crossover, both radiation-pressure comparisons, and distinct thermal and density-driven relativity scales. The curves are limiting-model comparisons, not sharp phase boundaries or a calibrated complete equation of state. Partial ionization, molecular physics and interactions modify the low-temperature regions indicated on the plot.

At sufficiently high temperature, electron-positron thermal pair abundance can become important. The pair abundance depends on density and chemical potential as well as temperature; $k_{B}T=m_e{c}^2$ is not a universal onset line. In the dilute ultrarelativistic limit, both pair species together add energy density $7a_r{T}^4/4$ to the photons' $a_r{T}^4$, and have pressure one third of their energy density. While pairs are being created, thermal energy is spent on rest mass, which can reduce the stellar adiabatic exponent below $4/3$ and contribute to pair-instability supernova physics.

At high mass density and low temperature, Interactions governed by Coulomb's law invalidate the noninteracting-ion approximation. The ionic Coulomb coupling parameter $\Gamma =Z^2{e}^2/(4\pi {ϵ}_0{a}_i{k}_{B}T)$, where $a_i=(3/(4\pi n_i){)}^{1/3}$, grows as ${\rho }^{1/3}/T$. Corrections become significant when $\Gamma$ is of order one, and a sufficiently strongly coupled plasma can crystallize. At still greater mass density, electron capture alters ${\mu }_e$ and nuclear matter replaces the ideal electron-ion model; neutron star interiors require strong-interaction and relativistic equations of state. These further regimes lie beyond the simple pressure curves plotted here. A useful stellar EOS is thermodynamically consistent across the crossovers, rather than just the maximum of unrelated pressure laws.

Use Coulomb's law in vacuum, $F=q_1{q}_2/(4\pi {ϵ}_0{r}^2)$. The constant $4\pi$ is dimensionless, so the vacuum-permittivity dimension isThus the physical dimension of vacuum permittivity follows directly from the dimensions of electric charge, force, and separation.

Write $R=|\mathbf{p}|>a$. In SI units, the exterior electrostatic potential solveswith the point-charge singularity $q/(4\pi {\epsilon }_0|\mathbf{x}-\mathbf{p}|)$. The image charge for a grounded conducting sphere choiceslie inside the sphere. On $|\mathbf{x}|=a$, direct squaring gives $|\mathbf{x}-{\mathbf{p}}^{\prime }|=(a/R)|\mathbf{x}-\mathbf{p}|$. Thereforesatisfies the source equation, grounded boundary value and decay at infinity. The Uniqueness of the Dirichlet problem, applied to the harmonic difference with its singularity removed, identifies this with the physical exterior potential. Thus its gradient gives the required exterior electric field.

The charge feels the induced field, equivalently the field of the image charge, excluding its own singular field. Their separation is $(R^2-a^2)/R$, so Coulomb's law givesIts magnitude is $q^{2}aR/[4\pi {\epsilon }_0(R^2-a^2{)}^2]$ and its direction is toward the sphere's center, independent of the sign of $q$. No factor one-half belongs in this force; that factor occurs in induced electrostatic energy instead.

The three fixed charges are related by rotations through $2\pi /3$ about the origin. Their three Coulomb's law forces on a charge placed at the origin have equal magnitudes and directions separated by $2\pi /3$, so their vector sum vanishes. Thusis an equilibrium point.

The first term is radial kinetic energy from the Laplacian, the second is the attractive Coulomb potential, and the third is the centrifugal energy from the angular-momentum eigenvalue $L^2=l(l+1){\hslash }^2$. Single-valued spherical harmonics require

Here $N=2$ and $M=1$. At $(x,0,0)$, the $z$-components of the fields of the two outer charges cancel by reflection symmetry. By Coulomb's law, the remaining field isMultiplying by the test charge $-Q$ givesThe quantity in parentheses is positive. Thus for $x>0$ the force points in the negative $x$ direction, while for $x<0$ it points in the positive $x$ direction. It is always attractive toward the origin, as in the transverse force from a collinear electric quadrupole.

Put $C=q^2/(4\pi {ϵ}_0)>0$. The spring exerts the force $k(\ell -x)$ by Hooke's law, while Coulomb's law gives the attractive force $-C/x^2$. Newton's second law therefore gives the equation of motionFor $x>0$, the derivative is $f^{\prime }(x)=-k+2C/x^3$. Thus $f$ increases up to $x_{\ast }=(2C/k{)}^{1/3}$ and decreases thereafter. Also $f(x)\to -\infty$ at both ends of the positive half-axis. Its unique maximum isThere are therefore two equilibrium points precisely whenWriting them $x_{-}<x_{\ast }<x_{+}$, the linearization $m\ddot{\eta }=f^{\prime }(x_{\pm })\eta$ has exponentially growing solutions at $x_{-}$ and oscillatory solutions at $x_{+}$. Equivalently, the potential energy has a strict local maximum at $x_{-}$ and a strict local minimum at $x_{+}$. Hence the smaller equilibrium point is an unstable equilibrium; the larger is a stable equilibrium, with no damping to make it attracting.

At equality $\ell ={\ell }_1$ there is one degenerate equilibrium point $x_{\ast }$; it is unstable, since a displacement to its left gives a leftward force. For $\ell <{\ell }_1$ there are no equilibrium points.

At a large separation, the two like net charges give the familiar leading repulsion $k{Q}_1{Q}_2/R^2$, where $k=(4\pi {ϵ}_0{)}^{-1}$ and $R$ is centre separation. Each conductor also changes the other's surface charge distribution through electrostatic induction. The facing and remote surface charge densities need not remain equal, and an external insulating suspension prevents net charge transfer without preventing redistribution within either sphere.

Like net charges do not guarantee repulsion for arbitrary conducting spheres at every separation. Sufficiently unequal charges can give an attractive resultant after polarization. This does not contradict Coulomb's law: the law still applies to the distributed charges, while the point-charge approximation has lost the relevant induced contribution.

A useful controlled example takes a small conducting sphere of radius $a$ and fixed positive charge $Q_2$ in the nearly uniform field of a much more strongly charged distant sphere. For $a\ll R$, the small sphere has electric polarizability $\alpha =4\pi {ϵ}_0{a}^3$ and induced dipole moment $p=\alpha E$. Its leading interaction energy isDifferentiation gives the outward radial forceThe second term is attractive. With $\eta =Q_2/Q_1>0$, this leading model gives $F_R<0$ when $\eta <2(a/R{)}^3$. One can choose a sufficiently small positive $\eta$ while retaining $a/R\ll 1$, so the example does not require a neutral or earthed sphere. It is an asymptotic model, not an exact finite-sphere threshold.

An exact analysis confirms that close-range attraction is possible for like-charged conducting spheres and identifies a special contact-charge ratio for which the force remains repulsive.^[ref] This later analysis supplies a precise qualification to the elementary expectation; the source question gives neither a charge ratio nor a separation regime that would justify asserting universal repulsion.

For a point charge $q_j$, Coulomb's law gives ${\mathbf{E}}_j=q_j\hat{\mathbf{r}}/(4\pi {ϵ}_0{r}^2)$. The flux through an oriented surface element isHere $d\Omega$ is its signed solid angle seen from the charge. A closed surface subtends total oriented solid angle $4\pi$ for a point inside and zero for a point outside. Thus the flux of each inside charge is $q_j/{ϵ}_0$ and each outside charge has zero net flux. By superposition,This is Gauss's law; a surface passing through a point-charge singularity is excluded.

The integral Gauss law in vacuum is ${\oint }_{\partial V}\mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enc}}/{ϵ}_0$. A single isolated point charge has spherical symmetry, so its electric field is radial and constant in magnitude on a centred sphere. The flux is $4\pi r^2{E}_r$, giving Coulomb's lawIn electrostatic equilibrium, a conductor is an equipotential: a tangential electric field would move its mobile charges. The electric field inside the conducting material is zero. A thin pillbox straddling its surface therefore gives $E_{n}A=\sigma A/{ϵ}_0$, henceHere $\sigma$ is the charge density on the particular face and $\hat{\mathbf{n}}$ points into vacuum. For an isolated thin plate with total charge divided equally between two faces, each face instead has half that total areal density. The local boundary result must not be confused with the $\sigma /(2{ϵ}_0)$ field of an isolated nonconducting infinite charged sheet.

Place eight identical cubes around the charged corner to make one larger cube, with the point charge at its centre. Each large face has electric flux $q/(6{ϵ}_0)$ and consists of four equivalent small faces. Each of the original cube's three faces opposite the charged corner therefore hasOn each of its three faces containing the corner, the electric field of the point charge obeying Coulomb's law is tangent to the face at every regular point; its outward-normal component is zero. Defining the face integral by omitting the singular corner givesA boundary point charge does not satisfy the ordinary nonsingular hypotheses for simply counting it as an interior charge in Gauss's law. The $1/8$ result expresses the octant solid angle and the specified improper surface integrals. Moving a charge towards the corner from inside a cube before integrating is a different limiting operation; its total closed-surface electric flux stays $q/{ϵ}_0$ until it reaches the boundary.

The electric field of a signed point electric charge in vacuum follows Coulomb's law:For the conducting plane, the method of images replaces each charge $Q_j$ at height $h_j$ by an image charge $-Q_j$ at height $-h_j$. The real and image electric fields have cancelling tangential components and equal normal components at the surface. Thus the electric field at a conducting plane from axial point charges, just above the point directly below them, isNumerically,The surface value means the air-side limit. Inside the ideal conductor the electrostatic electric field is zero; its discontinuity across the surface is supported by induced surface charge. Omitting the image charges would give half the external answer.

The method of images replaces electrical conductors by fictitious image charges outside the physical region. Choose the image charges so that the physical electric charges and their images produce the prescribed constant electric potential on every conducting boundary. The uniqueness theorem for the electrostatic potential then identifies this constructed electric potential with the physical solution in the charge-containing region. Several conducting planes can require several, or infinitely many, image charges; a single reflected image is sufficient for one infinite plane.

Put the plane at $z=0$ and the point charge at $(0,0,d)$. For $z>0$, use an image charge $-q$ at $(0,0,-d)$:This vanishes on the plane. An arbitrary constant can be added if a different reference electric potential is chosen. Excluding the point charge's own electric field, Coulomb's law givesThe separation from the image charge is $2d$, and the force is attractive for either sign of $q$. No imposed external electric field is included.

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.

In SI units, Maxwell's equations in vacuum, allowing charge density $\rho$ and electric current density $\mathbf{J}$, areThe source-free specialization sets $\rho =0$ and $\mathbf{J}=0$; it cannot describe the charged objects at their locations. Applying the divergence theorem to the first equation gives Gauss's law:For an isolated point charge $q_1$, spherical symmetry makes the electric field radial and constant in magnitude on a concentric sphere. Therefore $4\pi r^2{E}_r=q_1/{ϵ}_0$. Multiplying by a second point charge gives Coulomb's law,The force points away from the first charge for like signs and toward it for opposite signs.

For the spherical capacitor, a Gaussian sphere between the shells encloses $q$, so $E_r=q/(4\pi {ϵ}_0{r}^2)$ for $a<r<b$. Integrating the electric field inward from the outer shell yieldsAn arbitrary constant $V(b)$ remains unless a reference is specified. The capacitance is determined by the potential difference:For the subsequent charging operation, take the initial outer shell to be earthed, or interpret the stated $V$ as the inner-to-outer voltage with the outer reference zero. Isolating the inner shell fixes its electric charge at $q=CV$. Removing the outer shell leaves an isolated conducting sphere. Its electric field and electric potential, now with zero at infinity, areConsequentlyThese changes of electric potential are compatible with constant electric charge. Without the initial reference convention the absolute stated $V$ does not determine $q=CV$: for example, if the outer shell were initially isolated and neutral, the absolute inner electric potential would instead be $q/(4\pi {ϵ}_{0}a)$.

Finally take the connected spheres to be sufficiently far apart that the long wire's capacitance and their mutual influence can be neglected. Their common electric potential requiresso $q_a=q/3$ and $q_{2a}=2q/3$. The magnitudes of the surface electric fields areThis is the usual widely separated-sphere approximation. A wire enforces an equipotential, but alone does not make this ratio exact for arbitrary arrangements. As a concrete counterexample, if the second shell enclosed the first concentrically and they were connected, all net electric charge would reside on the exterior and the inner sphere's surface electric field would be zero.

Write $k=1/(4\pi {ϵ}_0)$. The separation of the two point charges is $2l\sin \varphi$, so Coulomb's law gives the horizontal repulsive force $F=k{Q}^2/(4l^2{\sin }^2\varphi )$. Resolving the rod's tension vertically and horizontally gives $T\cos \varphi =mg$ and $T\sin \varphi =F$. HenceThe left side increases from zero to infinity for $0<\varphi <\pi /2$, so this determines a unique outward equilibrium. For a small angle, ${\varphi }^3\simeq Q^2/(16\pi {ϵ}_{0}mg{l}^2)$. With $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, direct numerical solution gives