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The electrostatic boundary conditions at a conductor relate the one-sided total electric fields at a smooth metal–vacuum interface. Choose a unit normal $\mathbf{n}$ from metal into the adjacent vacuum, let ${\mathbf{E}}_{\mathrm{metal}}$ and ${\mathbf{E}}_{\mathrm{vac}}$ be the two limiting electric fields, and let $\sigma$ be the signed surface charge density. A thin Gaussian pillbox with cap area $A$ has outward cap normals $+\mathbf{n}$ on the vacuum side and $-\mathbf{n}$ on the metal side. As its height tends to zero, its side electric flux tends to zero for bounded local electric fields, leavingThe minus sign comes from the second outward cap normal, not from the sign of the electric charge. A shrinking rectangular loop across the interface and the zero closed line integral in electrostatics give continuity of the tangential component:For a conductor in electrostatic equilibrium, ${\mathbf{E}}_{\mathrm{metal}}=0$, henceThus positive surface charge density gives an electric field away from metal and negative surface charge density gives an electric field toward metal. The derivative uses the same chosen unit normal, from the vacuum side. The ideal surface layer has no unique electric field value on the layer itself; these are limiting values. Ordinary surface electric charge preserves continuity of electric potential; an ideal electric dipole sheet would require a different model. ^[ref].

Here “exterior” means just outside the metal, including the vacuum within a cavity in a conductor. On a spherical inner wall of radius $a$, its unit normal from metal into the cavity in a conductor is $-\hat{\mathbf{r}}$. A central positive point charge $q$ gives ${\mathbf{E}}_{\mathrm{vac}}=q\hat{\mathbf{r}}/(4\pi {ϵ}_0{a}^2)$ there, soThe outward normal of the cavity in a conductor as a volume is instead $+\hat{\mathbf{r}}$; confusing these two unit normals reverses the inferred sign. For a stipulated spherical enclosure with inner radius $a$, outer radius $b$, no external sources and total isolated metal electric charge $Q_c$, spherical symmetry and Gauss's law give the analytical modelIts outer surface charge density is $(Q_c+q)/(4\pi b^2)$. Grounding to $V_c=0$ with zero electric potential at infinity instead gives $Q_c=-q$ and zero exterior electric field, while preserving the same nonzero electric field inside the charged cavity in a conductor.

As a numerical sign check, stipulate $q=1\mathrm{nC}$, $a=0.10\mathrm{m}$, $b=0.20\mathrm{m}$ and $1/(4\pi {ϵ}_0)\simeq 8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. The inner electric field is approximately $899N/C$ along $+\hat{\mathbf{r}}$, but ${\mathbf{E}}_{\mathrm{vac}}\cdot \mathbf{n}=-899N/C$ and ${\sigma }_{\mathrm{inner}}\simeq -7.96nC/{\mathrm{m}}^2$. An isolated neutral enclosure has $+1\mathrm{nC}$ on the outer wall, approximately $+1.99nC/{\mathrm{m}}^2$ and $225N/C$ just outside. Multiplying each surface charge density by its spherical surface area recovers the signed wall electric charges.

The local result uses the total electric field, including all induced and imposed sources. The factor $\sigma /{ϵ}_0$ follows from the zero metal-side electric field; a symmetric isolated sheet in vacuum instead has magnitude $|\sigma |/(2{ϵ}_0)$ on each side. ^[ref]. At sharp corners, limiting electric fields can be singular and the smooth local pillbox argument needs care. In material outside the metal, use the normal free-charge jump of the electric displacement field and the material's permittivity, as in electrostatic interface boundary conditions. ^[ref].

A long coaxial cylindrical capacitor of length $L$ and radii $a,b$ has a dielectric-filled overlap of length $x$ and relative permittivity ${ϵ}_r$. Neglecting end and interface fringing, its filled and vacuum portions are in parallel because both share the same two conductors. The electric potential difference $V$ produces $E_r=V/[r\ln (b/a)]$ in either portion; the electric displacement field and electrode charge differ by the local permittivity. Adding their capacitances gives the title formula. The dielectric boundary is axial while the interior electric field is radial and tangential to that interface, consistent with the electrostatic interface boundary conditions.

Let the electrolyte permittivity be $\epsilon$ and let ${\kappa }_D=1/{\ell }_D$ be the inverse Debye–Hückel screening length. We reserve $k$ for the imposed lateral wavevector. In the Debye–Hückel approximation, linearization about an electrically neutral bulk electrolyte givesaway from fixed charges. For ionic species of valence $z_i$ and bulk number density $n_i$, ${\kappa }_D^2={\sum }_i{n}_i{z}_i^2{e}^2/(\epsilon k_{B}T)$; the approximation requires $|z_{i}e\varphi |\ll k_{B}T$. Substitution of a lateral cosine into this screened Poisson equation leaves a vertical decay constantThe screened sinusoidal surface-charge mode thus decays faster than a laterally uniform charge mode. A nonzero $k$ retains a finite decay length even in the zero-salt limit.

The exterior media and their boundary conditions are not specified in the paper. Specifying the two surface charge densities alone does not fix the normal derivative on the inside unless the outside response is also fixed. We first take the confined-gap idealization, with the prescribed charge supplying all the displacement flux into the electrolyte between the sheets. The electrostatic interface boundary conditions are thenThis is, for example, the limit of negligible exterior displacement admittance. The electrostatic potential solving these conditions and the Debye–Hückel approximation equation isDifferentiation checks the two surface signs directly. This solution allows a sine as well as a cosine lateral component when the phase is nonzero.

In linear screened electrostatics, the quadratic charging free energy is $\frac{1}{2}\int \sigma \varphi dA$. It is equivalently $\frac{\epsilon }{2}\int [(\nabla \varphi {)}^2+{\kappa }_D^2{\varphi }^2]dV$ with the stated boundary conditions. The second term represents the linearized ionic response; integrating only the bare electric field energy would omit it. If $\mathcal{A}$ denotes membrane area, the wavelength-averaged energy density isUsing $⟨{\cos }^2(kx){⟩}_x=1/2$ and $⟨\cos (kx)\cos (kx+\theta ){⟩}_x=\cos \theta /2$ gives the phase registration of screened charged sheets energyThe positive cosine coefficient shows that, for nonzero charge amplitude and finite separation,At the minimum the electrostatic potential simplifies to $\varphi =\alpha \cos (kx)\sinh (pz)/[\epsilon p\cosh (pd/2)]$. Positive and negative charge patches face opposite signs across the gap. The two charge patterns therefore favor a half-wavelength lateral offset rather than like-charge alignment.

For comparison, if identical electrolyte fills all of space on both sides of infinitesimally thin charge sheets, the electrostatic interface boundary conditions are continuity of $\varphi$, decay at infinity, and the charge-induced derivative jumpsSuperposing the two screened sinusoidal surface-charge modes gives, in particular between the sheets,Its full-space continuation replaces each vertical distance by the corresponding absolute distance. Evaluating the electrostatic potential on both sheets now yieldsThe constant term is the two isolated-sheet self energies, and the cosine term is their screened interaction. The energy depends on the exterior convention, but both stated physical idealizations give the same phase registration of screened charged sheets: unlike charge patches oppose each other. In the full-space convention the optimum interaction energy is $-{\alpha }^2{e}^{-pd}/(4\epsilon p)$, giving an attractive normal force density $-{\alpha }^2{e}^{-pd}/(4\epsilon )$. Phase sensitivity becomes exponentially weak for $pd\gg 1$; if $\alpha =0$ or the separation tends to infinity there is no selected phase.

One can also display how an exterior dielectric response interpolates between these cases. Let $Y_o\ge 0$ be its normal displacement admittance for this lateral Fourier mode: $Y_o=0$ for the confined-gap idealization, $Y_o=\epsilon p$ for identical exterior electrolyte, and $Y_o={\epsilon }_o|k|$ for an ion-free exterior dielectric. DefineThe symmetric and antisymmetric surface-charge combinations giveSince $D_s<D_a$, its cosine coefficient is positive. This makes the phase-minimizing conclusion robust while exposing the boundary information needed for an absolute electrostatic energy.

In electrostatics, the Maxwell equations in matter areHere ${\rho }_f$ is free charge density and $\mathbf{P}$ is polarization density. For a homogeneous isotropic linear dielectric, the constitutive equation is $\mathbf{D}={ϵ}_0{ϵ}_r\mathbf{E}$. Integrating the first equation over a thin pillbox straddling an interface, with unit normal $\mathbf{n}$ from medium 1 to medium 2, gives $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$. Integrating the second around a vanishingly narrow rectangular loop gives $\mathbf{n}\times ({\mathbf{E}}_2-{\mathbf{E}}_1)=0$. These are the electrostatic interface boundary conditions. When there is no free surface electric charge, the normal electric displacement field and tangential electric field are continuous; the normal electric field generally is not.

Neglect end fringing field of a capacitor. In either axial segment of the coaxial cylindrical capacitor, a cylindrical Gaussian surface gives $D_r={\lambda }_f/(2\pi r)$, where ${\lambda }_f$ is free electric charge per length. Integrating the radial electric field between the conductors givesThe liquid and air segments share the same conductor electric potential difference, so their capacitances add in parallel. Thus the partially filled coaxial capacitor hasHere the question's $ϵ$ is relative permittivity, and air's relative permittivity is approximated by one.

Write $x$ for the rise above the external level. The wide reservoir keeps that level approximately fixed. At constant applied voltage, one must include work done by the source: the appropriate electrical contribution to the mechanical potential is $-C{V}^2/2$. The upward dielectric insertion force at fixed voltage is consequentlyThe excess hydrostatic pressure $\rho gx$ acts over the annular area $A=\pi (b^2-a^2)$. Alternatively its gravitational potential energy is $\rho gA{x}^2/2$, since liquid initially at the reservoir level must be raised through all heights from zero to $x$. Balancing the derivatives givesThis assumes $h+x<L$, a communicating liquid reservoir, negligible capillarity, and a sufficiently long capacitor that end effects are small. Differentiating gives the dielectric-rise voltmeter's sensitivity at a bias $V$:To obtain its numerical value in $\mu \mathrm{m}{\mathrm{V}}^{-1}$, multiply the right-hand side in SI units by $10^6$. A convenient parametric scale, with $\beta =b/a$, isFor a concrete estimate, choose $a=1\mathrm{mm}$, $b=2\mathrm{mm}$, mass density $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$, relative permittivity $ϵ=2$ and operating voltage $V=100\mathrm{V}$. These are illustrative choices, not supplied apparatus data. They giveThus this millimetre-scale example requires fine optical displacement measurement for volt-level changes. The parametric formula states exactly how the estimate changes with the chosen geometry, liquid and bias.

The displacement is quadratic in voltage, cannot distinguish its sign, and has zero first-order sensitivity at zero bias. Smaller gaps and larger relative permittivity increase sensitivity, but capillarity, leakage, dielectric breakdown, temperature-dependent liquid properties, and finite available rise then limit performance. A nonconducting liquid avoids a sustained conduction load, although charging still draws electric charge and leakage is possible. A high voltage can give a readable displacement; a low-voltage instrument would require fine displacement measurement and careful calibration.

At the hemispherical surface $r=a$, the outward normal from the conductor into the gap is $\hat{\mathbf{r}}$. The tangential field is zero there andThe electrostatic interface boundary conditions give surface charge density $\sigma ={\epsilon }_0{E}_n$, because the interior electric field is zero. HenceIt is negative for a positive upper-plate voltage, greatest in magnitude at the apex, and zero at the equator. Its units are coulombs per square metre.

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.

Let $\mathbf{n}$ point from medium 1 to medium 2. Integrate $\nabla \cdot \mathbf{D}={\rho }_f$ over a thin Gaussian pillbox crossing the interface. As its thickness tends to zero, the side contribution vanishes and any free surface charge density ${\sigma }_f$ givesApply $\oint \mathbf{E}\cdot d\mathbf{\ell }=0$ to a thin rectangular loop spanning the same interface. Its short sides vanish in the limit, so the tangential electric field is continuous:For a clean dielectric interface with no free surface charge density, these electrostatic interface boundary conditions reduce to $D_{1n}=D_{2n}$ and $E_{1t}=E_{2t}$. In particular ${ϵ}_1{E}_{1n}={ϵ}_2{E}_{2n}$; the normal electric field itself need not be continuous. Bound polarization charge is already incorporated into $\mathbf{D}$.

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.

Two equally modulated charge sheets in identical full-space electrolyte have charging energy per area ${\alpha }^2[1+e^{-pd}\cos \theta ]/(4\epsilon p)$, where $p=\sqrt{k^2+{\kappa }_D^2}$. Its phase-dependent interaction favors a half-wavelength offset, with opposite charge patches facing each other. For an electrolyte gap with negligible exterior displacement response, the energy is instead ${\alpha }^2[\coth (pd)+\cos \theta /\sinh (pd)]/(2\epsilon p)$, with the same minimizing phase. Absolute energies require specified exterior electrostatic interface boundary conditions.

For a plane bounding a symmetric electrolyte in $x>0$, assume no displacement field on the other side, bulk potential zero, and $|ze\psi |\ll k_{B}T$. The Debye–Hückel approximation gives ${\psi }^{\prime \prime }={\kappa }^2\psi$, whose decaying solution is ${\psi }_0{e}^{-\kappa x}$. The electrostatic interface boundary conditions give $\sigma =-\epsilon {\psi }^{\prime }(0^{+})=\epsilon \kappa {\psi }_0$. Equivalently, its diffuse charge density is ${\rho }_e=-\epsilon {\kappa }^2\psi$ and its integral is $-\sigma$. Here $\epsilon ={\epsilon }_0{\epsilon }_r$ in SI units and ${\kappa }^2=2z^2{e}^2{n}_{\infty }/(\epsilon k_{B}T)$ when $n_{\infty }$ counts each ionic species. An isolated sheet with identical electrolyte on both sides has two diffuse layers, so the total sheet charge is $2\epsilon \kappa {\psi }_0$ instead. A surface geometry and the field on its other side must therefore be specified.