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Set the grounded plane at $z=0$ and the real point charge at $(0,0,h)$. The image charge $-q$ at $(0,0,-h)$ makes the electrostatic potential vanish throughout the plane. The uniqueness theorem for the electrostatic potential then identifies the physical solution in $z>0$:The normal electric field is $E_z=-{\partial }_z\Phi$. Differentiation givesand thereforeFor $q>0$ the electric field points into the electrical conductor, consistently with negative induced surface charge density.

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

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

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

Let the unit normal point from layer 1 to layer 2. The electrostatic electric field boundary condition is $\mathbf{n}\times ({\mathbf{E}}_2-{\mathbf{E}}_1)=0$, while Gauss's law for the electric displacement field gives $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$. Here ${ϵ}_i$ denotes absolute permittivity; replace it by ${ϵ}_0{ϵ}_{ri}$ if dielectric constants are relative.

In the steady conducting state, charge conservation makes the normal current density the same in both layers:Writing $\mathcal{D}={\sigma }_2{d}_1+{\sigma }_1{d}_2$, the steady fields in two leaky dielectric layers areThe signed interfacial free surface charge density is the jump of $D_n$:The total free-plus-bound sheet charge instead follows from the vacuum form of Gauss's law,The bound contribution is ${\sigma }_{\mathrm{total}}-{\sigma }_f$. These are signed quantities for the chosen normal and voltage direction; their magnitudes are obtained by absolute values. They describe the long-time DC state, not the initial capacitive transient. Positive conductivities suffice for the derivation; if both vanish, conductivity alone does not determine the field division, which also depends on the initial free interfacial charge.

Images and uniqueness. The method of images replaces a conducting boundary by fictitious charges outside the physical field region. Choose them so that the resulting electric potential satisfies the required conductor potentials. The Uniqueness of the Dirichlet problem then guarantees the same physical exterior field. For example, a point charge $q$ at height $h$ above a grounded plane has an image $-q$ at height $-h$: the plane is an equipotential at zero, and the charge is attracted toward it by $q^2/(16\pi {\epsilon }_0{h}^2)$. The image is a computational device; the physical conductor carries a continuous induced surface charge density.

Potential of a line. Surround an infinite line charge $\lambda$ by a coaxial Gaussian cylinder of radius $r$ and length $L$. Gauss's law gives $E_r(2\pi rL)=\lambda L/{\epsilon }_0$, hence $E_r=\lambda /(2\pi {\epsilon }_{0}r)$. Integrating $E_r=-dV/dr$ givesWriting an additive constant instead is equivalent. The reference length makes the logarithm dimensionless; an isolated infinite line cannot be assigned a finite zero potential at infinity.

Locating the two lines. Put the grounded plane at $x=0$ and the physical cylinder at $(x,y)=(d,0)$, with axis along $z$ and radius $a<d$. Set $c=\sqrt{d^2-a^2}$, and place line charges $+\lambda$ and $-\lambda$ at $(c,0)$ and $(-c,0)$. If $R_{+}$ and $R_{-}$ are their distances from an observation point, the cylinder-plane image charges giveOn $x=0$, $R_{-}=R_{+}$, so $V=0$. To check the cylinder, define $q=(d+c)/a$. The level set $R_{-}=q{R}_{+}$ is a circle: completing the square givesSince $d^2-c^2=a^2$, its centre is $d$ and radius is $a$. The positive line lies inside the cylinder; the negative line lies behind the plane. There is consequently no fictitious charge in the exterior physical region. A Gaussian contour around the cylinder encloses $+\lambda$ only, so the required physical charge per unit length is also correct. These boundary conditions and the far-field behaviour determine the physical solution by Uniqueness of the Dirichlet problem.

Cylinder voltage. The constant distance ratio on the cylinder immediately givesIts capacitance per unit length is $C^{\prime }=2\pi {\epsilon }_0/\mathrm{arcosh}(d/a)$. In the thin-cable limit, $\mathrm{arcosh}(d/a)\simeq \ln (2d/a)$.

Cable attraction. Because the cable is idealized as infinite, the meaningful answer is force per unit length. For $a=0.020\mathrm{m}$, centre height $d\simeq 10\mathrm{m}$ and $V_c=1.1\times 10^4\mathrm{V}$,The electrostatic attraction per unit length of a cylinder above a plane isFor example, this follows without any ambiguity about moving image positions from $F_d=\frac{1}{2}{V}_c^{2}d{C}^{\prime }/dd$ at fixed voltage, including the work of the voltage source. Its sign is negative because capacitance decreases with height. A cable length $L$ experiences approximately $fL$ downward when end effects are negligible. This is the electrostatic contribution only.

An infinite sheet of surface charge density $\sigma$ gives an electric field of magnitude $E_s=\sigma /(2{\epsilon }_0)$ on either side, normal to the sheet and directed away from positive charge. Use the superposition principle for the two sheets. Choose normals $\hat{\mathbf{y}}$ for $y=0$ and $\mathbf{n}=(-\hat{\mathbf{x}}+\hat{\mathbf{y}})/\sqrt{2}$ for $y=x$. Away from the charged planes,The minus sign on the second contribution makes its field point toward the negatively charged sheet. Each sector has a uniform electric field; its direction changes on crossing a sheet.

For example, above both sheets the field points right and slightly upward, whereas in the sector $y>0$ but $y<x$ it points left and steeply upward. The other two sectors follow by inversion through the origin. The normal field jumps by the appropriate surface charge divided by ${\epsilon }_0$; a single field value on an ideal zero-thickness charged sheet is not required.

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.

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 $\mathbf{D}$.

The method of images solves certain electrostatics problems by replacing a conductor by fictitious electric charges outside the region where the solution is required. The replacement must satisfy the actual boundary values of the electric potential as well as the prescribed sources. The uniqueness of Poisson equation then establishes that the constructed electric potential equals the physical one in that region. The image locations are mathematical devices; they are not real charges inside the metal.

For a point charge $q$ at $(0,0,a)$ above an infinite grounded conducting plane $z=0$, put an image charge $-q$ at $(0,0,-a)$. In a cylindrical coordinate system with coordinates $(s,z)$ the potential in the physical region isThis vanishes at $z=0$, has the required singularity at the real charge, and tends to zero at infinity. With electric field $\mathbf{E}=-\nabla \varphi$, the induced surface charge density follows from the normal field immediately above the conductor:The force on $q$ is the field of the image evaluated at the real charge:The electrostatic energy relative to infinite separation is $U=\frac{1}{2}q{\varphi }_{\mathrm{image}}=-q^2/(16\pi {ϵ}_{0}a)$. The factor $1/2$ accounts for the fact that the induced charge changes as the real charge is assembled; treating the image as an independently fixed physical charge would give the wrong energy.

A second standard geometry is a grounded conducting sphere of radius $b$ with a charge $q$ a distance $d>b$ from its centre. Its exterior field is generated by an image $q^{\prime }=-qb/d$ at distance $b^2/d$ from the centre on the line towards the real charge. On the sphere the distances to the real and image charges are in the ratio $d/b$, making their potentials cancel. For an isolated neutral sphere, an additional central image $+qb/d$ makes the total induced charge zero and leaves the surface an equipotential. The construction is powerful for suitable symmetric boundaries, but arbitrary boundaries do not in general admit a finite image system.

A magnetic dipole is the leading magnetic source moment of a localized steady current distribution. For a small planar loop, $\mathbf{m}=I\mathbf{A}$ with area orientation set by the right-hand rule. A fixed permanent moment in a slowly varying external field has energy $U=-\mathbf{m}\cdot \mathbf{B}$ and force $\mathbf{F}=\nabla (\mathbf{m}\cdot \mathbf{B})$.

For the given purely azimuthal vector potential, the spherical curl givesThese are the magnetic dipole field components outside the localized source.

The conducting shell has $Q=4\pi {ϵ}_{0}aV$ and surface charge density $\sigma =Q/(4\pi a^2)$. A latitude strip of area $2\pi a^2\sin \theta d\theta$ carries current $dI=\sigma \omega a^2\sin \theta d\theta$. Its loop area is $\pi a^2{\sin }^2\theta$, givingIn vector form the magnetic moment of a rotating charged spherical shell isThe positive-z expression in the question takes $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. “Clockwise” needs a viewing direction: this rotation looks clockwise when viewed looking along $+z$ from below, and counterclockwise when viewed from the positive-z side. Looking from the other end reverses the verbal label, not the vector relation.

Since the positive charge and the loop current circulate in the same stated sense, their dipole moments are parallel when judged from a common viewpoint. At $h\gg R$, the loop's axial field magnitude is $B\simeq {\mu }_{0}I{R}^2/(2h^3)$. With the sphere small compared with that variation scale,The magnetic force is attractive, pulling the sphere toward the loop. Reversing only one circulation would reverse that magnetic force. The calculation treats the rotating charge as a prescribed moment and uses the stated separated length scales; it does not add unrelated induction or electric interactions to the dipole model.

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.

One method is to give every particle a sufficiently large surface charge density of the same sign. For example, attach ionizable sulfonate groups covalently to the surface. Dissociation leaves negatively charged surface groups and releases mobile positive counterions into the water. A surfactant with a strongly bound head group can provide another charging route, provided it does not desorb during the experiment.

Each charged particle is surrounded by an electric double layer. As two such particles approach, overlap of these ionic clouds raises the electrostatic and ionic free energy, producing repulsion. In a dilute electrolyte, the range is set by the Debye–Hückel screening length,For identical weakly charged spheres a representative screened Coulomb pair potential isIts sum with the Van der Waals force attraction is the DLVO theory pair potential. Choose the charging level so that the repulsive barrier is many $k_{B}T$ high; particles then rarely reach the primary attractive well during the observation time. This is kinetic stabilization rather than removal of all attraction.

Deionized water is useful because low added ionic strength gives a long-ranged repulsion. It does not mean that the charged dispersion contains no ions: the surface's own counterions are unavoidable and contribute to screening. Surface dissociation, contamination and particle concentration therefore matter even without added salt. An alternative valid route is a dense, firmly attached layer of water-soluble polyethylene glycol, producing steric stabilization without relying on a large electrostatic barrier.

The particle carries positive surface charge density $\sigma =Q/(4\pi a^2)$. Its surrounding electric double layer has an excess of negative chloride counterions and a deficit of positive magnesium coions. Thus $\varphi (r)>0$ decreases to zero, $n_{\mathrm{Cl}}(r)$ decreases towards its bulk value, and $n_{\mathrm{Mg}}(r)$ increases towards its bulk value. The diffuse charge density is negative and tends to zero.

If $c$ denotes the bulk number of dissolved salt formula units per unit volume, the far-field number densities are $n_{\mathrm{Mg},\infty }=c$ and $n_{\mathrm{Cl},\infty }=2c$. The bulk is neutral even though these number densities differ. In the dilute mean-field picture the total diffuse electric charge is $-Q$, ensuring overall electroneutrality. A molecularly bound compact layer would require additional surface-chemistry information and is not assumed in the supplied potential.

For an infinite circular cylinder of radius $R$, assume the electrostatic potential is independent of the axial coordinate. An angular Fourier mode $s_n\cos n\theta$, or $s_n\sin n\theta$, in the prescribed jump ${\partial }_r\Phi {|}_{R^{+}}-{\partial }_r\Phi {|}_{R^{-}}=-\sigma (\theta )$ produces the same angular mode in the potential, multiplied byHere $r_{<}=\min (r,R)$ and $r_{>}=\max (r,R)$ in the abbreviated header formula. Separation of variables selects $r^n$ inside by regularity and $r^{-n}$ outside by decay. Continuity equates their boundary values, and the derivative jump is minus $2n/R$ times that value. A nonzero angular mean of the surface charge density introduces a logarithmic term, so zero mean is necessary for a potential decaying at infinity in this two-dimensional reduction.

A lateral sinusoidal surface charge density with wave number $k$ in the Debye–Hückel approximation generates vertical exponential or hyperbolic functions with inverse length $p=\sqrt{k^2+{\kappa }_D^2}$. In a homogeneous electrolyte filling all space, an isolated sheet at $z_s$ produces $\varphi ={\sigma }_k(x)e^{-p|z-z_s|}/(2\epsilon p)$. Boundary conditions determine the normalization when the electrolyte occupies only a gap.

For two planar homogeneous layers in series, charge conservation makes the steady normal current density the same in both layers. Combining Ohm's law with $E_1{d}_1+E_2{d}_2=V$ determines the fields when both conductivities are positive. The free interfacial surface charge density is ${ϵ}_2{E}_2-{ϵ}_1{E}_1$, whereas total free-plus-bound charge is ${ϵ}_0(E_2-E_1)$. The two notions of charge are distinct. This DC state can differ from the immediate response to switching the voltage; interfacial free charge accumulates during the transient.

Electric charge concentrated on a surface. Its local amount per unit area is the surface charge density $\sigma$; in vacuum the jump of the outward normal electric field is $\sigma /{ϵ}_0$ by Gauss's law.

For $a<r<a(1+\delta )$, a charge density ${\rho }_0{a}^2{h}^{\prime }(r/a)/(\delta r^2)$ has total electric charge $4\pi {\rho }_0{a}^3[h(1+\delta )-h(1)]/\delta$. As $\delta \to 0$, it tends to uniform surface charge on $r=a$ with surface charge density $\sigma ={\rho }_{0}a{h}^{\prime }(1)$. The limiting normal electric field has jump $\sigma /{ϵ}_0$, although every finite-thickness layer has continuous electric field at its inner edge. Taking the thin-layer limit and taking a one-sided spatial limit need not commute.