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Reflecting repeatedly in the two parallel planes creates an infinite method of images. For ${\mathbf{r}}^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$, introduceThe Dirichlet Green function between parallel planes isEach bracket is $O(n^{-2})$, so the paired series converges absolutely away from the sources. At $x=-1$, reflection pairs the positive image indexed by $n$ with the negative image indexed by $-n$. At $x=1$ it pairs that positive image with the negative image indexed by $1-n$. The distances agree in each pairing, making both boundary values zero. This can be checked with symmetric finite partial sums first; the unmatched end terms tend to zero, so no subtraction of separately divergent series is needed. Only ${\mathbf{r}}_0^{+}={\mathbf{r}}^{\prime }$ lies within the slab, so ${\Delta }_{\mathbf{r}}{G}_{\mathrm{slab}}=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$. This is the usual Green function decaying at large transverse separation; such a condition selects the physical solution in the unbounded region.

Let $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$ be the reflection of the source in the equatorial plane. For a source in the upper half-ball, the Dirichlet Green function inside a hemisphere isBoth terms vanish on the spherical boundary. On the flat boundary $z=0$, the two source distances are equal, and the distances to their spherical image points are also equal, so the difference vanishes there too. The reflected physical source is in the lower half-ball and both inverted image points are outside the sphere. Therefore the only source singularity in the desired region is the original one, with coefficient $+1$ in $\Delta G$. This combines reflection with the spherical method of images.

Let $s=|{\mathbf{r}}^{\prime }|>0$ and $\mathbf{p}={\mathbf{r}}^{\prime }/s^2$. For $|\mathbf{r}|=1$, expanding both squared Euclidean distances givesTaking nonnegative square roots provesThis is the distance identity behind inversion in a circle and the method of images for a circular boundary.

For a source in the exterior, its inverse point $\mathbf{p}$ lies strictly inside the excluded disc. The method of images construction of the bounded exterior disk Green function therefore allows an opposite logarithmic source there without adding a singularity to the domain. The candidate Dirichlet Green function isIts Laplacian in the exterior is exactly $\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$, since the image singularity is outside that domain. The identity in (d) makes $G=0$ on the unit circle. Moreoverso it obeys the required boundedness condition. These checks establish the stated logarithm-of-a-ratio expression with $A=1/(2\pi )$. Bounded exterior Dirichlet uniqueness for the Poisson equation follows by inversion in a circle and removal of the bounded singularity at the origin.

For the isolated line vortex, use polar coordinates $z=r{e}^{i\theta }$ to obtainThus for nonzero $\Gamma$ the streamlines are circles and their counterclockwise circulation is $2\pi r{u}_{\theta }=\Gamma$. If $\Gamma =0$ the fluid is stationary and the streamline description degenerates.

Place an opposite image vortex at $z=-1$. The method of images givesOn $z=iy$, the distances to the two vortex positions agree, so $\psi =-(\Gamma /2\pi )\log (|z-1|/|z+1|)=0$. Hence the wall is a streamline. Directly the wall velocity has $u=0$ andFor the stipulated steady flow, Bernoulli's equation in the irrotational fluid outside the vortex core givesThe force exerted by the fluid on this wall points in the $-x$ direction for positive pressure. Relative to uniform pressure $p_{\infty }$, however, its change is in the $+x$ direction. Using the supplied integral, the force change per unit span isThus the wall is attracted towards the vortex, with this magnitude if the other side remains at $p_{\infty }$. Subtracting the uniform pressure is essential for an infinite wall. A freely moving vortex would translate parallel to the wall under its image-induced velocity; the calculation here is for the stated steady vortex configuration, with its position held fixed.

A point vortex of circulation $\Gamma$ at $z_0$ contributes the complex potential of a two-dimensional potential flow $-i\Gamma \log (z-z_0)/(2\pi )$ with the stated convention. The method of images places a vortex of the opposite circulation at ${\overline{z}}_0$ below the wall, outside the fluid domain. Superposition with the stagnation flow givesFor real $z=x$, the distances $|x-z_0|$ and $|x-{\overline{z}}_0|$ agree. The two vortex contributions to the stream function are $-\Gamma \log |x-z_0|/(2\pi )$ and $+\Gamma \log |x-{\overline{z}}_0|/(2\pi )$, which cancel. The background stream function is also zero there, so $y=0$ remains an impermeable streamline. The image changes the boundary condition without introducing another singularity in the upper fluid region.

Take the conducting plane as $y=0$, with the real point charges $Q$ at $(0,b)$ and $-2Q$ at $(a,b)$. The method of images replaces the plane by image charges $-Q$ at $(0,-b)$ and $+2Q$ at $(a,-b)$. This makes the plane an equipotential; the physical field above it is the sum of the real- and image-charge fields. No factor of $1/2$ enters the force on a real charge.

Set $k_e=1/(4\pi {ϵ}_0)$ and $d=\sqrt{a^2+4b^2}$, and use positive $x$ from $Q$ toward $-2Q$, positive $y$ upward. The horizontal forces combine the mutual attraction and the force from the other charge's image:For the vertical components, the own-image attractions act downward; each cross-image term acts upward. ThusThese are the electrostatic forces; the external forces holding the charges fixed are their negatives. The smaller positive charge's vertical force need not always point downward, because the other charge's induced surface charge also contributes. The construction assumes the usual ideal large-plane limit with no imposed background field.

First use the usual idealization of a grounded infinite conducting plane at $z=0$, with the real point charge $+q$ at $(0,0,d)$. The method of images replaces the metal, for purposes of calculating the exterior electric field, by an image charge $-q$ at $(0,0,-d)$. The exact electric potential in $z>0$ isIt vanishes on the plane, and the uniqueness theorem for the electrostatic potential identifies it with the physical exterior solution. The tangential electric field cancels on the plane, while the normal components add:The electric field inside the conductor is zero, so the electrostatic boundary condition gives $\sigma ={ϵ}_0{E}_z(0^{+})$. Integrating the surface charge density over the facing plane yieldsThe exact image expression is essential near the charge: using the far-field electric dipole approximation at $\rho$ comparable with $d$ would be invalid.

Electrical boundary condition. The source does not specify grounding or the initial total electric charge of the slab. The boxed total is the grounded infinite-plane answer and the large-plane approximation to the charge on its nearby face. An initially neutral, finite, electrically isolated slab retains zero total electric charge; an approximately $-q$ accumulation near the real charge is then balanced by approximately $+q$ on its remote face and edges. This gives a concrete counterexample to interpreting $-q$ as the unconditional total charge of every large metal slab. The local image formula assumes the infinite-plane limit with no applied background electric field.

The method of images replaces a conducting boundary by fictitious point charges placed outside the physical region being solved. Choose their positions and strengths so that the resulting electric potential has the required constant value on each conductor. The uniqueness theorem for the electrostatic potential then guarantees that this solution equals the physical solution in the allowed region, provided the other boundary conditions are also satisfied.

For a plane held at zero electric potential, a real charge $q$ a height $d$ above the plane has an image $-q$ the same distance below it. The pair's electric potential vanishes everywhere on the plane. Its tangential electric field consequently vanishes there, as required in electrostatic equilibrium. The image is a mathematical representation of the induced surface charge, not a physical charge placed inside the conductor. Forces on the real charge can be computed from the image field, but total physical electrostatic energy must account for the conductor's induced response rather than simply counting a full real-plus-image pair energy.

The method of images replaces a conductor's induced surface charge by suitable fictitious image charges located outside the physical region in which the electric field is sought. Choose them so that the combined electric potential satisfies the specified conductor boundary values, has the required behaviour at infinity, and has the correct physical sources. The uniqueness theorem for the electrostatic potential then identifies this constructed electric potential with the physical solution in that region.

For an infinite grounded conducting plane at $z=0$, a point charge $q$ at $(x,y,z)$ with $z>0$ has an image charge $-q$ at $(x,y,-z)$. Distances to the pair are equal on the plane, so their electric potentials cancel there. This construction determines the electric field above the plane; below it the actual conductor has constant electric potential and zero electric field, rather than the fictitious image's continuation.

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.

Electrical contact equalizes electric potential and preserves total electric charge. The usual isolated-sphere approximation assigns the two conducting spheres the capacitances $4\pi {ϵ}_{0}R$ and $8\pi {ϵ}_{0}R$. It then givesThese are accurate when the conducting spheres are far apart and joined by a thin wire. Literal touching changes the answer: each conducting sphere polarizes the other, so its electric potential is not its isolated value. The source does not specify an approximation which would remove this effect.

For completeness, the exact result for ideal externally tangent conducting spheres can be obtained by the method of images. Let their radii be $a,b$, their centre separation $D=a+b$, and their common electric potential $V$. Begin with central image charges $4\pi {ϵ}_{0}Va$ and $4\pi {ϵ}_{0}Vb$. Reflect each external image charge $q^{\prime }$ at centre-distance $s$ into the other conducting sphere of radius $a$: the new image charge is $-a{q}^{\prime }/s$ at centre-distance $a^2/s$ toward the original. Repeat, alternately, in the two conducting spheres. Pairing successive opposite-sign terms gives the convergent net electric chargesEach reflection cancels a nonconstant surface electric potential; the remaining central term fixes $V$. The total internal image charge equals the exterior electric flux divided by $1/{ϵ}_0$, so these sums give the physical net electric charges.

For $a=R,b=2R$, multiply both dimensionless sums by the common factor $4\pi {ϵ}_{0}V(2R/3)$. Their values follow by writing reciprocals as integrals and setting $t=u^3$:Dividing by $S_a+S_b$ enforces conservation of electric charge and yieldsThis concrete touching-sphere result disproves applying the isolated $1:2$ sharing rule to literal contact.

A large electrical conductor between the point charges provides electrostatic shielding: one must not retain their original mutual Coulomb force through an ideal infinite plate. In the usual broad-plate approximation, let $\sigma$ denote the unperturbed surface charge density on each face. The electrostatic boundary condition gives an outward electric field $E_0=\sigma /{ϵ}_0$. Neglecting local electrostatic induction, the rod's tension then satisfies $mg\tan \varphi =Q{E}_0$, givingThere are two physical qualifications. First, if $\sigma$ means the total charge per projected plate area, shared equally between the two faces, then $E_0=\sigma /(2{ϵ}_0)$ and the same approximation gives $\varphi \simeq 0.989^{\circ }$. The factor of two is a convention about the specified surface charge density, not a different Gauss's law.

Second, a point charge near an electrical conductor makes its surface charge density nonuniform. For an infinite plane with prescribed background electric field $E_0$, the method of images adds an attractive force $k{Q}^2/(4x^2)$, where $x=l\sin \varphi$. The more accurate equilibrium equation isIts larger, stable root is approximately $1.96^{\circ }$ for the per-face convention, or $0.877^{\circ }$ for the total-area convention. This ideal-plane refinement is not an exact answer for an unspecified finite plate: its dimensions and the meaning of the quoted surface charge density would be needed. In particular, adding a uniform-sheet electric field to the original mutual repulsion would describe a different, nonshielding model.

Take the grounded plane as $z=0$. The method of images replaces it, in $z>0$, by an image charge $-q$ at $z=-d$ and an image charge $+q$ at $z=-2d$. These make the electric potential zero on the plane; the uniqueness theorem for the electrostatic potential establishes the solution in the physical half-space.

At $z=3d/2$, both real point charges contribute upward electric fields, each of magnitude $kq/(d/2{)}^2$. The nearer image charge contributes downward and the farther one upward. HenceThis is upward for $q>0$; retaining signed $q$ automatically handles the opposite sign. No self-electric field is involved because the evaluation point is distinct from every point charge.

For image charges in a sixty-degree conducting wedge, use polar coordinates in a cross-section normal to the planes' intersection, putting the physical wedge at $0<\theta <\pi /3$ and the real point charge at $(r,\theta )=(d,\pi /6)$. Repeated reflection in the two grounded faces produces five image charges, all at radius $d$ and the same coordinate along the intersection:Equivalently, the positive copies are at $\theta =\pi /6+2k\pi /3$ and the negative copies at $\theta =-\pi /6+2k\pi /3$, for $k=0,1,2$. Only the original point charge lies inside the physical wedge.

For any observation point on either face, reflection pairs every positive point charge with an equidistant negative one. Their Coulomb potentials cancel, so both grounded faces have zero electric potential. The electric potential also tends to zero at infinity. The uniqueness theorem for the electrostatic potential then identifies this method of images construction with the physical solution.

The method of images replaces conductors by fictitious charges chosen to reproduce their boundary potentials in the physical domain. The uniqueness theorem for the electrostatic potential then identifies the resulting physical exterior field; the fictitious interior field is not the field inside the conductor.

For a source at $d\hat{\mathbf{z}}$, let $r^{\prime }=a^2/d$ and $q^{\prime }=-qa/d$. At a point on $r=a$, the distances $R$ and $R^{\prime }$ to source and image obey $R^{\prime }=(a/d)R$. Hence $q/R+q^{\prime }/R^{\prime }=0$ everywhere on the sphere. This proves the image charge for a grounded conducting sphere construction.

With real charges at $\pm d\hat{\mathbf{z}}$, superpose two images of charge $-qa/d$ at $\pm a^2\hat{\mathbf{z}}/d$. For the right-hand charge, take positive force away from the centre. Its electrostatic force isPutting $x=d/a$, the equal-charge balance around a grounded sphere requiresFor large $x$, the right side is approximately $2/x^3$, so $x\simeq 8$. This is an approximate equilibrium, not an exact cancellation at $x=8$.

To hold the sphere at electric potential $V$, add a central image $Q_0=4\pi {ϵ}_{0}aV$. The grounded pair already contributes zero potential on its surface. ThereforeAt $d=8a$ this is $F_0(8a)+qV/(64a)$, or approximately $qV/(64a)$ if the small grounded residual is neglected as in the preceding approximation. The charge on the sphere is $Q_0-2qa/d$; fixing its potential is different from independently fixing its net charge.

For the image force on a line charge above a conducting plane, with no imposed external electric field, use the method of images: replace the plane by a parallel image line of charge $-\lambda$ at height $-h$. The image is $2h$ from the real line, and its field at the real line has magnitude $\lambda /(4\pi {ϵ}_{0}h)$. ThereforeHere $\hat{\mathbf{z}}$ points upward from the plane. The force is attractive and normal to the conductor, for either sign of $\lambda$. The image reproduces the induced surface charge field; it is not an extra physical charge, and no factor $1/2$ is inserted in the force on the actual line.

The method of images replaces induced charges or currents on a boundary by fictitious sources outside the physical region, chosen to satisfy that boundary's conditions. For example, a point charge $q$ a height $d$ above a grounded conducting plane has an image $-q$ at the reflected point. The sum has zero potential on the plane, and the electrostatic uniqueness theorem identifies it with the physical solution in the upper half-space. Images do not describe a second physical charge below the metal.

For the superconducting plane, use the different boundary condition $B_z=0$ at its surface: magnetic flux cannot enter the ideal bulk. Choose the real ring moment $\mathbf{m}=m\hat{z}$, where $m=I\pi R^2$, at $z=h$. Its image has moment $-m\hat{z}$ at $z=-h$. This superconducting plane image dipole makes the normal fields cancel at $z=0$, while their tangential fields add. The image is a computational continuation; the physical field inside the plane is zero. The original sketch shows the dipole-scale field, appropriate to $h\gg R$; the fine field close to the ring wire is outside that approximation.

In the vacuum region the magnetic scalar potential isAt the surface $z=0$, it reduces to $-mh/[2\pi ({\rho }^2+h^2{)}^{3/2}]$. Its tangential derivative therefore givesBoth $B_z$ and $B_{\varphi }$ vanish there; the sign is inward radially for the chosen positive moment. The expression is the leading dipole result, not an exact finite-ring formula at arbitrary $R/h$.

For the force, calculate the external image field on the axis near the real dipole:Differentiate this field with the image location held fixed, then evaluate at $z=h$. The dipole levitation force above a superconducting plane isdirected upward. Differentiating a substituted separation $2h$ while moving both dipoles in the force derivative would incorrectly double the result.

To levitate ring mass $M$, balance this force against $Mg$:A light, thin wire and the smallest practical gap reduce current. At fixed gap and wire cross-section $s$, $M=2\pi Rs{\rho }_w$ gives $I_{\mathrm{lev}}\propto h^2\sqrt{s{\rho }_w}/R^{3/2}$, favouring a larger ring until the dipole-limit constraint on $R/h$ matters. There is no unique least current without geometric and material constraints. As a geometrically consistent order-of-magnitude example, take $R=1\mathrm{mm}$, $h=10\mathrm{mm}$, wire radius $50\mu \mathrm{m}$ and an assumed metallic-scale density ${\rho }_w=6.0\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. Then $M=2.96\times 10^{-7}\mathrm{kg}$ and $I_{\mathrm{lev}}\simeq 2.8\times 10^2\mathrm{A}$. The large value reflects the weak $h^{-4}$ force in the stipulated far-field geometry. Uniformly shrinking this geometrical example reduces current as size to the power $3/2$, but increases required electric current density; arbitrarily thin wire is therefore not a free optimization.

This balance estimate is not a verified superconducting design. The actual wire must carry that current without quenching, its self-field counts toward the limit, the plane must remain in its assumed flux-excluding state, and leads or supports add load. Transport critical-current measurements specify temperature, applied field and the voltage criterion rather than a material-independent limiting current.^[ref]

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.