Method of images

For real point charges $Q$ and $-2Q$ at $(0,b)$ and $(a,b)$ above the plane $y=0$, the image charges are $-Q$ and $+2Q$ at the reflected positions. The image charges enforce constant electric potential on the plane. The physical forces follow by summing ordinary Coulomb's law forces on each real charge; the electrostatic-energy factor of one half for induced charges does not multiply this force. The smaller positive charge can have an upward total force component, even though its own image attracts it downward.

An ideal electric dipole at height $a$ above a grounded plane has an image electric dipole moment with reversed tangential component and unchanged normal component. If $\theta$ is its angle to the normal, its conductor-response potential energy is the displayed expression. The factor one half in $U=-\mathbf{p}\cdot {\mathbf{E}}_{\mathrm{im}}/2$ accounts for induced surface charge. Therefore its downward force magnitude is $3p^2(1+{\cos }^2\theta )/(64\pi {ϵ}_0{a}^4)$, and its orienting torque follows from $-{\partial }_{\theta }U$. The two normal orientations are stable against tilting; the parallel orientation is unstable. The net induced plane electric charge is zero for a neutral source, although its local surface charge density is nonzero.

For two grounded conducting half-planes meeting at angle $\pi /3$, place a point charge $q$ at radius $d$ on the bisector $\theta =\pi /6$. The method of images closes after six copies: positive copies at angles $\pi /6+2k\pi /3$ and negative copies at $-\pi /6+2k\pi /3$, for $k=0,1,2$. Only the first positive copy is real. Opposite-sign reflected pairs make the electric potential vanish on both faces; the uniqueness theorem for the electrostatic potential proves the construction. Summing the five image charge electric fields gives the electrostatic force $\mathbf{F}=-q^2(5/4-1/\sqrt{3}){\hat{\mathbf{e}}}_r/(4\pi {ϵ}_0{d}^2)$, towards the intersection. The coefficient follows from distances $d$, $\sqrt{3}d$, and $2d$ to the adjacent negative, nonadjacent positive, and opposite negative copies.

A real infinite line charge $\lambda$ lies at distance $b>a$ from the axis of a grounded conducting cylinder of radius $a$. An image line $-\lambda$ at radius $a^2/b$ lies inside the excluded conductor. On the circle, real and image distances obey $r_1/r_2=b/a$, so $\varphi =(\lambda /(2\pi {ϵ}_0))\ln [(b/a)r_2/r_1]$ vanishes there. The image field solves the physical exterior problem with bounded far potential. Its additive constant matters; the interior image field is not a physical field inside the conductor.

For the real line at radius $b>a$ and polar angle measured toward it, the exterior radial field gives $\sigma (\theta )=-\lambda (b^2-a^2)/[2\pi a(b^2+a^2-2ab\cos \theta )]$. Its integral around the circumference is $-\lambda$ per unit axial length. At $b=2a$, this becomes $-3\lambda /[2\pi a(5-4\cos \theta )]$. The nearer surface has larger induced charge magnitude.

With no imposed external field, a line of charge density $\lambda$ at height $h$ above a conducting plane has an image $-\lambda$ a distance $2h$ away. The force per unit length is $-{\lambda }^2\hat{\mathbf{z}}/(4\pi {ϵ}_{0}h)$. It is always attractive. The image field gives the force directly; a factor one-half used in some image-energy calculations does not multiply this force.

For a positive integer $N$, a point source in a grounded planar wedge of opening $\pi /N$ can be supplemented by $2N-1$ images. Keep positive sources at $z_0{e}^{2\pi ik/N}$ and negative sources at $\overline{z_0}{e}^{2\pi ik/N}$, $k=0,\dots ,N-1$, counting the actual source as the first positive one. Reflection in either boundary pairs equal distances with opposite signs, so the logarithmic electrostatic potential vanishes on both rays. Every image lies outside the open physical wedge.

The identity ${\prod }_k(z-z_0{e}^{2\pi ik/N})=z^N-z_0^N$ compresses the image sum toHere the normalization is that of a two-dimensional point source, equivalently a charge per unit length in a translationally invariant three-dimensional problem. For $N=3$, a source on the bisector with unit radius has $z_0^3=i$, giving the ratio $(z^3-i)/(z^3+i)$. The method of images works finitely here because successive boundary reflections generate a finite rotation group; arbitrary wedge angles need not have a finite image construction.

For $b>0$, the Poisson equation $\Delta V=q{\delta }^{(2)}(x,y)$ in $-b<x<b$ with homogeneous Neumann boundary conditions on both edges admits the even-in-$y$ method of images solution in the title, where $r_n^2=(x-2nb{)}^2+y^2$. All image strengths are $q$. Pairing the images at $\pm 2nb$ and subtracting their constant logarithms makes each summand $O(n^{-2})$, giving local uniform convergence away from the source. The unrenormalized sum of logarithms does not converge.

Differentiating givesThe corresponding $x$-derivatives cancel in reflected pairs on $x=\pm b$. For $y>0$, integration across the strip yields a telescoping arctangent sum:One closed form, up to an additive constant, is $V=(q/4\pi )\log [\cosh (\pi y/b)-\cos (\pi x/b)]$. Its near-source logarithmic singularity has coefficient $q/(2\pi )$. Conditions at large $|y|$ or a symmetry class must be specified for a uniqueness claim; side Neumann boundary conditions alone allow additional homogeneous harmonic functions.

An infinite conducting cylinder of radius $a$ at centre height $d>a$ above a grounded plane has the same exterior electric potential as opposite line charges $\pm \lambda$ at heights $\pm c$. If $R_{+}$ and $R_{-}$ are distances from the positive and negative lines, $V=\lambda \ln (R_{-}/R_{+})/(2\pi {\epsilon }_0)$. The plane has $R_{-}=R_{+}$; the cylinder has $R_{-}/R_{+}=(d+c)/a$. Its voltage is therefore $\lambda \mathrm{arcosh}(d/a)/(2\pi {\epsilon }_0)$, giving the displayed capacitance per unit length. The Uniqueness of the Dirichlet problem fixes the exterior field. A Gaussian contour surrounding the cylinder encloses the positive image line, so $\lambda$ is also its physical charge per unit length.

For the cylinder-plane image charges geometry, the attractive force is the displayed quantity per unit length. One derivation uses the effective energy including the voltage source: at fixed cylinder voltage $V$, the force toward increasing centre height is $F_d=V^2(d{C}^{\prime }/dd)/2$. Since $C^{\prime }=2\pi {\epsilon }_0/\mathrm{arcosh}(d/a)$ decreases with $d$, this force points toward the plane. For $d\gg a$, $\lambda \simeq 2\pi {\epsilon }_{0}V/\ln (2d/a)$ and $f\simeq \pi {\epsilon }_0{V}^2/[d{\ln }^2(2d/a)]$. A total force additionally requires the cable length.

For a current dipole in a homogeneous insulating sphere of radius $a$, let $\lambda =r^2/a^2$ and let $\Psi$ be its primary electric potential. The displayed representation solves the interior Neumann boundary condition up to an additive constant. It rewrites the spherical Kelvin transform correction and a distributed dipole correction using the dipole line-integral identity. The image parameters depend on the observation radius; this is an exact parametrized representation rather than a fixed collection of image source positions.

Reflecting an interval repeatedly across its endpoints maps a reflected path to a free path in an image interval. A return trajectory from $q\in (0,L)$ ends at $q+2rL$ after an even number of wall reflections, or $-q+2rL$ after an odd number. An infinite square well has reflection amplitude $-1$, so the corresponding Dirichlet heat kernel on an interval is the difference of direct and mirrored free kernels. Individual straight paths illustrate the geometry; the path integral sums all time-sliced paths with the same endpoints.