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The original PDF places this complete question before B6 without a visible question number. It is included here under a descriptive heading.

Why images work. In the method of images, place fictitious image charges outside the region where the physical electric field is required. Choose them so that the combined electric potential satisfies the conductor boundary conditions. The Uniqueness of the Dirichlet problem then proves that this construction gives the physical exterior field. The images do not describe the actual surface-charge locations.

Two examples are a charge $q$ above a grounded conducting plane, represented by $-q$ at the mirror point, and the image charge for a grounded conducting sphere used below. In each case the image lies inside the excluded conductor and makes its surface an equipotential of the required value.

Potential of the two charges. By the law of cosines, the distances from the observation point to charges at $x_j$ are $s_j=(r^2+x_j^2-2r{x}_j\cos \theta {)}^{1/2}$. The superposition principle and Coulomb's law therefore give

The grounded sphere. Set $r=a$ and require $s_1=\alpha s_2$ at every polar angle. Matching the coefficients of $\cos \theta$ and the constant terms givesThe useful noncoincident solution isIndeed $s_1=(a/x_2)s_2$ on the sphere, so the two potential terms cancel there. A Gaussian surface enclosing the sphere but not the external charge measures the total image charge in the exterior representation. Thus the physical induced net charge is

An isolated neutral sphere. The grounded image would give nonzero net charge. Add a central image $q_0=+aq/x_2$ to cancel it. A central image changes the surface potential by a constant, preserving the equipotential condition, whereas an arbitrary off-centre image would generally spoil it. The image charges for a neutral conducting sphere giveThe potential is measured relative to zero at infinity for each finite $x_2$.

Fixed applied field and dipole moment. For positive $q$ on the positive $x$ axis, the applied central field is ${\mathbf{E}}_0=-q\hat{\mathbf{x}}/(4\pi {\epsilon }_0{x}_2^2)$. The central image contributes no electric dipole moment, soIts units are coulomb metres. Keeping $E_0$ fixed makes $q/x_2^2$ constant, so this induced moment is independent of the distance.

The distant-charge limit. As $x_2\to \infty$ with $q/x_2^2$ fixed, the applied field becomes uniform over any fixed neighbourhood of the sphere. The neutral pair of opposite images approaches a point dipole of the finite moment above. The external potential contains a growing spatial constant $q/(4\pi {\epsilon }_0{x}_2)$, so the infinity-referenced conductor potential does not have a finite limit. Subtracting this irrelevant constant gives the usual induced charge on a conducting sphere in a uniform electric field; the exterior potential becomesThis has zero potential on the sphere, but a uniform applied field cannot simultaneously have its potential vanish at spatial infinity.

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.

In electrostatic equilibrium a conductor has $\mathbf{E}=0$ throughout its bulk and constant electric potential on each connected component. Otherwise free charge would continue moving. Gauss's law then excludes bulk excess charge; it resides on surfaces. Just outside a conductor the tangential field vanishes and the normal jump isfor a vacuum interface with outward normal $\hat{\mathbf{n}}$. Surface charge rearranges in an external field to enforce these boundary conditions. Greater curvature often concentrates charge, but the actual distribution is a global boundary-value problem.

An empty closed cavity inside a conductor has constant potential and zero field, even in an external electrostatic field. If a charge $q$ is placed in the cavity, a Gaussian surface in the metal shows that the cavity wall carries total induced charge $-q$; its detailed distribution depends on geometry. An initially neutral isolated conductor then has $+q$ on its exterior surface. A grounded conductor can exchange charge with Earth, so this isolation condition no longer applies.

A useful example is a charge $q$ at height $d$ above an infinite grounded plane. In the physical half-space, the method of images replaces the induced charges by an image $-q$ at height $-d$, giving the attractive force $q^2/(16\pi {ϵ}_0{d}^2)$. The image is a mathematical construction, not an actual charge behind the conductor. Equipotential surfaces and charge conservation determine conductor fields together.

The method of images replaces conductors by fictitious charges outside the physical solution domain. Choose images so the real-plus-image potential obeys the specified conductor boundary values. The electrostatic uniqueness theorem then identifies that potential in the physical domain; fields inside the replaced conductor are not physical predictions.

Put the cylinder axis at the origin, real line at $(2a,0)$ and image at $(a/2,0)$. Let $r_1,r_2$ be distances from these lines. The exterior potential can be writtenOn the surface, $r_1^2=a^2(5-4\cos \theta )$ and $r_2^2=r_1^2/4$, so $\varphi =0$ everywhere as required. The additive constant is essential for a grounded cylinder; the two-dimensional potential need not also be zero at infinity.

The physical line feels the image field at separation $3a/2$. Its force per unit length is attractive:Measure $\theta$ from the radius toward the real line, as in the original diagram. The radial field just outside the surface isSince the interior field is zero,Integrating $a\sigma (\theta )$ around the circumference gives total induced charge $-\lambda$ per unit length.

Let $\zeta =g(z)$ be a conformal map, and let a point vortex of circulation $\Gamma$ lie at $z=c$. Suppose the mapped complex potential of a two-dimensional potential flow iswhere $R$ is a holomorphic function near $g(c)$ and represents the other point vortices, boundaries, and any imposed flow. The physical complex velocity is $w^{\prime }=g^{\prime }{W}^{\prime }$. The point vortex follows its regular velocity, found by removing $\Gamma /[2\pi i(z-c)]$ before taking the limit. A Taylor expansion proveswhich gives the title formula. The complex conjugate occurs because the complex velocity convention is $w^{\prime }=u-iv$, whereas $\dot{c}=u+iv$.

For a mapped upper half-plane with no imposed flow, the method of images gives $R(\zeta )=-\Gamma \log (\zeta -\overline{g(c)})/(2\pi i)$, henceThe first term is essential: excluding a mapped singularity and then transforming its regular velocity does not in general equal subtracting the physical singularity after transformation.

In three dimensions the Poisson kernel for the upper half-space isIt gives the harmonic function $\varphi (x,y,z)={\int }_{{\mathbb{R}}^2}{P}_z(x-\xi ,y-\eta )f(\xi ,\eta )d\xi d\eta$ for compatible boundary data. The method of images constructs the zero-boundary Dirichlet Green function by subtracting the reflected-source Green function of the Laplacian from the source kernel. On the boundary plane the outward unit normal is $-e_z$, and $-{\partial }_{n}G=P_z$, giving the integral formula through Green's second identity.

The kernel is positive and has integral one: in polar coordinates this is $z{\int }_0^{\infty }r(r^2+z^2{)}^{-3/2}dr=1$. Its mass outside any fixed disk tends to zero as $z\downarrow 0$. It is therefore an approximate identity, recovering bounded boundary data at continuity points. At a jump across a smooth curve, the vertical limit gives the average of the two adjacent values; assigning a different value only on that curve does not change the harmonic extension.

For a conducting sphere of radius $R$ at center height $h>R$ above a grounded infinite plane,The distance $h$ is measured to the center, so the surface gap is $h-R$. Relative to the capacitance of an isolated conducting sphere, the enhancement is $c(R/h)=C(R,h)/(4\pi ϵR)$.

The method of images gives a constructive proof. At fixed sphere potential $V_s$, start with $q_0=4\pi ϵR{V}_s$ at height $z_0=h$ and place its opposite image below the plane. Cancel that image's nonconstant sphere potential by the successive chargeswith opposite mirror charges below the plane at every step. The resulting sphere potential is $V_s$ and the plane potential is zero. Since $q_n=4\pi ϵR{V}_s\sinh \alpha /\sinh ((n+1)\alpha )$, summing the sphere charges gives the displayed capacitance. In particular $c(y)=1+y/2+O(y^2)$ for small $y$.

The exact sphere-plane boundary-value problem is also treated in Behunin and colleagues' analysis of sphere-plane electrostatics.

On three noncompact spatial dimensions times a circle of circumference $L$, the method of images gives ${\sum }_{j\in \mathbb{Z}}[r^2+j^2{L}^2{]}^{-1}=\pi \coth (\pi r/L)/(Lr)$ at equal compact coordinates. This interpolates from $1/r^2$ at short distances to $\pi /(Lr)$ at long distances. A gravitational circle-radius scalar can change the long-range coefficient if it remains unstabilized.

For a particle with mass $m$ on $(0,L)$, put $K_0(x;\beta )=\sqrt{m/(2\pi {\hslash }^2\beta )}{e}^{-m{x}^2/(2{\hslash }^2\beta )}$. The method of images gives $K_D(q_f,q_i;\beta )={\sum }_r[K_0(q_f-q_i+2rL;\beta )-K_0(q_f+q_i+2rL;\beta )]$. The diagonal integral isThe reflected intervals tile the real line, giving the subtraction $1/2$. The Poisson summation formula converts this to ${\sum }_{n\ge 1}{e}^{-\beta {\hslash }^2{\pi }^2{n}^2/(2m{L}^2)}$, proving equality between the image and energy eigenstate calculations.

In a vacuum domain with specified charge density, fixed boundary electric potential and the appropriate condition at infinity, two solutions of the Poisson equation have difference $w$ satisfying ${\nabla }^{2}w=0$ with zero boundary data. Integration by parts gives $\int |\nabla w{|}^{2}dV=0$; hence their electric fields coincide, and their potentials coincide after fixing the constant. This is the Dirichlet case of the more general dielectric uniqueness result. It justifies the exterior field constructed by the method of images.