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For radii $a<b$ and a point charge $q$ at $r_0{\mathbf{e}}_z$, $a<r_0<b$, set $k=b/a$. The method of images gives the two families $(q{k}^n,r_0{k}^{2n}{\mathbf{e}}_z)$ and $(-qa{k}^n/r_0,a^2{k}^{2n}{\mathbf{e}}_z/r_0)$ for $n\in \mathbb{Z}$. The first family's $n=0$ term is the physical point charge. All other image charges lie outside the gap. Sum their Coulomb's law electric potentials with factor $1/(4\pi {ϵ}_0)$.

For a short proof, reflection of $(Q,z)$ in a grounded sphere of radius $R$ gives $(-RQ/z,R^2/z)$. On $r=a$ this pairs first-family index $n$ with second-family index $-n$; on $r=b$ it pairs $n$ with $1-n$. Each pair has zero boundary electric potential. The negative-index electric potential tails are $O(k^n)$ and the positive-index tails are $O(k^{-n})$, uniformly on the closed gap after excluding the physical singularity. These geometric series establish absolute convergence, boundary cancellation, and termwise differentiation away from the physical point charge. Uniqueness of Poisson equation identifies the solution. A spherical Gaussian surface just outside $a$ encloses the first family with $n\le -1$ and the second with $n\le 0$, so Gauss's law and the geometric series give $Q_a=-qa(b-r_0)/(r_0(b-a))$.

The image charge for a grounded conducting sphere is obtained by the Kelvin transform:It lies inside the electrical conductor, outside the region where it represents the physical electric field. The electric potential isAt the physical point charge, the electric potential has its usual Coulomb's law singularity. There is no physical singularity at the image charge: the interior expression is identically zero.

For $|\mathbf{r}|=a$, direct expansion of squared distances givesConsequently the two terms cancel exactly, proving the zero Dirichlet boundary condition on the sphere. At infinity each term is $O(r^{-1})$, and more explicitlyThus the Dirichlet boundary condition at infinity also holds. The candidate solves the Poisson equation with the one physical point charge; uniqueness of Poisson equation, with these Dirichlet boundary conditions, identifies it as the required electric potential.

Put $k=b/a>1$ and ${\mathbf{e}}_z=(0,0,1)$. The complete construction of image charges between concentric grounded spheres consists of two families, indexed by every integer:The term $(q_0,{\mathbf{r}}_0)$ is the physical point charge; every other term is an image charge. In the first family $n\le -1$ lies inside $r=a$ and $n\ge 1$ lies outside $r=b$. In the second family $n\le 0$ lies inside $r=a$ and $n\ge 1$ lies outside $r=b$. Thus no image charge lies in the physical gap.

To derive and verify these families, reflection of an axial point charge $(Q,z)$ in a grounded sphere of radius $R$ replaces it by the image charge $(-RQ/z,R^2/z)$. Reflection in $a$ sends $(q_n,{\mathbf{r}}_n)$ to $({\tilde{q}}_{-n},{\tilde{\mathbf{r}}}_{-n})$; reflection in $b$ sends it to $({\tilde{q}}_{1-n},{\tilde{\mathbf{r}}}_{1-n})$. Successive reflections therefore generate exactly the listed families. The method of images givesThe physical point charge itself is excluded when evaluating this formula. On $r=a$, pair the first-family term $n$ with the second-family term $-n$; their electric potentials cancel by the single-sphere distance identity. On $r=b$, pair $n$ with $1-n$ instead. These reindexings are justified by absolute convergence, as follows.

For $n\to -\infty$, both source positions tend to the origin while their electric charges have size $O(k^n)$. On the closed gap their distances are eventually at least $a/2$, so both electric potential terms are $O(k^n)$. For $n\to +\infty$, both distances are at least a positive constant times $k^{2n}$ and the electric charges are $O(k^n)$, giving $O(k^{-n})$ terms. Both tails are bounded by convergent geometric series. This proves absolute convergence and uniform convergence of the tails on the closed gap. On any compact set avoiding the physical point charge, the differentiated tails obey the same type of bounds: for example, the large positive tail of the electric field is $O(k^{-3n})$. Thus termwise differentiation is valid and the sum is a harmonic function except at the physical point charge, where it has the correct Coulomb's law singularity. The boundary cancellations and uniqueness of Poisson equation complete the construction.

Translate the concentric spheres so that their centre is $(0,0,-a)$, take $b=a+d$, and keep the point charge fixed at $(0,0,h)$. Its distance from the centre is $r_0=a+h$. As $a\to \infty$, the upper portions of the inner and outer spheres approach the planes $z=0$ and $z=d$. The preceding total induced electric charge becomesThe outer sphere has total induced electric charge $Q_b=-q-Q_a$: a Gaussian surface within its conducting material has zero electric flux, and the grounded exterior is field-free by uniqueness of Poisson equation. Therefore $Q_b\to -qh/d$.

One can justify the total-charge limit without assuming that local convergence of electric fields preserves their surface integrals. Use induced charge as harmonic measure: if a harmonic function $w$ is one on the selected grounded boundary and zero on the other, Green's second identity with the electric potential givesIndeed, ${\nabla }^2\varphi =-q{\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)/{ϵ}_0$, $\varphi =0$ on the boundaries, and their unit normal outward from the gap satisfies $\sigma ={ϵ}_0{\partial }_n\varphi$. For the spheres the weighting harmonic function is $w_a(r)=a(b-r)/(r(b-a))$, exactly the factor already obtained. For the slab the two weighting harmonic functions are $w_0=1-z/d$ and $w_d=z/d$. Applying the same identity to the decaying grounded-slab electric potential, whose boundary term at lateral infinity vanishes, establishes directlyTheir sum is $-q$, and their relative sizes measure how close the point charge lies to each grounded plane.

The method of images uses the real line charge at the origin and the opposite image charge at $(2d,0)$. LetSuperposing their logarithmic electrostatic potentials gives, in the physical region $x<d$,At $x=d$, $r_1=r_2$, so $\varphi =0$. Its tangential derivatives also vanish on that plane, giving $E_y=E_z=0$, exactly the electrostatic boundary conditions at a conductor. The image lies outside the physical domain, so it creates no extra physical source there. Together with the appropriate behaviour at infinity, uniqueness of Poisson equation identifies this field with the grounded-conductor solution.

Suppose ${\varphi }_1,{\varphi }_2$ solve the Dirichlet problem and put $u={\varphi }_1-{\varphi }_2$. Then ${\nabla }^{2}u=0$ in $V$ and $u=0$ on $S$. The divergence theorem applied to $u\nabla u$ gives Green's identityThus $\nabla u=0$, so $u$ is constant; its zero boundary value makes it zero. The Dirichlet solution is therefore unique.

For homogeneous Neumann data the same calculation again shows that $u$ is constant, but the boundary condition does not determine that constant. Hence, whenever a Neumann solution exists, adding any constant produces another solution. This is the standard uniqueness of Poisson equation distinction.

Find the potential from $\mathbf{E}=-\nabla \varphi$ and $\nabla \cdot (ϵ\nabla \varphi )=-{\rho }_f$. One must specify boundary data: a potential on each Dirichlet boundary, a normal displacement on each Neumann boundary, or an appropriate mixture. Conductors are equipotentials; at dielectric interfaces $\varphi$ is continuous and the normal displacement jumps by the specified free surface charge. With no singular dipole sheet, tangential electric field is continuous.

To establish uniqueness of Poisson equation, subtract two solutions with the same charge and boundary data, writing their difference as $u$. Integration by parts givessince $\nabla \cdot (ϵ\nabla u)=0$. The right side vanishes for identical Dirichlet data, identical Neumann data, or their usual mixed combination. Interface terms cancel under the homogeneous difference conditions. With positive permittivity, $\nabla u=0$: the field is unique, and the potential is unique if any suitable reference potential is fixed. For a pure Neumann problem an additive constant remains, and the prescribed outward displacement flux must equal the enclosed free charge. This theorem justifies image-charge solutions once their boundary data have been checked.

An electrostatic field calculation begins by specifying the charge information, material response and boundary conditions. In vacuum, prescribed volume and surface electric charge densities can be superposed directly using Coulomb's law. For a volume distribution,Differentiating this electric potential gives the electric field. This is useful for known source distributions, but in the presence of conductors the induced surface electric charge is usually an unknown rather than an independently specified input.

Gauss's law is especially effective when symmetry fixes the direction and magnitude of the electric field on a suitable closed surface. A spherical charge distribution gives a radial field, an infinitely long uniform line gives a cylindrical field, and a uniform infinite sheet gives a planar field. In an arbitrary geometry, the flux law remains true but does not alone determine the pointwise electric field. A multipole expansion is another useful approximation: far from a localized charge distribution, successive monopole, dipole and higher terms separate the effects of total charge and its spatial structure.

For a boundary-value formulation, use the electrostatic equationsThey imply Poisson equation ${\nabla }^2\varphi =-\rho /{ϵ}_0$, or Laplace equation where $\rho =0$. A conductor in electrostatic equilibrium is an equipotential with zero internal electric field. Its exterior normal electric field fixes the surface electric charge through $E_n=\sigma /{ϵ}_0$. One must specify whether each conductor is grounded, maintained at a prescribed voltage, or isolated with prescribed total electric charge.

Uniqueness of Poisson equation justifies methods that construct a convenient candidate solution. With fixed boundary electric potential and suitable behaviour at infinity, a solution satisfying the field equation is the solution. With normal-derivative boundary conditions, the source and flux must be compatible, and the electric potential is determined only up to an additive constant.

For example, a point charge $q$ a distance $h$ above a grounded infinite plane has the same upper-half-space electric potential as $q$ and a mirrored image charge $-q$ below it. Their potentials cancel on the plane. The image is a mathematical source in the excluded region, not a second physical charge. Separation of variables similarly constructs solutions suited to rectangular, cylindrical or spherical boundaries; matching coefficients enforces the prescribed boundary data.

In dielectrics, the distinction between total and free charge is essential. In a linear isotropic material,Only where permittivity is constant may this be reduced to ${\nabla }^2\varphi =-{\rho }_f/ϵ$. At an ordinary interface, $\varphi$ and the tangential electric field are continuous, while the jump in normal electric displacement field is the free surface charge. Applying Gauss's law to total charge with ${ϵ}_0$, or its displacement-field form to free charge, gives consistent descriptions; mixing their sources and permittivities does not.

A useful interface example is the image charge at a dielectric interface. If $q$ lies in medium 1, the source-side potential is represented bywhere $r_1$ is distance to the real charge and $r_2$ to its mirrored position. The other-side potential can be written ${\varphi }_2=q_t/(4\pi {ϵ}_2{r}_1)$, where $q_t=2{ϵ}_{2}q/({ϵ}_1+{ϵ}_2)$. Continuity of $\varphi$ and of normal $\mathbf{D}$ fixes both coefficients. The equal-permittivity limit gives no image; the conductor limit ${ϵ}_2\to \infty$ gives $q^{\prime }\to -q$.

For a sphere of radius $a$ and relative permittivity ${ϵ}_r$ in a uniform external electric field ${\mathbf{E}}_0$, matching the spherical solutions gives a uniform internal electric field and an external induced electric dipole moment:This illustrates how a material boundary modifies an otherwise simple applied electric field.

For complex boundaries or spatially varying materials, finite difference methods or the finite element method solve the same boundary value problem numerically. A sound calculation checks its boundary conditions, charge or flux balance and convergence with mesh refinement. The governing equations, source classification and uniqueness argument remain the organizing principles even when an analytic formula is unavailable.

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.