Method of Images

In 1845, a twenty-year-old William Thomson explained to Joseph Liouville how to solve a problem that looks hopeless. A point charge hangs near a grounded conductor; the conductor's surface crawls with induced charge arranged in some unknown distribution that depends on the very field it helps create. Coulomb's law is circular here, and the honest boundary-value machinery — series, grids — is heavy. Thomson's move: throw the conductor away. Replace it with one or two fictitious "image" charges hidden in the region where the conductor used to be, chosen so that the potential of real charge plus fictions vanishes exactly where the conductor's surface was. In the physical region, this cartoon solves the same Poisson equation with the same boundary values as the real problem — and the uniqueness theorem, freshly minted in the previous article, does the rest: the cartoon is the answer. That audacity is the method of images, the most elegant weapon in the boundary-value arsenal. This article is about the method itself: why it is a theorem and not a trick, the strict rules that keep it legal (images must live outside the region you care about; the solution is fiction inside the old conductor), the two general theorems of the trade — the total induced charge always equals the total image charge, and the energy of a charge near grounded metal is half the naive charge-image energy — and the catalog of geometries where images exist, from the single mirror charge of a plane to the three-image chorus of a right-angle corner and the infinite hall-of-mirrors between parallel plates. The two headline constructions get their own full-length features next: the grounded plane and the conducting sphere, each worked to the induced-charge map. Here you learn the method; there you watch it perform.

Key takeaways

The method: replace conductors by fictitious image charges placed outside the physical region, tuned so the total potential meets the boundary conditions; by uniqueness, the resulting $V$ is exact in the physical region.
The rules: images never sit inside the region being solved (that would change $\rho$ ); the solution is valid only in the physical region — inside the replaced conductor the real field is zero, not the image's.
Theorem of totals: the induced charge on a grounded conductor equals the total image charge — provable with one Gaussian surface, before any integration of $\sigma$ .
Energy theorem: a charge near grounded conductors stores $W = \tfrac{1}{2}\,qV_{\text{im}}(\mathbf{r}_q)$ — half the naive charge–image interaction energy, because induced charge is linear response, not a fixed partner.
The catalog: plane (1 image) · sphere (1 image, next articles) · right-angle corner (3 images) · wedge of angle $180°/n$ ( $2n-1$ images) · parallel plates (infinite ladder) · dielectric half-space (partial image).
KEY IDEA: uniqueness converts a well-chosen fiction into the exact solution — any image arrangement outside the region that matches the boundary data solves the problem, complete with real forces, real induced charge, and half-price energy.

The core insight

Replace the conductor by image charges outside the physical region so that the combined potential satisfies the boundary conditions; uniqueness then guarantees this fictional configuration produces the true field, the true induced charge (totaling the image charge), and the true force — with the system energy equal to half the naive charge-image value.

Where this lives today

Images never retired; they industrialized. Fast capacitance extractors for chip interconnects seed their Green functions with image ladders over the substrate's conducting planes; antenna engineers model any monopole over a ground plane as half of a dipole — the image supplying the other half — which is why AM towers are half the length the wavelength seems to demand; and plasma physicists compute the pull of a tokamak wall on a dust grain, or of a sheath on an ion, with the same mirror charge Thomson mailed to Liouville. In scanning-probe microscopy, the tip–sample force at nanometer range is dominated by exactly the image attraction of worked example 2, corner geometry and all. The most striking modern cameo is quantum mechanical: an electron outside a metal surface feels its own image potential $-e^2/16\pi\varepsilon_0 z$ (the plane article's energy, read as a potential well) — and that well supports bound states. These image-potential states, hydrogen-like levels hovering angstroms above metal surfaces with binding energies scaling as the Rydberg over sixteen, are routinely photographed by two-photon photoemission and were an early playground of time-resolved surface science. A fictitious charge, certified by a nineteenth-century uniqueness theorem, binding real electrons into a measurable spectrum: the method of images does not merely compute physics — occasionally it is the physics.

Origins · 1 minWilliam Thomson (Lord Kelvin) and how this was discovered

William Thomson (Lord Kelvin)

1824–1907 · Glasgow and Cambridge

Thomson invented electrical images in 1845, aged twenty-one, in letters to Joseph Liouville written while the young Scot toured Paris after his Cambridge degree — Liouville published extracts in his own journal, and the sphere result (the $-qR/a$ image the second application article derives) circulated across Europe before Thomson wrote it up properly. The same trip produced his other great act of scientific matchmaking: having hunted down the obscure 1828 Essay of George Green, he had it reprinted where the continent's mathematicians would finally read it — images and Green functions entered the canon holding hands. Thomson's later career consumed Victorian physics whole: thermodynamics' absolute temperature scale bears his title (he was ennobled as Baron Kelvin for the transatlantic telegraph cable, physics' first megaproject), and his knack for turning theorems into instruments — mirror galvanometers, cable theory, tide predictors — made him the century's model of the engineer-mathematician. Images are pure early Thomson: a problem everyone agreed was hard, dissolved by one artfully placed fiction and a theorem to certify it.

History

1828

— Green's Essay constructs the half-space potential with what is, in modern eyes, the plane image hiding inside a Green function

1845

— Thomson, writing to Liouville from Paris, invents electrical images and solves the grounded sphere

1848

— Thomson's collected papers on "electrical images" systematize the method: planes, spheres, and inversions

1873

— Maxwell's Treatise devotes a chapter to images, with the famous engraved field-line plates that taught generations to see the mirror charge

The physical picture

Stand in front of a mirror. Optics in the room is as if a duplicate of you stood behind the glass — light paths in your half of the world are indistinguishable from that fiction. The fiction has strict limits: there is no one behind the mirror, and no light there either. The method of images is this idea with the electric field: a grounded conducting plane "reflects" a charge $+q$ as an image $-q$ at mirror-image position, and in your half of space the two stories — real charge plus messy induced surface charge, versus real charge plus one tidy image — produce identical fields. Behind the conductor's surface, the image story is pure fiction: the real field there is zero.

Figure

A fiction certified by uniqueness — same region, two stories, one field

Why is the trick legal rather than merely persuasive? Because of what the previous article proved. The physical region — the vacuum above the plane, say — is governed by three facts: Poisson's equation with the real charge $q$ as its only source, $V=0$ on the conductor surface, $V\to0$ at infinity. The image configuration reproduces all three in that region: the image sits outside, so it adds no source inside; its position and strength are tuned to zero the potential on the old boundary surface; and a finite set of charges dies off at infinity. Same equation, same data — the first uniqueness theorem admits exactly one solution, so the cartoon's field is the field. Everything physical follows by ordinary rules: the force on $q$ is the force the image exerts (that is the field of the induced charge, evaluated at $q$ ); the surface charge is harvested from $\sigma = -\varepsilon_0\,\partial V/\partial n$ ; even the energy follows, with one famous subtlety we derive below. The method's Achilles heel is geometry. An image solution exists only when the boundary shape is generous enough to be "mirrored away" by finitely many (or a tractable infinity of) point charges: planes, spheres, cylinders, wedges whose angle divides 180°, and combinations. A right-angle corner of two grounded planes needs three images — mirror in each plane, then mirror the mirrors through the corner. A wedge of angle $180°/n$ needs $2n-1$ . Two parallel grounded planes trap the reflections in an infinite regress — image of image of image — yielding a convergent infinite series, the electrostatic hall of mirrors. A wedge of, say, $70°$ (which $180°$ does not divide) defeats point images entirely, and you are back to separation of variables. The method is a precision instrument, not a universal solvent — but where it strikes, nothing is faster.

Figure

The corner and the hall of mirrors — how image sets close, or don't

The method as a theorem, and the two theorems of the trade

The recipe, stated as the checkable proposition it is:

V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\left[\sum_{\text{real }i}\frac{q_i}{|\mathbf{r}-\mathbf{r}_i|} + \sum_{\text{images }j}\frac{q'_j}{|\mathbf{r}-\mathbf{r}'_j|}\right], \qquad \mathbf{r}'_j \notin \Omega \qquad (1)

where: $\Omega$ — the physical region; $q_i, \mathbf{r}_i$ — the genuine charges inside $\Omega$ ; $q'_j, \mathbf{r}'_j$ — image charges, all placed strictly outside $\Omega$ . If this $V$ reproduces the prescribed boundary values on $\partial\Omega$ , it is the unique solution in $\Omega$ . The exclusion $\mathbf{r}'_j\notin\Omega$ is not pedantry — an image inside $\Omega$ would add a delta-function source and you would have solved a different problem, exactly (Derivation A dwells on this). The first general theorem converts image bookkeeping into physical charge:

Q_{\text{induced}} = \sum_j q'_j \qquad (2)

where: $Q_{\text{induced}}$ — the total charge drawn onto the grounded conductor's surface. No integration of $\sigma$ over the surface is needed to know the total — one Gaussian surface does it (Derivation B), and the detailed maps in the next two articles will confirm it by honest integration. The second is the energy fine print:

W = \tfrac{1}{2}\,q\,V_{\text{im}}(\mathbf{r}_q) \qquad (3)

where: $V_{\text{im}}(\mathbf{r}_q)$ — the potential of the images alone, evaluated at the real charge's location. The naive guess — treat charge and image as an ordinary pair, $W_{\text{naive}} = qV_{\text{im}}$ — is exactly a factor of two too large. The half is the signature of linear response: the induced charge is not a pre-existing partner that was assembled at cost; it is conjured by $q$ itself, in proportion to $q$ . Derivation B proves the half cleanly; forces, by contrast, carry no factor of one-half — the force on $q$ is the full image force, because force is a property of the actual field at $q$ , not of assembly history.

Derivation AWhy the method works — and the two clauses that keep it legal

We elevate the recipe to a theorem: if an image configuration (1) matches the boundary data of a grounded-conductor problem, its potential is exact throughout the physical region $\Omega$ . The proof is a checklist against the first uniqueness theorem — deliberately pedantic, because each clause is a way real calculations go wrong. Strategy: verify that the image potential and the true potential solve the same boundary-value problem, then let uniqueness collapse them.

Step 1. Sources inside $\Omega$ : the true problem has charge density $\rho_\Omega$ (the real charges $q_i$ ). The image potential's sources are the $q_i$ and the $q'_j$ — but every $q'_j$ sits outside $\Omega$ by construction, so within $\Omega$ the image potential satisfies exactly $\nabla^2V = -\rho_\Omega/\varepsilon_0$ . Same equation. Why the exclusion clause is load-bearing: place an image inside $\Omega$ and this step fails — you have added a source, and uniqueness will faithfully deliver the solution of the wrong problem. When the sphere article puts its image at $R^2/a < R$ , the legality rests on the physical region being outside the sphere; solving the interior of a charged shell needs images outside it, and so on. Always ask: which region am I solving, and are all fictions banished from it?

Step 2. Boundary data: by tuning (position, strength) the images were chosen so $V=0$ on the grounded surfaces (or $V=V_0$ on held surfaces — add the appropriate constant-generating images or terms). Verifying this is the whole creative content of any image solution; it must be checked on the entire surface, not at convenient points. Same data.

Step 3. The boundary at infinity: a finite set of charges gives $V\to0$ at least as fast as $1/r$ ; the true localized problem demands the same. Same infinity clause. All hypotheses of the first uniqueness theorem hold, so in $\Omega$ :

V_{\text{image}} \equiv V_{\text{true}}.

Step 4. Harvest with a clear conscience — every physical quantity that lives in $\Omega$ or on its boundary is now exact: the field $-\nabla V$ , the force on each real charge (exerted, equivalently, by the induced surface charge or by the images — they have the same field in $\Omega$ ), the surface density $\sigma = -\varepsilon_0\,\partial V/\partial n$ , capacitances, all of it.

Step 5. And state the fiction clause with equal force: outside $\Omega$ — inside the replaced conductor — the image potential describes nothing. The real field there is zero (conductor in equilibrium); the image's smoothly continuing field is the price of the trick, paid in a region where nobody lives. Plotting the image field inside the metal, or computing "the energy stored in the image's field" there, are the two classic ways to be confidently wrong (the energy theorem below dodges the second).

Go deeper · Derivation B · 2 minThe two theorems of the trade — total induced charge, and the half-price energy

Derivation BThe two theorems of the trade — total induced charge, and the half-price energy

Two results hold for every image solution, whatever the geometry, and proving them once here saves the applications from re-deriving them. The totals theorem uses a single Gaussian surface and the fact that the image and true fields agree in $\Omega$ . The energy theorem uses linear response — the same argument that puts the $\tfrac12$ in a capacitor's $\tfrac12CV^2$ — and explains why forces carry no such factor.

Step 1. Totals. Enclose the grounded conductor in a closed surface $S$ lying entirely within the physical region $\Omega$ . In the real system, $S$ encloses the conductor with its induced surface charge $Q_{\text{ind}}$ (the conductor is otherwise neutral, grounded through a thin wire whose charge is negligible): flux $= Q_{\text{ind}}/\varepsilon_0$ .

Step 2. In the image system, the same surface $S$ encloses instead the image charges $\sum_j q'_j$ (they occupy the space where the conductor was): flux $= \sum_j q'_j/\varepsilon_0$ . But the two systems have identical fields on $S$ — $S$ lies in $\Omega$ , where the potentials coincide by Derivation A — so the two fluxes are equal:

Q_{\text{ind}} = \sum_j q'_j.

Why this is such a useful audit: it is available before any hard work. The plane's single image $-q$ announces total induced charge $-q$ ; the sphere's image $-qR/a$ announces the less obvious total $-qR/a$ ; the corner's three images announce $-q(2 - \text{corner factor})$ ... rather, $-q-q+q = -q$ . When the next articles integrate their $\sigma$ maps, this theorem is the answer key they must hit.

Step 3. Energy, by charging up. Fix the real charge's position $\mathbf{r}_q$ and dial its magnitude from $0$ to $q$ via $\lambda q$ , $\lambda: 0\to1$ . Grounded conductors respond linearly: at parameter $\lambda$ , the induced charge — hence the image set, hence the induced potential at $\mathbf{r}_q$ — is $\lambda$ times its final value: $V_{\text{ind}}^{(\lambda)}(\mathbf{r}_q) = \lambda V_{\text{im}}(\mathbf{r}_q)$ . The work to add the next sliver $\mathrm{d}(\lambda q)$ against the induced potential is $V_{\text{ind}}^{(\lambda)}\,\mathrm{d}(\lambda q)$ , so

W = \int_0^1 \lambda\,V_{\text{im}}(\mathbf{r}_q)\;q\,\mathrm{d}\lambda = \tfrac{1}{2}\,q\,V_{\text{im}}(\mathbf{r}_q).

Why half, in one sentence: the partner you are paying to approach is being created by the payment itself — averaged over the charging, you pay half the final rate. It is exactly the $\tfrac12$ of $\tfrac12CV^2$ and of induced-dipole energies $-\tfrac12\alpha E^2$ .

Step 4. Cross-check against the fixed-charge route: alternatively drag the full charge $q$ in from infinity along a path, paying $-\int\mathbf{F}\cdot\mathrm{d}\boldsymbol{\ell}$ against the image force. The image trails along as you move, so the force en route is position-dependent in precisely the way that makes this integral also come out to $\tfrac12 qV_{\text{im}}$ — the plane article performs this integral explicitly and lands on the same half, a satisfying two-method agreement in the spirit of the uniqueness article's warning.

Step 5. Forces, no half: at the final configuration, the force on $q$ is $q$ times the actual field of the induced charges at $\mathbf{r}_q$ — which equals the image field there, full strength. Energy bookkeeping halves; instantaneous forces never do. Conflating the two — e.g., computing force as $-\partial W/\partial d$ but forgetting that $W$ already contains the half — is self-correcting only if done consistently: indeed $F = -\partial W/\partial d$ with the halved $W$ gives the correct full image force in the plane problem, because the image distance moves with the charge. The applications keep these ledgers separate.

Limiting cases

Geometry	Images required	Where treated
Grounded infinite plane	1 (mirror charge $-q$ )	Image Charge for a Grounded Plane
Grounded sphere, exterior charge	1 ( $q'=-qR/a$ at $R^2/a$ )	Image Charges for a Conducting Sphere
Isolated/charged/held sphere	2 (add one at center)	sphere article's extensions
Right-angle corner (two planes)	3 (mirror, mirror, double-mirror)	worked examples below
Wedge of angle $180°/n$	$2n-1$	integer $n$ only — else no point-image solution
Two parallel grounded planes	infinite ladder (convergent series)	hall-of-mirrors regress
Dielectric half-space	partial image $q' = -q\,\tfrac{\varepsilon_2-\varepsilon_1}{\varepsilon_2+\varepsilon_1}$	E3 chapter; conductor limit $\varepsilon_2\to\infty$ gives $-q$

Worked examples

1 · Building the corner solution (confidence builder)

A point charge $q = 10\,\mathrm{nC}$ sits at $(x,y) = (3.0, 4.0)\,\mathrm{cm}$ from two grounded, semi-infinite conducting planes occupying the $x=0$ and $y=0$ half-planes (a right-angle corner). Construct the image set, verify the boundary conditions, and evaluate $V$ at the point $P = (1.0, 1.0)\,\mathrm{cm}$ .

Step 1. Mirror systematically. Reflect in the plane $x=0$ : image $-q$ at $(-3,4)\,\mathrm{cm}$ . Reflect in $y=0$ : image $-q$ at $(3,-4)\,\mathrm{cm}$ . Each reflection fixes its own plane but spoils the other — so reflect the reflections: both send a charge to $(-3,-4)\,\mathrm{cm}$ , with sign flipped twice: $+q$ . Three images, alternating signs around the corner; the set is closed (further reflections reproduce existing charges).

Step 2. Verify $V=0$ on the plane $x=0$ : points there are equidistant from $q$ at $(3,4)$ and $-q$ at $(-3,4)$ — pairwise cancellation — and equidistant from $-q$ at $(3,-4)$ and $+q$ at $(-3,-4)$ — cancellation again. By symmetry the same pairing kills $V$ on $y=0$ . Both images sit outside the physical quadrant ✓, sources in the quadrant unchanged ✓, $V\to0$ far away ✓ — uniqueness certifies the construction.

Step 3. Evaluate at $P=(1,1)\,\mathrm{cm}$ , with $kq = (8.99\times10^{9})(10^{-8}) = 89.9\,\mathrm{V\cdot m}$ and the four distances $r_1 = \sqrt{2^2+3^2} = 3.61\,\mathrm{cm}$ (from $q$ ), $r_2 = \sqrt{4^2+3^2} = 5.00\,\mathrm{cm}$ , $r_3 = \sqrt{2^2+5^2} = 5.39\,\mathrm{cm}$ , $r_4 = \sqrt{4^2+5^2} = 6.40\,\mathrm{cm}$ :

V(P) = 89.9\left[\frac{1}{0.0361} - \frac{1}{0.0500} - \frac{1}{0.0539} + \frac{1}{0.0640}\right] = 89.9\,(27.7 - 20.0 - 18.6 + 15.6) = 4.3\times10^{2}\,\mathrm{V}.

Step 4. Audit with the totals theorem: images sum to $-q-q+q = -q$ , so the two planes together carry induced charge $-10\,\mathrm{nC}$ — all of $q$ 's flux ends on the corner. $V(P) \approx 430\,\mathrm{V}$ , and the induced-charge total came free.

2 · Force and energy in the corner (exam level)

For the same configuration, find the force on $q$ and the electrostatic energy of the system, and exhibit the factor-of-two discipline.

Step 1. The force on $q$ is the full image force (no halves). With $kq^2 = (8.99\times10^{9})(10^{-8})^2 = 8.99\times10^{-7}\,\mathrm{N\cdot m^2}$ : from $-q$ at $(-3,4)$ , separation $6.0\,\mathrm{cm}$ , attraction along $-\hat{x}$ : $F_1 = -kq^2/(0.060)^2\,\hat{x} = -2.50\times10^{-4}\,\hat{x}\,\mathrm{N}$ . From $-q$ at $(3,-4)$ , separation $8.0\,\mathrm{cm}$ : $F_2 = -1.40\times10^{-4}\,\hat{y}\,\mathrm{N}$ . From $+q$ at $(-3,-4)$ , separation $\sqrt{6^2+8^2} = 10.0\,\mathrm{cm}$ , repulsion along $(0.6, 0.8)$ : $F_3 = (8.99\times10^{-5})(0.6, 0.8) = (5.39, 7.19)\times10^{-5}\,\mathrm{N}$ .

Step 2. Total:

\mathbf{F} = (-2.50 + 0.54)\times10^{-4}\,\hat{x} + (-1.40 + 0.72)\times10^{-4}\,\hat{y} = (-1.96\,\hat{x} - 0.69\,\hat{y})\times10^{-4}\,\mathrm{N},

magnitude $|\mathbf{F}| = \sqrt{1.96^2 + 0.69^2}\times10^{-4} = 2.1\times10^{-4}\,\mathrm{N}$ , pointing into the corner — the charge is attracted to its own induced screen, more strongly than a single plane at either distance would manage alone... check: single plane at 3 cm would pull $kq^2/(0.06)^2 = 2.5\times10^{-4}\,\mathrm{N}$ ; the corner's net $2.1\times10^{-4}\,\mathrm{N}$ is slightly less, because the far positive image pushes back. Images correct intuition as well as arithmetic.

Step 3. Energy by the half-price theorem: the image potential at $q$ 's location (distances $6.0$ , $8.0$ , $10.0\,\mathrm{cm}$ , $kq = 89.9\,\mathrm{V\cdot m}$ ):

V_{\text{im}} = 89.9\left[-\frac{1}{0.060} - \frac{1}{0.080} + \frac{1}{0.100}\right] = 89.9\,(-16.67 - 12.5 + 10.0) = -1.72\times10^{3}\,\mathrm{V},

W = \tfrac{1}{2}qV_{\text{im}} = \tfrac{1}{2}(10^{-8})(-1.72\times10^{3}) = -8.6\times10^{-6}\,\mathrm{J}.

Step 4. Read the ledger: $|W| = 8.6\,\mu\mathrm{J}$ is the work needed to drag the charge from the corner region out to infinity; the naive charge-image sum would claim $17.2\,\mu\mathrm{J}$ — the fictitious cost of assembling images that were never assembled. Force: $0.21\,\mathrm{mN}$ into the corner, full strength; energy: $-8.6\,\mu\mathrm{J}$ , half price.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §3.2 · Jackson, Classical Electrodynamics, 3rd ed., §2.1–2.5 · Purcell & Morin, Electricity and Magnetism, 3rd ed., §3.4 · Zangwill, Modern Electrodynamics, §8.3. Key results to keep in view: for the grounded plane, $V=0$ on $z=0$ and integrating $\sigma$ over the surface yields total induced charge $-q$ ; for the grounded sphere, imposing $V=0$ on the sphere surface gives image charge $q'=-qR/d$ located at $b=R^2/d$ ; the plane problem's force and energy satisfy the factor-of-two relation between $F$ and $W$ ; and the dielectric half-space image factor is $(\varepsilon_1-\varepsilon_2)/(\varepsilon_1+\varepsilon_2)$ . All numerical constants are exact.

Builds on: Uniqueness Theorems for the Potential · Boundary Conditions for the Potential · Laplace's Equation (E2) · Poisson's Equation (E2)

Leads to: Image Charge for a Grounded Plane · Image Charges for a Conducting Sphere · Separation of Variables in Spherical Coordinates

Is the induced surface charge real, or is it as fictional as the image?

Completely real — measurable with a proof plane, bankable coulomb by coulomb. The fiction is only the bookkeeping: the smooth sheet of genuine induced charge on the conductor's surface produces, throughout the physical region, exactly the field that one tidy point image would. The image is a compression algorithm for the induced charge's external field. That is also why harvesting $\sigma = -\varepsilon_0\,\partial V/\partial n$ from the image potential is legitimate: the potential at the surface is real; only its continuation past the surface is theater.

How do I find the image configuration in the first place? The derivations only verify guesses.

Honestly: pattern, symmetry, and a short catalog. Planes suggest mirror reflection; spheres suggest inversion ( $\mathbf{r} \to R^2\mathbf{r}/r^2$ — Kelvin's own route); wedges suggest iterated reflection until the set closes. The verification-only structure is not a weakness but the method's deal with uniqueness: any discovery procedure is admissible because certification is separate. When no finite closed set of reflections exists (the 70° wedge, two spheres in general position — which generate an infinite, still-useful image series), the honest conclusion is that this instrument doesn't fit, and separation of variables or a grid takes over.

Why is the force on the charge the full image force, but the energy only half the image energy?

Because they answer different questions. Force is instantaneous: the charge feels the field that actually exists at its location now, and in the physical region that field is exactly the image's — no history, no halves. Energy is biographical: it asks what work was done to reach this configuration, and the partner (induced charge) grew from zero during the approach, so the average bill runs at half the final rate. The two ledgers reconcile perfectly: differentiate the halved energy with respect to the charge's position and — because the image moves and scales in sympathy — you recover the full, unhalved force.

What changes if the conductor is not grounded but isolated, or held at a fixed potential?

Add images that fix the new bookkeeping without disturbing the boundary shape. For a sphere: grounded needs one image; held at $V_0$ adds a charge $4\pi\varepsilon_0RV_0$ at the center (a constant-potential generator on the sphere); isolated with total charge $Q$ adds at the center whatever charge makes the totals theorem read $Q$ . The second uniqueness theorem is the license for the isolated case — total charge specified, distribution free. The sphere article works all three variants; the plane, being infinite, only comes grounded (an isolated infinite plane has pathological bookkeeping).

Does the method survive dielectrics, where nothing is grounded and no surface is equipotential?

In half-space form, beautifully: a charge in medium 1 near a flat interface with medium 2 is solved by a partial image $q' = -q(\varepsilon_2-\varepsilon_1)/(\varepsilon_2+\varepsilon_1)$ for the field in medium 1 (plus a rescaled charge viewed from medium 2). The conductor is the limit $\varepsilon_2\to\infty$ : $q'\to-q$ , recovering the mirror. The matching conditions being two (potential and displacement continuity) instead of one ( $V=0$ ) is what makes the image partial rather than full. Curved dielectric boundaries are far less forgiving than curved conductors — the sphere's dielectric analogue has no exact point-image solution. The E3 chapter takes this up properly.

In the parallel-plate hall of mirrors, does the infinite image series actually converge?

Yes, conditionally but controllably. Charges alternate in sign and march off to infinity at spacings set by the plate separation, so the potential series behaves like an alternating harmonic series — convergent, and fast for points near the real charge. The force series converges absolutely ( $\sim 1/n^2$ terms). Numerically one sums pairs (image and its partner) to accelerate convergence; analytically the sum has closed forms in digamma functions. The regress is also conceptually tidy: each new image is required to fix the boundary its predecessor spoiled, and the ladder is the fixed point of that repair process.

Since the image reproduces the field only outside the conductor, what happens to all the field energy the image would have had inside?

It was never there — and the energy theorem is built to respect that. The true field energy is $\tfrac{\varepsilon_0}{2}\int E^2$ over the physical region only (the conductor's interior contributes zero, its field being zero). Naively integrating the image pair's energy over all space double-counts precisely the fictional interior contribution, and the factor-of-two discrepancy in $W$ is the visible symptom. The half-price formula $W = \tfrac12qV_{\text{im}}$ is the shortcut that gets the honest region-restricted integral right without performing it — the plane article checks the two routes against each other explicitly.