Image Charges for a Conducting Sphere

The grounded plane was solved by a mirror. A sphere is a curved mirror — and remarkably, the image trick survives the bending. William Thomson's 1845 discovery, mailed to Joseph Liouville from Paris: a point charge $q$ at distance $a$ from the center of a grounded conducting sphere of radius $R$ is exactly screened, everywhere outside the sphere, by a single fictitious charge — not the mirror twin $-q$ , but a shrunken image $q' = -qR/a$ , parked inside the sphere at the inverse point $b = R^2/a$ . Two eerily simple formulas, and behind them a beautiful piece of geometry: the sphere is the locus of points where distances to charge and image hold a constant ratio, so a weighted pair of Coulomb potentials can cancel on the entire curved surface at once. This article works the sphere the way the previous one worked the plane, full arithmetic included. The construction is derived, then verified on the whole sphere by the similar-triangles identity that makes it tick. The induced-charge map $\sigma(\theta)$ comes out in closed form — crowding toward the near pole in a ratio of $(a+R)^3/(a-R)^3$ , a factor of 27 in our worked numbers — and integrates to exactly $q'$ , which this time is not $-q$ : unlike the plane, a grounded sphere intercepts only the fraction $R/a$ of the charge's flux, and the rest escapes to infinity. Force and energy follow, with the plane's formulas recovered exactly in the close-approach limit. Then the payoff that makes the sphere the more versatile tool: because the construction is modular, one extra charge at the center converts the grounded solution into the isolated sphere, the charged sphere, or the sphere held at any potential — including the counterintuitive gem that a neutral isolated sphere attracts any nearby charge, the electrostatic root of why dust sticks to everything.

Key takeaways

Kelvin construction: charge $q$ at distance $a$ from the center of a grounded sphere of radius $R$ is imaged by $q' = -qR/a$ at the inverse point $b = R^2/a$ ; the two-charge potential vanishes on the whole sphere.
The geometric engine: for any point on the sphere, (distance to image) $= (R/a)\times$ (distance to charge) — similar triangles from $b/R = R/a$ — so $q/s + q'/s'$ cancels identically.
Induced density $\sigma(\theta) = -\dfrac{q(a^2-R^2)}{4\pi R\,(a^2+R^2-2aR\cos\theta)^{3/2}}$ , integrating to $q' = -qR/a$ : the sphere soaks up only the flux fraction $R/a$ ; the remainder runs to infinity.
Force $F = -\dfrac{q^2}{4\pi\varepsilon_0}\dfrac{Ra}{(a^2-R^2)^2}$ and energy $W = -\dfrac{q^2R}{8\pi\varepsilon_0(a^2-R^2)}$ ; as $a\to R$ both collapse onto the grounded-plane formulas.
Modular extensions by one center charge: isolated neutral sphere (add $+qR/a$ ), total charge $Q$ (add $Q+qR/a$ ), potential $V_0$ (add $4\pi\varepsilon_0RV_0$ ) — and a neutral sphere still attracts the charge, with far-field force set by the polarizability $\alpha = 4\pi\varepsilon_0R^3$ .
KEY IDEA: inversion geometry lets one interior image charge $-qR/a$ at $R^2/a$ solve the grounded sphere exactly — and a second charge at the center then buys every other spherical boundary condition on the market.

The core insight

A grounded sphere of radius $R$ screens an exterior charge $q$ at distance $a$ via the image $q'=-qR/a$ at the inverse point $R^2/a$ — the constant distance-ratio $R/a$ on the sphere makes the pair's potential vanish there — and adding one more image at the center extends the exact solution to isolated, charged, or fixed-potential spheres.

Where this lives today

Kelvin's sphere is the workhorse "tip model" of modern surface science: an AFM or Kelvin-probe tip apex is a sphere at distance $a$ from a plane or sample, and the force–distance curves instruments record are fit with image series built from exactly this article's construction (sphere-plane needs the iterated ladder — Kelvin image of the plane image of the Kelvin image — converging fast and coded into every probe-calibration package). Colloid science leans on the same machinery: the interaction of charged micron spheres in electrolytes starts from conducting-sphere images before screening corrections, and the celebrated modern result that two like-charged conducting spheres attract at close range — counterintuitive, experimentally real, and analyzed exhaustively via image expansions (Lekner's 2012 revival) — is worked example 2's polarization victory taken one step further: close enough, induced charge beats bare Coulomb repulsion even between like charges. The isolated-neutral force law, $F \propto \alpha/a^5$ with $\alpha = 4\pi\varepsilon_0R^3$ , is also the classical floor under a quantum tower: replace the point charge by a fluctuating dipole and the same polarizability produces van der Waals attraction; make both bodies fluctuate and Casimir physics appears. When atomic physicists quote a metal nanosphere's response to a trapped ion, or when levitated-optomechanics groups compute the pull of a chamber wall on a charged silica bead, the first line of the estimate is $q' = -qR/a$ at $b = R^2/a$ — a formula mailed to Paris in 1845 and never once off duty since.

Origins · 1 minJoseph Liouville and how this was discovered

Joseph Liouville

1809–1882 · Paris, Collège de France

Liouville is the mathematician to whom the method of images was literally addressed: Thomson's 1845 letters unveiling the sphere construction were sent to him, and he published extracts in his Journal de Mathématiques Pures et Appliquées — the legendary "Liouville's Journal" he had founded at twenty-six and would edit for four decades, rescuing along the way the manuscripts of the deceased Galois. His own theorem repaid the correspondence with interest: in notes appended to Monge's works, Liouville proved that in three dimensions the only conformal (angle-preserving) transformations are the similarities and sphere-inversions — establishing that Kelvin's inversion trick wasn't one clever choice among many but essentially the only geometric symmetry electrostatics had left to offer. The rigidity is deep: in 2D, conformal maps abound and image methods generalize riotously; in 3D, Liouville's theorem says the sphere is where the road ends. A prodigious analyst (Sturm–Liouville theory, the first transcendental numbers, the phase-space theorem mechanics students meet), he was the great connector of nineteenth-century mathematics — and, through two letters from a twenty-one-year-old Scot, godfather to this article.

History

1828

— Green's Essay computes the potential of a charge near a grounded sphere by direct expansion, images avant la lettre

1845

— Thomson's letters to Liouville debut the inverse-point construction:

-qR/a

R^2/a

; extracts appear in Liouville's Journal

1850

— Liouville proves 3D conformal maps are exhausted by similarities and inversions — Kelvin's trick is geometrically unique

1873

— Maxwell's Treatise systematizes spherical images, adding the center-charge extensions for charged and insulated spheres

2012

— Lekner's image-series analysis revives the sphere pair, proving like-charged conducting spheres generically attract at close range

The physical picture

Why should a curved conductor be solvable by a point fiction at all? The plane worked because reflection preserves distances; a sphere has a subtler symmetry — inversion. The map $r \to R^2/r$ turns the sphere inside out while fixing its surface, and Coulomb potentials transform under it with unreasonable grace (Liouville proved this is essentially the only angle-preserving map in 3D beyond the trivial ones — Kelvin's trick is not one option among many; it is the only trick there could have been). The physical residue of all that geometry is one proportion: slide along the sphere's surface and your distances to the charge (at $a$ ) and to the inverse point (at $R^2/a$ ) change together, locked in the fixed ratio $a/R$ . A charge at the inverse point, weighted by $-R/a$ , therefore cancels $q$ 's potential at every surface point simultaneously. One knob ( $q'$ ), one geometric identity, an entire curved boundary condition satisfied.

Figure

Kelvin's construction — the inverse point and the locked ratio

The picture's fine print explains the sphere's personality differences from the plane. The image is weaker than the charge ( $|q'| = qR/a < q$ ), so the sphere's screen holds only part of $q$ 's flux — the fraction $R/a$ , shrinking as the charge retreats — while the rest of the field lines sail past to infinity; a grounded sphere is a partial flux-catcher, not the total sink the infinite plane was. The image sits at the inverse point, which crowds toward the center as the charge retreats ( $b = R^2/a \to 0$ ) and rushes toward the surface as the charge approaches ( $a \to R \Rightarrow b \to R$ , $q' \to -q$ ): stand close enough and the sphere's horizon looks flat, the image looks like a mirror twin, and every plane formula must re-emerge — a limit the derivations check and the worked examples exploit. And the construction is modular. The Kelvin pair handles the boundary shape; whatever bookkeeping remains (the sphere's total charge, or its potential) can be fixed by one more image at the center, which adds a constant potential over the surface without disturbing the $V=0$ cancellation structure. Grounded, isolated, charged, held at $V_0$ — one geometry, one extra knob, four exact solutions. That modularity, plus the second uniqueness theorem to certify the isolated cases, makes the sphere the working physicist's favorite image problem.

Construction, map, force, energy

The claimed solution for the region outside the sphere ( $r \ge R$ ), charge on the $z$ -axis at distance $a > R$ :

V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{s} + \frac{q'}{s'}\right], \qquad q' = -\frac{R}{a}\,q, \qquad b = \frac{R^2}{a} \qquad (1)

where: $s = |\mathbf{r} - a\hat{\mathbf{z}}|$ — distance from the field point to the real charge; $s' = |\mathbf{r} - b\hat{\mathbf{z}}|$ — distance to the image, which sits inside the sphere (outside the physical region, as the method demands). Derivation A earns both parameters and verifies the boundary condition globally. The harvested induced-charge map, with $\theta$ the polar angle from the charge's direction:

\sigma(\theta) = -\frac{q\,(a^2 - R^2)}{4\pi R\,\left(a^2 + R^2 - 2aR\cos\theta\right)^{3/2}} \qquad (2)

where: the bracket is the squared distance from the surface point to the charge (law of cosines), raised to the $3/2$ — same anatomy as the plane's bell, wrapped around a sphere. Negative everywhere; densest at the near pole $\theta=0$ , sparsest at the far pole, with the memorable ratio $\sigma(0)/\sigma(\pi) = \left[(a+R)/(a-R)\right]^3$ . Its integral over the sphere is exactly $q' = -qR/a$ — the totals theorem again, and this time the total genuinely surprises: the grounded sphere does not soak up all of $q$ 's flux.

Figure

The wrapped bell — induced charge crowds the near pole cubically

Force and energy, from the image pair at separation $a - b = (a^2-R^2)/a$ :

F = \frac{1}{4\pi\varepsilon_0}\,\frac{q\,q'}{(a-b)^2} = -\frac{q^2}{4\pi\varepsilon_0}\,\frac{R\,a}{(a^2-R^2)^2}, \qquad W = -\frac{q^2\,R}{8\pi\varepsilon_0\,(a^2-R^2)} \qquad (3)

where: the minus signs mean attraction toward the sphere and negative binding energy; $W$ is the half-price value $\tfrac12qV_{\text{im}}$ , per the method article's energy theorem (checked numerically in worked example 1). Two limits calibrate both formulas: as $a \to R + d$ with $d \ll R$ , $F \to -q^2/16\pi\varepsilon_0d^2$ and $W \to -q^2/16\pi\varepsilon_0 d$ — the plane, recovered exactly; as $a \gg R$ , $F \to -q^2R/4\pi\varepsilon_0a^3$ — Coulomb attraction to the sphere's net image charge $-qR/a$ at distance $\approx a$ .

Derivation AFinding the image — two conditions, one geometric identity, whole-sphere verification

We derive $q'$ and $b$ rather than pull them from a hat. Strategy: assume one interior image on the axis (symmetry demands the axis; hope demands only one), spend the two unknowns $(q', b)$ on the boundary condition at the two axis points of the sphere, then — since two points prove nothing about a continuum — verify the result on the entire surface with the similar-triangles identity. The verification step is where the inverse point reveals why it, and only it, works.

Step 1. Impose $V = 0$ at the near pole (distance $a - R$ from the charge, $R - b$ from the image) and the far pole (distances $a + R$ and $R + b$ ):

\frac{q}{a-R} + \frac{q'}{R-b} = 0, \qquad \frac{q}{a+R} + \frac{q'}{R+b} = 0.

Step 2. Solve the pair. Dividing eliminates $q'$ : $(R-b)/(a-R) = (R+b)/(a+R)$ . Cross-multiplying, $(R-b)(a+R) = (R+b)(a-R)$ : $Ra + R^2 - ab - bR = Ra - R^2 + ab - bR$ , so $2R^2 = 2ab$ and

b = \frac{R^2}{a}, \qquad q' = -q\,\frac{R-b}{a-R} = -q\,\frac{R - R^2/a}{a-R} = -\frac{R}{a}\,q.

Step 3. Verify on the whole sphere. Take any surface point $P$ , at distance $s$ from the charge and $s'$ from the image. The triangles $(O, P, \text{image})$ and $(O, \text{charge}, P)$ share the angle at the center $O$ and have proportional flanking sides — $Ob/OP = (R^2/a)/R = R/a$ and $OP/O\text{charge} = R/a$ — so they are similar, and the third sides inherit the ratio:

\frac{s'}{s} = \frac{R}{a} \quad\text{at every point of the sphere.}

Therefore

V\big|_{\text{sphere}} = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{s} - \frac{qR/a}{(R/a)\,s}\right] = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{s} - \frac{q}{s}\right] = 0.

Why the inverse point is forced: a constant ratio of distances to two fixed points is exactly the classical definition of a sphere (the Apollonius sphere), and the pair (external point $a$ , internal point $R^2/a$ ) is the unique axis pair whose Apollonius sphere is ours. The image position is geometry's choice, not ours; only the weight $-R/a$ was physics.

Step 4. Certify: image inside the sphere = outside the physical region ✓; only source outside is $q$ ✓; $V\to0$ at infinity ✓; $V=0$ on the boundary (Step 3) ✓. By the first uniqueness theorem, (1) is the exterior potential. The same algebra runs inside-out: for a charge inside a grounded spherical shell at radius $a<R$ , the image sits outside at $R^2/a$ with the same $-qR/a$ weight — one formula, both problems, with the physical region always the one the fiction has vacated. The $a\to R$ limit hands back the plane's mirror ( $q'\to-q$ at depth $\to d$ ), and the two-point fit is confirmed by the all-point similar-triangles identity, closing the gap the two-point start left open.

Go deeper · Derivation B · 2 minThe harvest — sigma map, partial flux capture, force, energy, and the center-charge extensions

Derivation BThe harvest — sigma map, partial flux capture, force, energy, and the center-charge extensions

Now the physics. We differentiate the potential to get $\sigma(\theta)$ , integrate it to expose the sphere's partial flux capture, assemble force and energy with their limits, and prove the modular extension rule that upgrades the grounded solution to isolated, charged, and held spheres. Each output gets an independent check — the totals theorem, the plane limit, the half-price rule — because that redundancy is how image results are kept honest.

Step 1. The map. With the law-of-cosines distances $s^2 = r^2 + a^2 - 2ar\cos\theta$ and $s'^2 = r^2 + b^2 - 2br\cos\theta$ , differentiate (1) radially and evaluate at $r=R$ , using $b = R^2/a$ and the surface identity $s' = (R/a)s$ to consolidate the two terms over the common factor $s^{-3}$ :

\sigma(\theta) = -\varepsilon_0\frac{\partial V}{\partial r}\bigg|_{r=R} = -\frac{q\,(a^2-R^2)}{4\pi R\,(a^2+R^2-2aR\cos\theta)^{3/2}}.

Why the shape was predictable: at the near pole the bracket collapses to $(a-R)^2$ , so $\sigma(0) = -q(a+R)/4\pi R(a-R)^2$ — diverging as the charge closes in. Set $a = R+d$ and let $R\to\infty$ at fixed $d$ : $\sigma(0) \to -q\cdot2R/4\pi R d^2 = -q/2\pi d^2$ , exactly the plane's peak. The bell of the previous article is this formula's flat-earth limit.

Step 2. Total induced charge. Integrate over the sphere with ring elements $\mathrm{d}A = 2\pi R^2\sin\theta\,\mathrm{d}\theta$ and substitute $u = a^2+R^2-2aR\cos\theta$ , so $\sin\theta\,\mathrm{d}\theta = \mathrm{d}u/2aR$ and $u$ runs from $(a-R)^2$ to $(a+R)^2$ :

Q_{\text{ind}} = -\frac{q(a^2-R^2)}{4\pi R}\cdot 2\pi R^2 \cdot \frac{1}{2aR}\int_{(a-R)^2}^{(a+R)^2} u^{-3/2}\,\mathrm{d}u = -\frac{q(a^2-R^2)}{4a}\left[\frac{-2}{\sqrt{u}}\right]_{(a-R)^2}^{(a+R)^2}.

Evaluate the bracket: $-2\left[\dfrac{1}{a+R} - \dfrac{1}{a-R}\right] = \dfrac{2\cdot 2R}{a^2-R^2}$ , and the $(a^2-R^2)$ factors cancel:

Q_{\text{ind}} = -\frac{q(a^2-R^2)}{4a}\cdot\frac{4R}{a^2-R^2} = -\frac{qR}{a} = q'.

Why only a fraction: the totals theorem said this before the integral did — but the physics deserves the sentence. Field lines from $q$ that would have missed the sphere mostly still miss it; only part of $q$ 's flux ends on the metal — but the share is fixed by the grounding condition, not by how much of the sky the sphere covers. At $a = 2R$ the sphere's geometric shadow subtends just $6.7\%$ of the solid angle while it actually captures $R/a = 50\%$ : the induced charge bends field lines inward, so the sphere catches far more than it blocks. The plane caught everything because it shadowed half of space and infinity closed the other half.

Step 3. Force and energy. The induced charge's exterior field is the image's, so the force is the pair's Coulomb attraction across $a - b = (a^2-R^2)/a$ , giving (3); the energy follows from the half-price theorem $W = \tfrac12 qV_{\text{im}} = \tfrac12 q\,q'/4\pi\varepsilon_0(a-b)$ , giving (3)'s second formula. Check both limits: $a = R+d$ , $d\ll R$ : $(a^2-R^2) \approx 2Rd$ , so $F \approx -q^2R^2/4\pi\varepsilon_0(2Rd)^2 = -q^2/16\pi\varepsilon_0d^2$ ✓ and $W \approx -q^2R/8\pi\varepsilon_0\cdot2Rd = -q^2/16\pi\varepsilon_0d$ ✓ — the plane article's pair, recovered with their factor-of-two intact.

Step 4. The extension rule. Add a charge $q_c$ at the center. On the sphere it contributes the constant $q_c/4\pi\varepsilon_0R$ — constants don't disturb the equipotential character, so the surface remains an equipotential at $V_{\text{sphere}} = q_c/4\pi\varepsilon_0R$ , and the exterior solution is still exact (three point charges now). Choose $q_c$ to match the bookkeeping: grounded: $q_c=0$ . Held at $V_0$ : $q_c = 4\pi\varepsilon_0RV_0$ . Isolated with total charge $Q$ : the sphere's surface must carry $Q$ in total while the Kelvin image accounts for $-qR/a$ of it, so $q_c = Q + qR/a$ — certified by the second uniqueness theorem, since here it is the total charge, not the potential, that is specified. The isolated neutral sphere ( $Q=0$ , $q_c = +qR/a$ ) will star in worked example 2. The u-substitution used to integrate $\sigma(\theta)$ parallels the plane article's; both limits $a\to R$ and $a\gg R$ are recovered as shown above; and the extension rule for the isolated sphere is closed by the second uniqueness theorem.

Limiting cases

Case	Result	Meaning
Close approach, $a = R+d$ , $d\ll R$	$F \to -q^2/16\pi\varepsilon_0d^2$ , $q'\to-q$	the grounded plane, exactly recovered
Far retreat, grounded	$F \to -q^2R/4\pi\varepsilon_0a^3$	Coulomb pull of the net image $-qR/a$
Far retreat, isolated neutral	$F \to -2q^2R^3/4\pi\varepsilon_0a^5$	induced-dipole force; polarizability $\alpha=4\pi\varepsilon_0R^3$
Charge inside grounded shell ( $a<R$ )	image outside at $R^2/a$ , same weights	one construction, both regions
Flux bookkeeping	sphere captures fraction $R/a$ of $q$ 's flux	rest escapes to infinity — unlike the plane
Near/far pole densities	$\sigma(0)/\sigma(\pi) = \left[(a+R)/(a-R)\right]^3$	cubic crowding toward the charge

Worked examples

1 · The grounded sphere, fully harvested (confidence builder)

A charge $q = +8.0\,\mathrm{nC}$ sits $a = 6.0\,\mathrm{cm}$ from the center of a grounded conducting sphere of radius $R = 3.0\,\mathrm{cm}$ . Find the image, the pole densities, the total induced charge, the force, and the energy.

Step 1. Image parameters: $q' = -qR/a = -8.0\times(3/6) = -4.0\,\mathrm{nC}$ at $b = R^2/a = (0.030)^2/0.060 = 1.5\,\mathrm{cm}$ from the center. Image separation from the charge: $a - b = 4.5\,\mathrm{cm}$ .

Step 2. Pole densities from (2): near pole $\sigma(0) = -q(a+R)/4\pi R(a-R)^2 = -(8.0\times10^{-9})(0.090)/[4\pi(0.030)(0.030)^2] = -2.1\,\mu\mathrm{C/m^2}$ ; far pole $\sigma(\pi) = -q(a-R)/4\pi R(a+R)^2 = -0.079\,\mu\mathrm{C/m^2}$ . Ratio $27 = 3^3 = [(a+R)/(a-R)]^3$ ✓ — the screen crowds cubically toward the charge.

Step 3. Total induced charge: $-qR/a = -4.0\,\mathrm{nC}$ — only half of $q$ 's flux ends on the sphere ( $R/a = \tfrac12$ ); the other half runs to infinity.

Step 4. Force and energy, with $kqq' = (8.99\times10^{9})(8.0\times10^{-9})(-4.0\times10^{-9}) = -2.88\times10^{-7}\,\mathrm{N\cdot m^2}$ :

F = \frac{kqq'}{(a-b)^2} = \frac{-2.88\times10^{-7}}{(0.045)^2} = -1.4\times10^{-4}\,\mathrm{N}, \qquad W = \tfrac12\,q\,\frac{kq'}{a-b} = \tfrac12(8.0\times10^{-9})\frac{(8.99\times10^{9})(-4.0\times10^{-9})}{0.045} = -3.2\times10^{-6}\,\mathrm{J}.

Attraction of $0.14\,\mathrm{mN}$ , binding of $3.2\,\mu\mathrm{J}$ — and the energy is exactly half of $qV_{\text{im}} = 8.0\,\mathrm{nC}\times(-799\,\mathrm{V})$ , the half-price rule on live display.

2 · Cut the ground wire — the neutral sphere that still attracts (exam level)

Same geometry, but now the sphere is isolated and neutral. Find the new image set, the force on $q$ , and the sphere's potential; then extract the far-field law.

Step 1. Extension rule: keep the Kelvin pair ( $-4.0\,\mathrm{nC}$ at $1.5\,\mathrm{cm}$ ) and add $q_c = +qR/a = +4.0\,\mathrm{nC}$ at the center, restoring total sphere charge to zero. (License: second uniqueness theorem — total charge specified, distribution free.)

Step 2. Force — two image terms now, attraction to the Kelvin image minus repulsion from the center charge:

F = kq\left[\frac{q'}{(a-b)^2} + \frac{q_c}{a^2}\right] = (8.99\times10^{9})(8.0\times10^{-9})\left[\frac{-4.0\times10^{-9}}{(0.045)^2} + \frac{+4.0\times10^{-9}}{(0.060)^2}\right]

= 71.9\left[-1.98\times10^{-6} + 1.11\times10^{-6}\right] = -6.2\times10^{-5}\,\mathrm{N}.

Still attractive — $62\,\mu\mathrm{N}$ , about $44\%$ of the grounded pull — because the negative screen sits closer to $q$ than the compensating positive charge: induced polarization always wins at finite distance.

Step 3. Sphere potential: the Kelvin pair contributes zero on the surface; the center charge contributes $kq_c/R = (8.99\times10^{9})(4.0\times10^{-9})/0.030 = 1.2\,\mathrm{kV}$ . Equivalently $V_{\text{sphere}} = kq/a$ — the potential the bare charge would produce at the center, a slick identity worth keeping (the sphere floats at its center's unperturbed potential).

Step 4. Far field: expand the two-term force for $a\gg R$ — the leading monopoles cancel by construction, leaving

F \to -\frac{q^2}{4\pi\varepsilon_0}\cdot\frac{2R^3}{a^5},

the induced-dipole law with polarizability $\alpha = 4\pi\varepsilon_0R^3$ (a conducting sphere's textbook value). A neutral conductor attracts any charge, at $62\,\mu\mathrm{N}$ here and as $1/a^5$ afar — the electrostatics of why dust, toner, and charged insects stick to grounded and ungrounded metal alike.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §3.2.2–3.2.4 · Jackson, Classical Electrodynamics, 3rd ed., §2.2–2.5 · Purcell & Morin, Electricity and Magnetism, 3rd ed., §3.4 · Zangwill, Modern Electrodynamics, §8.3–8.4. The key self-consistency checks for this construction are: $V=0$ on $r=R$ follows from the distance-ratio identity $s'=(R/a)s$ ; the surface density $\sigma(\theta)$ is obtained from $-\varepsilon_0\partial_rV|_R$ and integrates to $-qR/a$ ; the force $F=-q^2Ra/4\pi\varepsilon_0(a^2-R^2)^2$ and energy $W=-q^2R/8\pi\varepsilon_0(a^2-R^2)$ reproduce the planar forms in the limit $a\to R$ , and yield the polarizability $\alpha=4\pi\varepsilon_0R^3$ in the $a\gg R$ limit.

Builds on: Method of Images · Image Charge for a Grounded Plane · Uniqueness Theorems for the Potential · Boundary Conditions for the Potential

Leads to: Separation of Variables in Spherical Coordinates · Legendre Polynomials · Laplace's Equation (E2)

Where does the "missing" charge go when the sphere only draws up −qR/a?

Down the ground wire — or rather, only $qR/a$ worth of electrons ever came up it. The sphere needs to terminate only the field lines that actually strike it, and from distance $a$ the sphere subtends a limited view of the charge's flux. Follow the lines: a fraction $R/a$ of them curve onto the metal; the rest bend past and head to infinity, where the "other plate" of the system lives. The plane caught everything only because nothing could get past it.

How would I ever guess the inverse point without knowing the answer?

Two honest routes. Historically: Kelvin knew inversion geometry — the map $r\to R^2/r$ fixes the sphere and transforms harmonic functions into harmonic functions (times a known factor), so a point charge maps to a point charge at the inverse point; the image is inversion applied to physics. Pedestrianly: assume one axial image with unknown $(q', b)$ , impose $V=0$ at the two axis poles (two equations, two unknowns — Derivation A), and let the similar-triangles identity upgrade the two-point fit to the whole sphere. The second route needs no genius, which is rather the point of the method: cheap guesses, expensive-looking certified results.

The image solution for a charge inside a grounded shell puts the image outside. Is the interior force also attractive toward the wall?

Yes — and it grows without bound as the charge approaches the wall, by the same close-approach logic (locally the wall is a plane). A charge at the exact center feels zero force (the image flies to infinity with vanishing weight... more precisely $b=R^2/a\to\infty$ while the force $\to 0$ by symmetry), so the center is an equilibrium — but an unstable one, consistent with Earnshaw's theorem from the E1 chapter: displace the charge slightly and the wall's attraction wins. Grounded cavities do not cradle charges; they swallow them.

Why does the isolated neutral sphere attract the charge instead of ignoring it?

Because "neutral" is a statement about the total, and forces are computed from the arrangement. The charge polarizes the sphere: negative induced charge crowds the near face (the Kelvin image, at distance $a-b$ ), positive charge is exiled to spread around the surface (the center image, at distance $a$ ). Attraction acts across the shorter lever. The effect weakens fast — $1/a^5$ versus the grounded sphere's $1/a^3$ — but it never cancels, and it is why electrostatic adhesion needs no grounding: any conductor, floating or not, pulls on any charge.

Can a like-charged sphere and point charge attract?

Yes, at close range — the sphere-version of a famous surprise. Give the sphere charge $Q>0$ and bring up $q>0$ : the force is $kq[q'/(a-b)^2 + (Q+qR/a)/a^2]$ , and as $a\to R$ the Kelvin term diverges like $-1/d^2$ while the center term stays finite — attraction always wins sufficiently close. The crossover radius (where repulsion hands over to attraction) is where induced polarization beats bare Coulomb; for sphere pairs the same analysis (Lekner 2012) shows like-charged conductors generically attract before contact, a result with real consequences for colloids and electrostatic self-assembly.

Does the a → R limit really contain the whole plane article?

Quantitatively, yes — that's the calibration that makes both articles trustworthy. Set $a = R + d$ and let $R\to\infty$ at fixed $d$ : the image charge $-qR/a \to -q$ ; its depth below the surface $R - b = R(a-R)/a \to d$ ; the force $\to -q^2/16\pi\varepsilon_0d^2$ ; the energy $\to -q^2/16\pi\varepsilon_0d$ ; even $\sigma(\theta)$ flattens into the plane's bell $-qd/2\pi s^3$ -tailed profile. A sphere seen from up close is a plane, and the formulas know it. Running limits like this is the cheapest full-strength check the images toolkit offers — use it on every new geometry.

Is there an image solution for a charge near a conducting sphere held at potential V₀ by a battery?

Yes — the modular extension: Kelvin pair plus a center charge $q_c = 4\pi\varepsilon_0RV_0$ , making the surface potential $V_0$ exactly (the pair contributes zero there, the center charge a constant). The battery's job in the physical world — supplying whatever charge keeps the potential pinned as $q$ moves around — is played mathematically by letting $q_c$ stay fixed while the Kelvin image dances. Note what is not free: with the potential pinned, the sphere's total charge $q_c - qR/a$ varies with $a$ ; with charge pinned (isolated case), the potential varies. One knob, spent one way or the other — never both, per the boundary-conditions article.