Separation of Variables in Spherical Coordinates

Drop an uncharged metal sphere into a uniform electric field and watch the field react: lines that marched in parallel formation bend, funnel into the sphere's poles, and meet the metal at right angles, while charge silently redistributes. Ask for the exact potential everywhere, the induced charge at every latitude, the dipole moment of the disturbance — and spherical separation of variables hands you all three in half a page. That calculation is where this article is headed. You arrive equipped. Laplace's Equation established that the potential in charge-free space is harmonic and — the crucial license — uniquely pinned by its boundary values. Separation of Variables in Cartesian Coordinates built the method: factor, split by the separation-constant argument, superpose, fit coefficients by orthogonality. What changes on a sphere is the cast: the radial equation becomes an Euler equation with power-law solutions $r^\ell$ and $r^{-(\ell+1)}$ — one regular at the origin, one at infinity — and the angular equation delivers the Legendre polynomials $P_\ell(\cos\theta)$ , an orthogonal alphabet for latitude. The machinery buys more than sphere problems: the exterior series built here is the multipole expansion in embryo, and the same projection integrals decompose Earth's magnetic field and localize the electrical sources of a thought inside a human head. Wherever nature hands you data on a sphere, this is the method that reads it.

Key takeaways

In spherical coordinates Laplace's equation separates under $V=R(r)\,\Theta(\theta)$ (azimuthal symmetry assumed); the separation constant is written $\ell(\ell+1)$ .
The radial ODE is an Euler equation: trying $R=r^s$ gives $s=\ell$ or $s=-(\ell+1)$ — pure power laws, because Laplace's equation has no built-in length scale; the angular ODE is the Legendre equation, solved on the sphere by $P_\ell(\cos\theta)$ .
Regularity does the triage — origin-containing regions keep only $r^\ell$ , infinity-reaching regions only $r^{-(\ell+1)}$ , shells both — and data become coefficients via orthogonality, $\int_{-1}^{1}P_\ell P_{\ell'}\,\mathrm{d}x = \tfrac{2}{2\ell+1}\delta_{\ell\ell'}$ .
Flagship: a grounded sphere in a uniform field has $V=-E_0(r-R^3/r^2)\cos\theta$ , $\sigma=3\varepsilon_0E_0\cos\theta$ , induced dipole $p=4\pi\varepsilon_0R^3E_0$ .
Key idea: the radial Euler equation gives $r^\ell$ and $r^{-(\ell+1)}$ , the angular equation gives $P_\ell(\cos\theta)$ , and regularity + orthogonality + uniqueness turns any spherical boundary condition into an exact answer.

The core insight

Posit $V=R(r)\Theta(\theta)\Phi(\phi)$ ; the radial equation gives power laws $r^\ell$ and $r^{-(\ell+1)}$ , the polar equation gives Legendre functions $P_\ell(\cos\theta)$ (or $P_\ell^m$ with azimuthal dependence), and orthogonality + uniqueness turns boundary data into expansion coefficients.

Where this lives today

Every EEG cap and MEG helmet runs this article's mathematics in its inverse gear. The forward model treats the head as concentric spheres — brain, skull, scalp, each with its own conductivity — around a neural current source; the potential in each shell is exactly a series $\sum(A_\ell r^\ell+B_\ell r^{-(\ell+1)})P_\ell(\cos\theta)$ , matched shell-to-shell by the boundary conditions. Epilepsy surgeons localizing a seizure focus fit measured scalp potentials to these expansions, refined by finite-element models whose validation baseline is still the analytic multi-shell sphere. The same projection integrals, run on satellite magnetometer data, produce the International Geomagnetic Reference Field — Gauss's 1839 analysis, continuously updated, steering smartphone compasses and directional drilling alike. And the flagship never retired: the factor-of-3 enhancement is a design floor in high-voltage engineering, and the polarizability $4\pi\varepsilon_0R^3$ opens every model of dielectrophoresis — steering biological cells with field gradients on a chip.

The physical picture

A sphere offers two kinds of direction: radially in-or-out, and angularly across the surface. Laplace's equation is a zero-sum curvature budget — whatever bending the potential does angularly must be repaid radially. Separation exploits the bookkeeping: choose a single angular shape, and ask what radial profile keeps the total curvature zero. The remarkable answer for Laplace's equation, derived below: always a pure power of $r$ , each angular shape owning a matched radial pair — $r^\ell$ swelling outward, $r^{-(\ell+1)}$ collapsing inward. The index $\ell$ is an angular wiggle counter: $\ell=0$ is featureless (a point charge's angular shape), $\ell=1$ one north–south lobe pair $\propto\cos\theta$ (a dipole), $\ell=2$ two nodal circles (a quadrupole), and so on up the ladder. The pairing of wiggle number with radial power is the headline: the finer the angular pattern, the faster $r^{-(\ell+1)}$ buries it with distance — why every charge distribution, seen from far enough away, looks first like a point charge, then like a dipole, and only reluctantly reveals more.

Figure

The anatomy of a spherical boundary-value problem

The figure encodes the strategy. Regularity is the bouncer at each door: inside, the collapsing $r^{-(\ell+1)}$ are turned away (nothing diverges at an empty origin); outside, the swelling $r^\ell$ (a localized arrangement's potential must fade at infinity). What remains is one free coefficient per $\ell$ per region, and the boundary data is exactly enough to fix them all. That the answer so built is the answer is the first uniqueness theorem — proved in Uniqueness Theorems for the Potential, invoked every time we stop calculating.

From the PDE to a solution catalogue

The method's engine room — why a product ansatz splits a PDE, why "a function of $r$ equals a function of $\theta$ " forces both constant, why superposition is legitimate, how orthogonality isolates coefficients — was built once and for all in Separation of Variables in Cartesian Coordinates. We redeploy it on new terrain: Laplace's equation in spherical coordinates, which under azimuthal symmetry (nothing depends on the longitude $\phi$ — true of everything in this article) reads

\frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial V}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\,\frac{\partial V}{\partial\theta}\right)=0 \qquad (1)

where: $V(r,\theta)$ — the potential in the charge-free region; $r$ — distance from the origin; $\theta$ — the polar angle from the $z$ -axis. The two terms are the radial and angular curvature budgets; equation (1) demands they cancel point by point. (The full Laplacian's third term, in $\partial^2V/\partial\phi^2$ , leads to the spherical harmonics — the next two articles.)

Derivation ASeparating the spherical Laplacian and solving the radial Euler equation

The goal is the complete set of building-block solutions of equation (1): factor $V=R(r)\,\Theta(\theta)$ , split the PDE into a radial and an angular ODE sharing a separation constant, then actually solve the radial one by the trial $R=r^s$ — an Euler equation, which the trial converts into algebra. The angular equation we identify and state but do not solve: its solutions are the Legendre polynomials, catalogued in their own article. Watch the naming of the separation constant as $\ell(\ell+1)$ — a choice the radial algebra vindicates instantly.

Step 1. Insert the product $V=R(r)\Theta(\theta)$ into (1), multiply through by $r^2$ , and divide by $R\Theta$ :

\frac{1}{R}\frac{\mathrm{d}}{\mathrm{d}r}\!\left(r^2\frac{\mathrm{d}R}{\mathrm{d}r}\right)+\frac{1}{\Theta\sin\theta}\frac{\mathrm{d}}{\mathrm{d}\theta}\!\left(\sin\theta\,\frac{\mathrm{d}\Theta}{\mathrm{d}\theta}\right)=0

The first bracket depends only on $r$ , the second only on $\theta$ , and they sum to zero everywhere. By the separation-constant argument (proved in the Cartesian article), each bracket is constant, the two constants opposite. Call the radial one $\ell(\ell+1)$ . Why that peculiar name: any real number can be written as $\ell(\ell+1)$ , so no generality is lost — and the radial equation is about to vindicate the choice.

Step 2. The radial equation is

\frac{\mathrm{d}}{\mathrm{d}r}\!\left(r^2\frac{\mathrm{d}R}{\mathrm{d}r}\right)=\ell(\ell+1)\,R

Every derivative arrives chaperoned by a factor of $r$ . Such equidimensional (Euler) equations are scale-free — replace $r$ by $2r$ and nothing changes — so they cannot support solutions with a characteristic length, like $e^{-r/\lambda}$ or $\sin(kr)$ . The only scale-free functions are power laws, so try $R=r^s$ :

\frac{\mathrm{d}}{\mathrm{d}r}\!\left(r^2\cdot s\,r^{s-1}\right)=s(s+1)\,r^s

Step 3. Matching against $\ell(\ell+1)\,r^s$ and cancelling $r^s$ leaves algebra:

s(s+1)=\ell(\ell+1) \quad\Longrightarrow\quad s=\ell \;\;\text{or}\;\; s=-(\ell+1)

The quadratic factors as $(s-\ell)(s+\ell+1)=0$ — two roots, two independent solutions, exactly what a second-order ODE owes us: $R(r)=A\,r^\ell+B\,r^{-(\ell+1)}$ . For $\ell=0$ the pair is a constant and $1/r$ — the point-charge potential and its gauge freedom, the bottom rung of the ladder.

Step 4. The angular equation, with the opposite constant, is

\frac{1}{\sin\theta}\frac{\mathrm{d}}{\mathrm{d}\theta}\!\left(\sin\theta\,\frac{\mathrm{d}\Theta}{\mathrm{d}\theta}\right)+\ell(\ell+1)\,\Theta=0

Substituting $x=\cos\theta$ turns this into Legendre's differential equation on $-1\le x\le1$ . We state the facts and defer every proof to Legendre Polynomials: for each non-negative integer $\ell$ there is exactly one solution finite at both poles — the Legendre polynomial $P_\ell(x)$ — while for non-integer $\ell$ , and for the second solution $Q_\ell$ at any $\ell$ , the solutions diverge at the poles and are disqualified. Boundedness on the closed sphere quantizes $\ell$ .

Superposition assembles the master formula — each $\ell$ contributes its radial pair times its angular shape:

V(r,\theta)=\sum_{\ell=0}^{\infty}\left(A_\ell\,r^\ell+\frac{B_\ell}{r^{\ell+1}}\right)P_\ell(\cos\theta) \qquad (2)

where: $A_\ell, B_\ell$ — constants fixed, per region, by the boundary conditions; $P_\ell(\cos\theta)$ — the Legendre polynomials. Equation (2) is the general azimuthally symmetric solution: completeness of the $P_\ell$ (banked in Legendre Polynomials) means any reasonable boundary function can be represented; uniqueness means that once it is, we are done. The first few members of the angular alphabet: $P_0=1$ (uniform — monopole shape), $P_1=x$ (one lobe pair — dipole), $P_2=\tfrac{1}{2}(3x^2-1)$ (quadrupole), $P_3=\tfrac{1}{2}(5x^3-3x)$ (octupole), with $x=\cos\theta$ . The tool that unlocks coefficients is orthogonality — Fourier's trick with sines swapped for Legendre polynomials, quoted in the normalization proved in Legendre Polynomials:

\int_{-1}^{1}P_\ell(x)\,P_{\ell'}(x)\,\mathrm{d}x=\frac{2}{2\ell+1}\,\delta_{\ell\ell'} \qquad (3)

where: $\delta_{\ell\ell'}$ — the Kronecker delta. Multiply any Legendre series by $P_{\ell'}$ and integrate: every term dies except the one you wanted, which surrenders its coefficient wearing the factor $2/(2\ell'+1)$ . Often you will not need the integral — data that is already a short polynomial in $\cos\theta$ matches term by term, since the $P_\ell$ are linearly independent. The integral is the sledgehammer; inspection is the scalpel.

The flagship problem: a grounded sphere in a uniform field

Now the payoff promised in the overview. A grounded conducting sphere of radius $R$ sits in a field that, before it arrived, was uniform: $\mathbf{E}_0=E_0\hat{z}$ . We want the exact potential outside, the induced charge, and the strength of the disturbance. This is Griffiths's Example 3.8 — the problem that shows the method thinking, as the boundary conditions execute almost the entire series (2), leaving two terms.

Figure

The payoff — field lines around a grounded conducting sphere in a uniform field

Go deeper · Derivation B · 2 minThe grounded conducting sphere in a uniform field, end to end

Derivation BThe grounded conducting sphere in a uniform field, end to end

The goal is the exterior potential $V(r,\theta)$ for a grounded sphere of radius $R$ in an asymptotically uniform field $E_0\hat z$ , plus the induced charge $\sigma(\theta)$ and dipole moment $p$ . The strategy is the two-boundary squeeze: write the general exterior series (2), impose the behavior at infinity to fix the $A_\ell$ , then impose $V=0$ on the sphere to fix the $B_\ell$ . The subtlety is at infinity — the potential does not vanish there, because a uniform field fills all space; the correct condition is that $V$ approaches the uniform-field potential, which grows with $r$ . That growth lets one $A_\ell$ survive.

Step 1. Establish the far-field condition. A uniform field $E_0\hat z$ has potential $V=-E_0z+\text{const}=-E_0r\cos\theta$ (constant chosen so the equatorial plane, at the sphere's potential by symmetry, sits at $V=0$ ). Since the sphere's influence must fade far away,

V(r,\theta)\;\to\;-E_0\,r\cos\theta \qquad (r\gg R)

Why the sphere cannot change the far field: its induced charges are a localized, neutral arrangement whose potential dies off — as $1/r^2$ , we are about to prove.

Step 2. Spend the far-field condition on the general exterior series (2) — both branches admissible a priori, the region excluding the origin. As $r\to\infty$ every $B_\ell$ term dies, leaving $\sum A_\ell r^\ell P_\ell(\cos\theta)$ , which must equal $-E_0\,r\,P_1(\cos\theta)$ . Linear independence matches term by term: $A_1=-E_0$ , every other $A_\ell=0$ .

Step 3. Spend the surface condition: grounded means $V(R,\theta)=0$ for every $\theta$ :

0=-E_0R\,P_1(\cos\theta)+\sum_{\ell=0}^{\infty}\frac{B_\ell}{R^{\ell+1}}\,P_\ell(\cos\theta)

Linear independence (or: multiply by $P_{\ell'}$ and integrate with (3)) forces each angular coefficient to vanish separately: the $\ell=1$ slot gives $B_1=E_0R^3$ ; every other $B_\ell=0$ . The series has collapsed to two terms.

Step 4. Assemble the exterior potential:

V(r,\theta)=-E_0\!\left(r-\frac{R^3}{r^2}\right)\cos\theta \qquad (r\ge R)

The first term is the undisturbed uniform field; the second is exactly a point dipole at the center. Comparing with $p\cos\theta/(4\pi\varepsilon_0r^2)$ identifies $p=4\pi\varepsilon_0R^3E_0$ , along the applied field.

Step 5. Extract the surface charge: the conductor boundary condition (Boundary Conditions for the Potential) gives $\sigma=-\varepsilon_0\,\partial V/\partial r$ at $r=R$ :

\frac{\partial V}{\partial r}=-E_0\!\left(1+\frac{2R^3}{r^3}\right)\cos\theta \;\;\xrightarrow{\,r=R\,}\;\; -3E_0\cos\theta \qquad\Longrightarrow\qquad \sigma(\theta)=3\varepsilon_0E_0\cos\theta

Positive on the hemisphere centered on the $+\hat{z}$ direction ( $\theta<\pi/2$ ), negative on the opposite hemisphere, zero on the equator — and $\cos\theta$ integrates to zero over the sphere, so the grounded sphere draws no net charge from the earth. The surface field at the poles is $\sigma_{\max}/\varepsilon_0=3E_0$ : the applied field, concentrated threefold. Quick checks: $V(R,\theta)=0$ identically; evaluating $-\varepsilon_0\,\partial V/\partial r$ at $r=R$ returns $\sigma=3\varepsilon_0E_0\cos\theta$ exactly; and the $B_1/r^2$ exterior term at $r=0.3$ m agrees exactly with the dipole formula $p=4\pi\varepsilon_0R^3E_0$ .

Two results deserve display. The induced charge,

\sigma(\theta)=3\,\varepsilon_0E_0\cos\theta \qquad (4)

where: $\theta$ — measured from the field direction; positive where field lines exit, negative where they enter, zero net. And the induced dipole moment,

p=4\pi\varepsilon_0R^3E_0 \qquad (5)

where: $\alpha=4\pi\varepsilon_0R^3$ is the polarizability of a conducting sphere — proportional to its volume, a scaling that survives into the molecular physics of dielectrics. Both responses inherit the $P_1$ shape of the drive: a uniform field induces a pure dipole and nothing else.

Reading data off a boundary: Griffiths Example 3.6

The flagship was solved by inspection because its data was already a single $P_\ell$ . The general situation — an arbitrary $V_0(\theta)$ painted on a shell — needs the orthogonality machinery, and Griffiths's Example 3.6 is the canonical drill.

Go deeper · Derivation C · 1 minExtracting coefficients by orthogonality for a sphere at a prescribed surface potential

Derivation CExtracting coefficients by orthogonality for a sphere at a prescribed surface potential

The goal is the exact interior and exterior potential of a shell held at $V_0(\theta)=k\sin^2(\theta/2)$ , charge-free space on both sides. The strategy is the standard recipe in full: regularity selects the surviving radial branch in each region, the boundary condition at $r=R$ becomes a Legendre expansion of $V_0$ , and orthogonality — or, faster, a trigonometric identity — reads off the coefficients. Watch the half-angle disguise dissolve into a two-term series.

Step 1. Regularity sets up both regions. Inside: $V_{\text{in}}=\sum_\ell A_\ell\,r^\ell P_\ell(\cos\theta)$ . Outside: $V_{\text{out}}=\sum_\ell B_\ell\,r^{-(\ell+1)}P_\ell(\cos\theta)$ . Both equal $V_0(\theta)$ at $r=R$ .

Step 2. Undress the data with the half-angle identity $\sin^2(\theta/2)=\tfrac{1}{2}(1-\cos\theta)$ :

V_0(\theta)=\frac{k}{2}\left(1-\cos\theta\right)=\frac{k}{2}\,P_0(\cos\theta)-\frac{k}{2}\,P_1(\cos\theta)

Why this is the whole game: the data is a finite Legendre series, so only $\ell=0$ and $\ell=1$ can appear. Transcendental data would need Step 3's integral for every $\ell$ .

Step 3. Extract the interior coefficients — the sledgehammer once, to show it works. Multiply $\sum_\ell A_\ell R^\ell P_\ell = V_0$ by $P_{\ell'}(\cos\theta)\sin\theta$ , integrate over $x=\cos\theta$ , and apply (3):

A_\ell=\frac{2\ell+1}{2R^\ell}\int_{-1}^{1}V_0(x)\,P_\ell(x)\,\mathrm{d}x

Running it with $V_0=\tfrac{k}{2}(1-x)$ gives $A_0=\tfrac{k}{2}$ and $A_1=-\tfrac{k}{2R}$ , every $\ell\ge2$ integral vanishing by orthogonality — the same numbers inspection of Step 2 delivers instantly.

Step 4. The interior solution:

V_{\text{in}}(r,\theta)=\frac{k}{2}\left(1-\frac{r}{R}\cos\theta\right) \qquad (r\le R)

At the center, $V=k/2$ — the boundary average, exactly as the mean-value property of harmonic functions (Laplace's Equation) demands: a free consistency check, passed.

Step 5. Matching at $r=R$ gives $B_\ell=A_\ell R^{2\ell+1}$ — here $B_0=\tfrac{k}{2}R$ , $B_1=-\tfrac{k}{2}R^2$ :

V_{\text{out}}(r,\theta)=\frac{k}{2}\left(\frac{R}{r}-\frac{R^2}{r^2}\cos\theta\right) \qquad (r\ge R)

Far away the $\ell=0$ term dominates: a point charge $Q_{\text{eff}}=2\pi\varepsilon_0kR$ , with a dipole correction dying one power faster — the multipole hierarchy, exhibited by an exact solution. Numerical check: for $k=100$ V, $R=0.10$ m, direct quadrature of the Step 3 projection integrals gives $A_0=50.0$ V, $A_1=-500.0$ V/m, and $A_2,A_3\approx10^{-13}$ (consistent with exact truncation at $\ell=1$ ); both the interior and exterior closed forms reproduce $V(R,60°)=25.0$ V.

Limiting cases

Every limit lands on physics we already trust:

case	result	meaning
flagship, $r\to\infty$	$V\to-E_0\,r\cos\theta$	the dipole whisper dies as $1/r^2$
flagship, $r=R$	$V=0$ for all $\theta$	one equipotential, as a grounded conductor must be
flagship, equator $\theta=\pi/2$	$\sigma=0$	the sign-change seam; field is tangent to the surface
shell problem, $r=0$	$V=k/2$	center = boundary average: the mean-value property, honored exactly
shell problem, $r\gg R$	$V\approx \tfrac{kR}{2r}$	monopole behavior: from afar, a point charge $2\pi\varepsilon_0kR$
only $\ell=0$ data ( $V_0=\text{const}$ )	$V_{\text{in}}=V_0$ , $V_{\text{out}}=V_0R/r$	the familiar spherical shell held at constant potential, the monopole at the bottom rung of the ladder

Common trapKeep both radial terms everywhere — surely the series is more general that way

The most natural-feeling mistake in the subject, and it is fatal: carrying both branches into every region leaves twice as many unknowns as the data can determine. The physics is the referee: $r^{-(\ell+1)}$ diverges at $r=0$ , so in a region containing the origin its coefficient must be exactly zero (no charge sits there to justify a singularity); $r^\ell$ diverges as $r\to\infty$ , so in a region reaching infinity it must be discarded — unless an applied field genuinely fills all space, as in the flagship problem, in which case the surviving growth is dictated, coefficient and all, by the far-field condition. Only in a shell between two spheres do both branches serve, the two boundary surfaces supplying the two conditions per $\ell$ that fix them. Region first, then series — never the reverse.

Worked examples

1 · A sphere in a laboratory field (confidence builder)

A grounded metal sphere, $R=5.0$ cm, sits in a uniform field $E_0=1.0\times10^5$ V/m. Find the maximum induced charge density, the surface field at the poles, the charge on the downstream hemisphere, and the induced dipole moment.

Step 1. Maximum charge density sits at the poles, where $\cos\theta=\pm1$ in equation (4): $\sigma_{\max}=3\varepsilon_0E_0=3\times(8.854\times10^{-12})\times(1.0\times10^5)=\boxed{2.66\ \mathrm{\mu C/m^2}}$ .

Step 2. The field just outside a conductor is $\sigma/\varepsilon_0$ , so at the poles $E_{\text{pole}}=3E_0=\boxed{3.0\times10^5\ \mathrm{V/m}}$ — the factor of 3 is geometry alone, independent of $R$ and $E_0$ ; the local field, not the applied one, triggers corona discharge.

Step 3. Integrating $\sigma$ over the downstream hemisphere: $q=3\pi\varepsilon_0E_0R^2=\boxed{+20.9\ \mathrm{nC}}$ , with $-20.9$ nC upstream — zero net, as the field-line picture promised.

Step 4. Induced dipole moment from equation (5): $p=4\pi\varepsilon_0R^3E_0=4\pi\times(8.854\times10^{-12})\times(1.25\times10^{-4})\times(1.0\times10^5)=\boxed{1.39\times10^{-9}\ \mathrm{C\cdot m}}$ . Sanity check: at $r=15$ cm, $\theta=60°$ , the exterior potential gives $-7.22$ kV versus $-7.50$ kV undisturbed — a 3.7% correction at three radii, dying as $1/r^2$ .

2 · The half-angle shell (exam level)

A hollow sphere of radius $R=10$ cm is held at $V_0(\theta)=k\sin^2(\theta/2)$ with $k=100$ V. Find the potential at the center, at $(r,\theta)=(5.0\ \mathrm{cm},\,60°)$ , at $(20\ \mathrm{cm},\,60°)$ , and the effective point charge seen from far away.

Step 1. From the derivation, $A_0=k/2=50$ V and $A_1=-k/2R=-500$ V/m (all others zero): $V_{\text{in}}=50\,(1-\tfrac{r}{0.1}\cos\theta)$ and $V_{\text{out}}=50\,(\tfrac{0.1}{r}-\tfrac{0.01}{r^2}\cos\theta)$ in volts.

Step 2. Center: $r=0$ kills the $\ell=1$ term, $V(0)=\boxed{50\ \mathrm{V}}$ — the boundary average, since $\sin^2(\theta/2)$ averages to $1/2$ .

Step 3. Interior point: $V=50\times(1-0.5\times0.5)=\boxed{37.5\ \mathrm{V}}$ . Surface check: at $r=R$ , $\theta=60°$ the formula gives $25.0$ V, and the data says $100\times\sin^2 30°=25.0$ V ✓.

Step 4. Exterior point: $V=50\times(0.500-0.125)=\boxed{18.75\ \mathrm{V}}$ — the monopole piece contributes 25 V, the dipole correction $-6.25$ V: a 4:1 dominance at two radii.

Step 5. Far away only the monopole survives, $V\approx\tfrac{kR}{2r}=\tfrac{5}{r}$ V·m: a point charge $Q_{\text{eff}}=4\pi\varepsilon_0\times5=\boxed{0.556\ \mathrm{nC}}$ . Cross-check at $r=1$ m on the equator (no dipole term there): exact $V_{\text{out}}=5.00$ V, and $Q_{\text{eff}}/4\pi\varepsilon_0r=5.00$ V ✓.

Origins · 1 minPierre-Simon Laplace and how this was discovered

Pierre-Simon Laplace

1749–1827 · Normandy to the Académie — potential theory born of planets

The machinery of this article was forged not for electricity but for the shape of the Earth. In the early 1780s the Académie's open problem was the attraction of spheroids — how does a flattened planet pull on a moon or a plumb bob? — and two men raced through it. Legendre struck first (1782), inventing the polynomials $P_\ell$ to expand the inverse distance; Laplace answered with memoirs that swallowed the tool whole, generalized the angular functions to the full sphere, and in 1787 wrote $\nabla^2V=0$ in spherical coordinates while analyzing Saturn's rings. Legendre complained, with justice, that Laplace cited him grudgingly or not at all — the polynomials nearly went down in history unnamed. The apparatus went into the Mécanique Céleste, computing the figure of the Earth and the tides; when Poisson took up electricity in 1812, the hard mathematics had been done for gravity a generation earlier.

History

1782

— Legendre's memoir on the attraction of spheroids introduces the polynomials that bear his name.

1785

— Laplace's spheroid memoirs generalize the expansion, seeding what became the spherical harmonics.

1787

— Analyzing Saturn's rings, Laplace writes the potential equation in spherical coordinates — equation (1) here.

1839

— Gauss projects Earth's magnetic field onto spherical harmonics: the coefficient recipe at planetary scale.

1929

— Berger records the first human EEG; source localization turns the head into a spherical boundary-value problem.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §3.3.2 · Jackson, Classical Electrodynamics, 3rd ed., §3.1–3.5 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 3 · Zangwill, Modern Electrodynamics, §7.6–7.8 All equations are standard results of separation of variables for Laplace's equation in spherical coordinates. Key landmarks to keep in view: the Euler-equation roots $s=\ell$ and $s=-(\ell+1)$ ; Legendre orthogonality $\int_{-1}^1 P_\ell P_{\ell'}\mathrm{d}x=2\delta_{\ell\ell'}/(2\ell+1)$ ; and the conducting-sphere results — exterior potential $-E_0(r-R^3/r^2)\cos\theta$ , induced charge $\sigma=3\varepsilon_0E_0\cos\theta$ , and dipole moment $4\pi\varepsilon_0E_0R^3$ .

Builds on: Laplace's Equation · Boundary Conditions for the Potential · Uniqueness Theorems for the Potential · Separation of Variables in Cartesian Coordinates · Legendre Polynomials Leads to: Associated Legendre Functions · Image Charges for a Conducting Sphere · Separation of Variables in Cylindrical Coordinates

Why is the separation constant written as ell times ell plus one instead of just a constant?

Pure foresight, costing nothing: any real number can be written $\ell(\ell+1)$ , so it is a relabeling, not an assumption. The payoff comes twice — the radial Euler roots become simply $s=\ell$ and $s=-(\ell+1)$ , and the angular equation turns out to have pole-regular solutions only for non-negative integer $\ell$ , where the label coincides with the Legendre polynomial's degree.

Where did the second angular solution go? A second-order ODE should have two.

It exists — the Legendre function of the second kind, $Q_\ell(x)$ — and is discarded for a physical reason: it diverges along the entire polar axis, and no charge-free region containing that axis can host an infinite potential. In problems that exclude the axis (between two coaxial cones, say) $Q_\ell$ returns to service, exactly as $r^{-(\ell+1)}$ returns when the origin is excluded. Which solutions survive is a question about the region, never the equation.

The grounded sphere acquires sigma equals 3 epsilon-0 E-0 cos theta — did charge appear from nowhere?

No. Locally the charge was already in the metal: conduction electrons drift upstream, leaving exposed positive lattice downstream, so the hemispheres carry exactly $\pm3\pi\varepsilon_0E_0R^2$ . Globally, $\cos\theta$ integrates to zero, so the ground wire delivers nothing in a purely uniform field. It still matters: it pins $V=0$ , and in a non-uniform field (or near a point charge, as in the image problems) the sphere genuinely draws net charge through it.

Do I always have to compute the projection integrals, or can I match coefficients by eye?

Match by eye whenever the data is a polynomial in $\cos\theta$ — most textbook problems by design. Rewrite it in $P_0,P_1,P_2,\dots$ (e.g. $\cos^2\theta=\tfrac{1}{3}P_0+\tfrac{2}{3}P_2$ ) and equate coefficients by linear independence. The projection integral is the fallback for non-polynomial data — hemispheres held at $\pm V_0$ genuinely need it, and get an infinite series. The two routes are the same mathematics: inspection is orthogonality in disguise.

What changes if my region is the shell between two concentric spheres?

Both radial branches survive, and that is a feature: the shell contains neither the origin (so $r^{-(\ell+1)}$ stays finite) nor infinity (so $r^\ell$ stays bounded). Two unknowns per $\ell$ — but two boundary spheres each supply one condition per $\ell$ , so the count closes: a $2\times2$ system per $\ell$ . The spherical capacitor of Laplace's Equation is the $\ell=0$ case of this setup.

How do I handle boundary data that depends on the longitude phi?

Restore the third factor $\Phi(\phi)$ in the ansatz. Single-valuedness quantizes it to $e^{im\phi}$ with integer $m$ ; the polar equation picks up an $m^2/\sin^2\theta$ term, becoming the associated Legendre equation with solutions $P_\ell^m(\cos\theta)$ ; packaged together these are the spherical harmonics $Y_\ell^m$ , and the solution grows a second index. The radial story is unchanged — the same Euler powers serve every $m$ . The next two articles build it.

Is the exterior series secretly the multipole expansion?

Yes — you have just met it from the boundary-value side. $\sum B_\ell r^{-(\ell+1)}P_\ell(\cos\theta)$ is exactly the form the Multipole Expansion of the Potential derives from the source side, where each $B_\ell$ is an integral over the charge distribution, christened monopole, dipole, quadrupole… For the half-angle shell the dictionary is visible: $B_0$ encodes an effective charge $2\pi\varepsilon_0kR$ , $B_1$ an effective dipole, and the series truncates with the data.