Poisson's Equation

The E1 article on Poisson's equation earned the law $\nabla^2V=-\rho/\varepsilon_0$ from the two field equations and showed that, in free space, the Coulomb integral solves it outright: know $\rho$ everywhere, integrate, done. So why does this chapter need a second article with the same equation on its masthead? Because "know $\rho$ everywhere" is a luxury the real world almost never grants. Put one grounded conductor anywhere near your charges and part of the source — the induced surface charge — becomes unknown until the problem is solved. The Coulomb integral hasn't become wrong; it has become circular. This article is about what replaces it: Poisson's equation as a boundary-value problem. The rescue turns on a structure theorem so simple it fits in one line — every solution is one particular solution plus a harmonic correction, $V = V_p + W$ . The particular piece handles the charge you know (often literally the Coulomb integral over it); the harmonic piece, governed by Laplace's equation and the previous article's entire toolkit, bends the solution to meet the boundaries. Physically the split is no bookkeeping trick: $W$ is the potential of the induced charges. The theorem converts every Poisson problem into a Laplace problem plus one integral, which is why the rest of this chapter can concentrate on harmonic functions with a clear conscience. The numerical story upgrades in one line as well: the relaxation stencil of the previous article simply acquires a source term, and grids with loads solve depletion layers, ion ducts, and plasma sheaths. By the end you will have solved Poisson-with-boundaries twice — once with a shell, once with a grid — and both times the answer arrives as known charge plus harmonic response.

Key takeaways

With conductors present, the free-space Coulomb integral is circular: induced surface charge is part of $\rho$ , and it is unknown until $V$ is known. Poisson's equation must be solved as a boundary-value problem.
Structure theorem: every solution of $\nabla^2V=-\rho/\varepsilon_0$ in a region is $V = V_p + W$ , where $V_p$ is any particular solution and $W$ is harmonic — the difference of any two solutions with the same source has the source cancel.
Strategy: build $V_p$ from the known charge (Coulomb integral, or a symmetric special case); then solve a Laplace problem for $W$ with corrected boundary data $W = V_{\text{prescribed}} - V_p$ on the rim.
Physically, $W$ is the potential of the induced/boundary charges — the part of the source the integral couldn't see.
On a grid, the Laplace stencil gains a load: in 2D, $V = \tfrac14\sum V_{\text{nbr}} + h^2\rho/4\varepsilon_0$ — relaxation solves charged regions as easily as empty ones.
KEY IDEA: Poisson with boundaries = one particular solution for the charge you know + one harmonic function for the boundaries you must satisfy; the chapter's Laplace machinery therefore solves charged problems too.

The core insight

Every solution of Poisson's equation splits as $V = V_p + W$ — a particular solution absorbing the known charge plus a harmonic function absorbing the boundary conditions — so boundary-value methods for Laplace's equation extend to charged regions intact.

Where this lives today

Poisson-with-boundaries is arguably the most-solved PDE on Earth. Every transistor simulation begins with it: the depletion region of a p–n junction is a charged layer between effectively equipotential contacts, and TCAD tools relax loaded grids exactly like worked example 2's, billions of nodes at a time (the junction's own analytic profile was the E1 article's exam example — the boundary-value machinery here is what generalizes it to real device geometry). Structural biologists solve the Poisson–Boltzmann variant — Poisson's equation with a $\rho$ that responds to $V$ — over the surface of proteins to compute solvation energies and binding affinities; the "boundary" there is the molecular surface, as intricate as boundaries get. And at the largest scales, cosmological N-body codes spend most of their cycles on gravity's identical twin, $\nabla^2\Phi = 4\pi G\rho$ : particle-mesh methods deposit billions of dark-matter particles onto a grid, solve the loaded stencil with fast transforms or multigrid, and differentiate for the forces — the structure of every galaxy simulation is worked example 2 with the sign of the coupling flipped and no negative charge to hide behind. One equation, from nanometer junctions to gigaparsec boxes.

Origins · 1 minGeorge Green and how this was discovered

George Green

1793–1841 · Sneinton windmill, Nottingham

Green is the most improbable figure in mathematical physics: a miller's son with one year of schooling who taught himself analysis from a subscription library while running the family windmill, then self-published An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism (1828) — fifty-one subscribers, most of whom cannot have understood a page. The Essay invented the potential-theory toolkit this chapter runs on: the integral identities now called Green's theorems, the systematic split of a potential into source and boundary contributions, and the object later christened the Green function — the potential of a unit charge adapted to the boundary, which packages $V_p$ and $W$ into a single kernel. He entered Cambridge at forty, died at forty-seven, and his Essay might have vanished had a young William Thomson not tracked down copies in 1845 and evangelized it across Europe. The mathematics of "known charge plus boundary response" is Green's windmill legacy, and its full Green-function form returns in the advanced electrodynamics tier.

History

1813

— Poisson adds the source term to Laplace's equation, tying the potential's curvature to local charge

1828

— Green's self-published Essay founds boundary-aware potential theory: source term plus boundary response, and the kernel later named for him

1845

— Thomson rediscovers and circulates Green's Essay, making its methods the common language of electrostatics

1923

— Debye and Hückel couple Poisson's equation to a thermal charge response, founding screening theory in electrolytes

1949

— Shockley's p–n junction theory turns the loaded Poisson equation into the design equation of the semiconductor age

The physical picture

Picture the potential as an elastic membrane and charge as a load pressing on it. In the previous article the membrane was empty and taut — the smoothest interpolation of its frame, all curvature banished. Poisson's equation adds weights: where positive charge sits, the membrane bulges (the potential rises above the average of its neighbors, $\nabla^2V<0$ ); the frame — the boundary conditions — still holds the edges. The final shape is manifestly a sum of two influences: the dents the load would make on an unbounded membrane, plus the smooth adjustment the frame imposes. That sum is the structure theorem before any algebra.

Figure

Load plus frame — the anatomy of a Poisson solution

Now sharpen the crisis that makes this necessary. Hold a charge $q$ near a grounded conductor and try to compute $V$ by the E1 integral. You integrate over $q$ — fine. You must also integrate over the induced surface charge $\sigma$ — but $\sigma = -\varepsilon_0\,\partial V/\partial n$ depends on the answer. The integral formula demands the very unknown it is supposed to deliver. Nothing is inconsistent; the free-space formula is simply the wrong tool where boundaries respond. The boundary-value formulation dodges the circle: impose what you control ( $V=0$ on the conductor), solve the PDE, and collect $\sigma$ afterwards. The structure theorem is the dodge made systematic. Linearity means solutions with the same source differ by something with no source — a harmonic function. So pick the easiest function that has the right charge (any particular solution, however badly it treats the boundaries), and repair the boundaries with a chargeless harmonic correction. The repair is exactly what the induced charges physically do: their potential is harmonic inside the region (they live on its rim), and nature tunes them until the conductor reads zero. Your mathematics re-enacts the electrostatics.

The structure theorem and the workflow

The law itself, inherited from E1 with its meaning intact:

\nabla^2 V = -\frac{\rho}{\varepsilon_0} \qquad (1)

where: $\rho$ — the charge density in the region where you are solving; charges on the boundary (induced surface charge, electrode charge) enter through boundary conditions instead. That division of labor — volume charge in the equation, boundary charge in the data — is the whole grammar of this chapter. The decomposition that runs the article:

V = V_p + W, \qquad \nabla^2 V_p = -\frac{\rho}{\varepsilon_0}, \qquad \nabla^2 W = 0 \qquad (2)

where: $V_p$ — any function with the right source, boundary manners optional; $W$ — a harmonic function chosen to fix the manners. Two candidates for $V_p$ dominate practice: the Coulomb integral over the known $\rho$ (always available in principle), and symmetric closed forms like the uniformly charged ball's parabola (recalled from E1) when the charge is simple. Nothing about $V_p$ is unique — add any harmonic function to a particular solution and it is still a particular solution; $W$ absorbs whatever choice you made. The boundary data for $W$ follow by subtraction. If the physical problem demands $V = g$ on the boundary, then

W = g - V_p \quad\text{on } \partial\Omega \qquad (3)

where: $g$ — the prescribed potentials (zero on grounded conductors). Equation (3) is a Laplace problem — the previous article's entire catalog (integration, images, series, relaxation) applies verbatim. One particular solution plus one Laplace solve: that is the standard workflow for every charged boundary-value problem you will meet, and the uniqueness of the assembled total is guaranteed by the theorems of Uniqueness Theorems for the Potential (proved there for Poisson's equation with charge present — not re-proved here).

Figure

The decomposition in one dimension — parabola plus straight line

Derivation AThe structure theorem — every Poisson solution is particular plus harmonic

We prove the decomposition (2)–(3) and, just as important, see why it cannot fail. The strategy is pure linearity: subtract two solutions and watch the source cancel. Nothing about uniqueness is needed or claimed — this theorem says every solution has this form; the uniqueness article separately proves there is only one. Keep the two results distinct: structure here, uniqueness there.

Step 1. Let $V$ be the true solution of the physical problem: $\nabla^2V = -\rho/\varepsilon_0$ in $\Omega$ , $V = g$ on the boundary. Let $V_p$ be any function you can manufacture with $\nabla^2V_p = -\rho/\varepsilon_0$ in $\Omega$ — no boundary demands at all.

Step 2. Define the correction $W \equiv V - V_p$ and subtract the two Poisson equations:

\nabla^2 W = \nabla^2 V - \nabla^2 V_p = -\frac{\rho}{\varepsilon_0} + \frac{\rho}{\varepsilon_0} = 0.

Why the cancellation is exact: both functions carry the same $\rho$ — the known volume charge. Whatever is unknown (induced boundary charge) never appeared in either equation; it lives on $\partial\Omega$ , outside the region where we subtracted.

Step 3. Read off $W$ 's boundary data: on $\partial\Omega$ , $W = V - V_p = g - V_p$ . So $W$ solves a completely specified Laplace problem — harmonic inside, known values on the rim — exactly the shape of problem the previous article's methods construct.

Step 4. Reassemble: $V = V_p + W$ satisfies the source (from $V_p$ ) and the boundary (by construction of $W$ ). Existence of a solution to the physical problem is thereby reduced to existence for the Laplace problem — a genuine simplification, since harmonic functions are the best-understood objects in the chapter.

Step 5. Interpret the pieces physically. $V_p$ (say, the Coulomb integral over $\rho$ ) is the potential the known charges would make if the boundaries did not respond. $W$ is everything the response adds: the potential of the induced surface charges. Indeed $W$ is harmonic precisely because those charges lie on the boundary, not inside $\Omega$ — the mathematics knows where the induced charge lives before you do. Worked example 2 exhibits this decomposition explicitly — the parabola is $V_p$ , the straight line is $W$ , and the sum meets both plates.

Go deeper · Derivation B · 1 minThe loaded grid — Poisson's stencil and why it is exact for parabolas

Derivation BThe loaded grid — Poisson's stencil and why it is exact for parabolas

The previous article derived the discrete Laplacian from Taylor expansion: the neighbor-average deficit measures $\nabla^2V$ with error $O(h^2)$ . We do not repeat that derivation — we impose a different right-hand side on the same identity and obtain the numerical method for charged regions, then extract a bonus fact (exactness on quadratic potentials) that worked example 2 will cash.

Step 1. Recall the identity earned in Laplace's Equation (E2), valid at any interior grid node with spacing $h$ (2D):

\sum_{\text{4 nbrs}} V - 4V = h^2\,\nabla^2 V + \frac{h^4}{12}\left(\partial_x^4 + \partial_y^4\right)V + \cdots

Step 2. Substitute the equation being solved — this time $\nabla^2V = -\rho/\varepsilon_0$ evaluated at the node:

V = \frac{1}{4}\sum_{\text{4 nbrs}} V + \frac{h^2\rho}{4\varepsilon_0} + O(h^4) \qquad (4)

Why the sign is friendly: positive charge at a node lifts the node above its neighbor average — the membrane bulges under a load — which is $\nabla^2V<0$ in the continuum and a positive added term on the grid. (In 3D the same steps give $\tfrac16\sum V_{\text{nbr}} + h^2\rho/6\varepsilon_0$ .)

Step 3. The algorithm is relaxation with a load: hold boundary nodes at their data, sweep, replace each node by neighbor-average plus $h^2\rho/4\varepsilon_0$ . Convergence machinery (Gauss–Seidel, over-relaxation, multigrid) transfers wholesale from the Laplace case; the load only shifts the fixed point.

Step 4. The bonus: look back at the error term. It contains only fourth derivatives of $V$ . If the true solution is a polynomial of degree $\le 3$ — in particular the parabolas that one-dimensional uniform charge produces — the error is identically zero: the discrete solution agrees with the continuum solution at the nodes exactly, at any $h$ . Coarse grids are not always crude.

Step 5. Structure theorem, grid edition: since (4) is linear, the discrete solution also splits as particular-plus-discrete-harmonic, and solvers exploit it — e.g. subtracting an analytic $V_p$ and relaxing only the smooth correction $W$ , which converges faster because it has no sharp source features to resolve. The exactness-for-cubics claim follows directly from the displayed error term; worked example 2 confirms it numerically — grid and analytic values coincide to all displayed digits.

Limiting cases

Case	Result	Meaning
$\rho = 0$ everywhere in $\Omega$	$V_p$ can be taken zero; $V = W$	the previous article is the special case
No boundaries (all space, $V\to0$ )	$W = 0$ ; $V = V_p$ = Coulomb integral	E1's free-space solving is the other special case
Symmetric charge, symmetric boundary	$W$ reduces to a constant or a few terms	worked example 1: the correction is one number
Sheet/line/point sources	$\rho$ enters as delta functions	handled by images and Green-function ideas ahead
Grid solution, quadratic true $V$	discrete = continuum at nodes, any $h$	stencil error carries only 4th derivatives
Linear dielectric present	$\nabla\cdot(\varepsilon\nabla V) = -\rho_{\text{free}}$	E3 chapter; the decomposition logic survives

Worked examples

1 · A charged core in a grounded shell (confidence builder)

Charge $Q = 1.0\,\mathrm{nC}$ is spread uniformly through a ball of radius $a = 2.0\,\mathrm{cm}$ , concentric with a grounded conducting shell of inner radius $b = 10\,\mathrm{cm}$ . Find $V$ at the ball's surface and center by the particular-plus-harmonic workflow.

Step 1. Particular solution: the free-space potential of the uniform ball, recalled from the E1 article (derived there, not re-derived): outside the ball $V_p = Q/4\pi\varepsilon_0 r$ ; at the surface $V_p(a) = Q/4\pi\varepsilon_0 a$ ; at the center $V_p(0) = \tfrac{3}{2}\,Q/4\pi\varepsilon_0 a$ . Numerically $Q/4\pi\varepsilon_0 a = (8.99\times10^{9})(1.0\times10^{-9})/0.020 = 449.5\,\mathrm{V}$ , so $V_p(a) = 449.5\,\mathrm{V}$ , $V_p(0) = 674.2\,\mathrm{V}$ .

Step 2. $V_p$ violates the boundary: $V_p(b) = 449.5\times(0.02/0.10) = 89.9\,\mathrm{V} \neq 0$ on the grounded shell. The correction $W$ must be harmonic throughout $r<b$ (all the charge is already in $V_p$ ), regular at the origin, and equal to $-89.9\,\mathrm{V}$ at $r=b$ . The only candidate ( $A + B/r$ with $B=0$ by regularity) is the constant

W = -\frac{Q}{4\pi\varepsilon_0 b} = -89.9\,\mathrm{V}.

Step 3. Assemble: $V = V_p - 89.9\,\mathrm{V}$ everywhere inside. So

V(a) = 449.5 - 89.9 = 359.6\,\mathrm{V} \approx 3.6\times10^{2}\,\mathrm{V}, \qquad V(0) = 674.2 - 89.9 = 584.3\,\mathrm{V} \approx 5.8\times10^{2}\,\mathrm{V}.

Step 4. Read the physics of $W$ : the shell's induced inner-surface charge $-Q$ (the shielding article's bookkeeping) spreads uniformly and contributes, everywhere inside a uniform shell, the constant potential $-Q/4\pi\varepsilon_0 b$ — precisely our harmonic correction. The field inside is untouched by the grounding (a constant shifts no gradient); only the potential ledger moved. The correction is one number here because the symmetry is perfect — the general case simply spends more effort on the same $W$ .

2 · Grounded plates with a uniform space charge, analytically and on a grid (exam level)

An ion duct is modeled as two grounded plates at $x=0$ and $x=d=2.0\,\mathrm{cm}$ with uniform charge $\rho = 5.0\,\mu\mathrm{C/m^3}$ filling the gap. Find $V(x)$ , the peak potential, the field at the plates — then verify with a four-interval grid.

Step 1. Solve $\mathrm{d}^2V/\mathrm{d}x^2 = -\rho/\varepsilon_0$ by the decomposition: particular $V_p = -\rho x^2/2\varepsilon_0$ (check: second derivative $-\rho/\varepsilon_0$ ✓); harmonic (1D) $W = Ax + B$ . Boundary $V(0)=0$ gives $B=0$ ; $V(d)=0$ gives $A = \rho d/2\varepsilon_0$ . Hence

V(x) = \frac{\rho}{2\varepsilon_0}\,x(d-x),

a parabola — dented membrane, pinned at both frames.

Step 2. Peak at midgap:

V_{\max} = \frac{\rho d^2}{8\varepsilon_0} = \frac{(5.0\times10^{-6})(0.020)^2}{8(8.854\times10^{-12})}\,\mathrm{V} = 28.2\,\mathrm{V}.

Field at the plates: $E(0) = -V'(0)\cdot(-1)$ ... compute $V'(x) = \tfrac{\rho}{2\varepsilon_0}(d-2x)$ , so $|E| = \rho d/2\varepsilon_0 = (5.0\times10^{-6})(0.010)/(8.854\times10^{-12}) = 5.6\times10^{3}\,\mathrm{V/m}$ at each plate, pointing into the plates — each collects induced charge $\sigma = -\varepsilon_0 E = -\rho d/2 = -5.0\times10^{-8}\,\mathrm{C/m^2}$ , together exactly $-\rho d$ : the plates swallow the gap's charge, half each, as symmetry demands.

Step 3. Grid check with $h = d/4 = 0.50\,\mathrm{cm}$ : the 1D loaded stencil is $V_i = \tfrac12(V_{i-1}+V_{i+1}) + h^2\rho/2\varepsilon_0$ , load term $= (0.005)^2(5.0\times10^{-6})/(2\times8.854\times10^{-12}) = 7.06\,\mathrm{V}$ . Solve the three-node fixed point (symmetry: $V_1 = V_3$ ): $V_2 = V_1 + 7.06$ and $V_1 = \tfrac12 V_2 + 7.06$ , giving $V_1 = 21.2\,\mathrm{V}$ , $V_2 = 28.2\,\mathrm{V}$ .

Step 4. Compare with the exact parabola at the nodes: $V(d/4) = \tfrac{\rho}{2\varepsilon_0}(0.005)(0.015) = 21.2\,\mathrm{V}$ ✓ and $V(d/2) = 28.2\,\mathrm{V}$ ✓ — exact agreement, not approximate, because the stencil's error carries only fourth derivatives and a parabola has none (Derivation B, Step 4). Peak potential $28.2\,\mathrm{V}$ ; plate field $5.6\,\mathrm{kV/m}$ ; a four-interval grid nails both to every displayed digit.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §2.3.1 and §3.1 · Jackson, Classical Electrodynamics, 3rd ed., §1.7–1.10 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 2–3 · Zangwill, Modern Electrodynamics, §7.1–7.3. Key results underlying this article: $\mathbf{E}=-\nabla V$ combined with $\nabla\cdot\mathbf{E}=\rho/\varepsilon_0$ yields $\nabla^2 V=-\rho/\varepsilon_0$ ; the Green-function identity $\nabla^2(1/r)=-4\pi\delta^3(\mathbf r)$ follows from the divergence theorem; the uniformly-charged-sphere potential inside is $V_{\text{in}}=\frac{\rho}{6\varepsilon_0}(3R^2-r^2)$ . Constant used throughout: $\varepsilon_0=8.854\times10^{-12}\,\mathrm{F/m}$ .

Builds on: Poisson's Equation (E1) · Gauss's Law in Integral Form · The Electrostatic Potential · The Laplacian and Second Derivatives

Leads to: Laplace's Equation (E2) · Uniqueness Theorems for the Potential · Method of Images · Harmonic Functions and the Mean-Value Theorem

What exactly is the division of labor between this article and the E1 Poisson's Equation article?

E1 owns the law: deriving $\nabla^2V=-\rho/\varepsilon_0$ from the field equations, showing the Coulomb integral solves it in free space, and solving symmetric charged regions outright (the uniform ball, the p–n junction). This article owns the boundaries: what breaks when conductors respond, the particular-plus-harmonic structure theorem, boundary data handed to a Laplace solve, and the loaded grid. If your problem has no responding boundaries, E1's methods finish it; the moment grounded metal appears, you are on this page.

Is the particular solution unique? Which one should I pick?

Not remotely unique — add any harmonic function to a particular solution and you have another one. That freedom is a feature: pick whichever $V_p$ is cheapest. Symmetric charge? Use the closed form (ball parabola, plate parabola). Arbitrary blob of charge? The Coulomb integral over it (ignoring all boundaries) always qualifies. A sloppy $V_p$ just transfers work to $W$ 's boundary data; the assembled total is the same either way — that is the content of uniqueness, proved next door.

Why is the induced-charge potential harmonic inside the region? The induced charge is real charge.

Real, but elsewhere. Harmonicity of a potential in a region only requires that its sources lie outside that region — and induced charge lives on the conductor surfaces, i.e. on the region's boundary, not in its interior. Inside $\Omega$ , the induced-charge potential therefore satisfies Laplace's equation exactly. This is why the structure theorem's split is physical rather than cosmetic: the mathematics segregates the potential by where its sources sit, and the unknown sources all sit on the rim.

How do I keep the sign of Poisson's equation straight?

Anchor it to a picture, not a mnemonic: a positive point charge makes a potential hill. A hill's value exceeds the average of its neighbors, and the Laplacian measures value-below-average, so $\nabla^2V$ must be negative where $\rho$ is positive: $\nabla^2V = -\rho/\varepsilon_0$ . (Gaussian-units texts write $-4\pi\rho$ ; same sign logic, different constant — check conventions before comparing formulas across books.)

Does the boundary really reshape V instantaneously everywhere, as elliptic equations imply?

Within electrostatics, yes — elliptic equations have no propagation speed, so wiggling a boundary potential updates the whole solution "at once." That is the static idealization speaking, not physics violating relativity: statics assumes all transients have died. The honest time-dependent story replaces Poisson's equation with the wave equation and retarded potentials (the E9 chapter), where boundary news travels at $c$ . Use elliptic instantaneity as a solving convenience, never as a causal claim.

Does the particular-plus-harmonic trick survive dielectrics and nonlinear responses?

Linear dielectrics: yes, with upgrades — the equation becomes $\nabla\cdot(\varepsilon\nabla V) = -\rho_{\text{free}}$ , and for piecewise-uniform $\varepsilon$ you decompose region by region and stitch with matching conditions (the E3 chapter's business). Nonlinear cases (Poisson–Boltzmann with its $\sinh V$ response, semiconductor drift-diffusion) break superposition, so the clean split fails; solvers instead iterate — linearize, decompose, correct, repeat — which is the structure theorem used as the inner loop rather than the whole answer.

With Neumann data (specified surface charge) instead of grounded conductors, does anything change?

The workflow is identical — build $V_p$ , hand the corrected data to a Laplace solve — but the corrected data are now normal derivatives, $\partial W/\partial n = \partial V/\partial n - \partial V_p/\partial n$ , and two caveats appear. The solution is determined only up to an additive constant (nothing pins the overall level), and the data must satisfy a global consistency check: total flux through the boundary must match the enclosed charge, or no solution exists at all. Both caveats are proved and dissected in Boundary Conditions for the Potential and Uniqueness Theorems for the Potential.