Poisson's Equation

Electrostatics, as built so far, is a theory with two faces. Coulomb's law and its superposition integrals compute fields directly from charges; Gauss's law and the curl-free condition describe the same fields locally, one point at a time. This article is where the two faces fuse into a single line. The curl-free condition — established honestly in The Curl of the Electrostatic Field — guarantees a potential with $\mathbf{E}=-\nabla V$ . Feed that into Gauss's law in differential form and the entire theory collapses to one scalar equation: $\nabla^2 V = -\rho/\varepsilon_0$ . Poisson's equation. It is hard to overstate what this buys. Three coupled field components, two vector field equations, and every superposition integral you have computed are now packaged as one linear partial differential equation for one function. Hand it a charge density and boundary conditions and it hands back the potential; one gradient later, the field. Every capacitor design code, every chip-level device simulator, every cosmological structure-formation run is, at bottom, solving this equation — sometimes for $V$ , sometimes for its gravitational twin $\Phi$ . There is also a conceptual payoff. Poisson's equation is local: the curvature of the potential at a point is fixed entirely by the charge density at that same point. Yet its solution — the Coulomb integral over all of space — is global. Watching how a local differential law and a global integral formula turn out to be the same statement is the best preparation you can get for the rest of field theory.

Key takeaways

Poisson's equation $\nabla^2 V = -\rho/\varepsilon_0$ follows in two lines: the curl-free field is a gradient, $\mathbf{E}=-\nabla V$ , and Gauss's law $\nabla\cdot\mathbf{E}=\rho/\varepsilon_0$ then reads $\nabla\cdot(-\nabla V)=\rho/\varepsilon_0$ .
It is local — the curvature of $V$ at a point is set by $\rho$ at that point — and linear, so solutions superpose exactly like the fields they encode.
The Coulomb integral $V=\frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\mathrm{d}\tau'$ is the solution of Poisson's equation for a localized source with $V\to0$ at infinity — the bridge is the delta identity $\nabla^2(1/r)=-4\pi\delta^3(\mathbf{r})$ from the E0 chapter.
Where $\rho=0$ the equation degenerates to Laplace's equation $\nabla^2V=0$ , whose rigid harmonic solutions are the next article's subject.
With finite boundaries the Coulomb integral alone is not enough: add a solution of Laplace's equation to meet the boundary data — the strategy behind all of E2.
Newtonian gravity obeys the structurally identical $\nabla^2\Phi = 4\pi G\rho_m$ , so every technique built here transfers verbatim.
Key idea: Poisson's equation is what you get when you feed $\mathbf{E}=-\nabla V$ into $\nabla\cdot\mathbf{E}=\rho/\varepsilon_0$ — the divergence of a gradient is the Laplacian, so $\nabla^2 V = -\rho/\varepsilon_0$ , and all of electrostatics fits in one scalar PDE.

The core insight

Poisson's equation is what you get when you feed $\mathbf{E}=-\nabla V$ into $\nabla\cdot\mathbf{E}=\rho/\varepsilon_0$ : the divergence of a gradient is the Laplacian, so $\nabla^2 V = -\rho/\varepsilon_0$ . It packages all of electrostatics into one scalar elliptic PDE.

Where this lives today

Ask what equation is solved most often, on the most computers, every second of every day, and Poisson's is a serious contender. Semiconductor device simulation (TCAD) solves it self-consistently with carrier statistics across every proposed transistor geometry before a mask is ever made — the worked junction above is the hand-calculable seed of that industry. Molecular-dynamics codes solve it under the name particle-mesh Ewald to get long-range electrostatics in proteins and electrolytes; the Poisson–Boltzmann variant, with mobile ions loading the right-hand side self-consistently, prices the electrostatic cost of drug binding and membrane transport in computational biochemistry. The gravitational twin runs the largest simulations in science. Cosmological structure-formation codes evolve billions of dark-matter particles by depositing mass on a grid, solving $\nabla^2\Phi=4\pi G\rho_m$ by FFT or multigrid at every timestep, and differentiating for the forces — the same operator, the same solvers, the same boundary subtleties as electrostatics, at box sizes of gigaparsecs. When a 2026 survey pipeline compares observed galaxy clustering against theory, a Poisson solve sits in the innermost loop of the theory side. One scalar elliptic PDE, written down as a correction to Laplace in 1813, is the shared engine of chip design, drug design, and cosmology.

The physical picture

Think of $V$ as the height of an elastic membrane. Charge density acts as a distributed load: positive charge tents the membrane up into a peak, negative charge dimples it into a well. The Laplacian — the averaging meter of Second Derivatives and the Laplacian — measures how much the membrane at a point sags below the average of its neighbours, and Poisson's equation says that this local sag is set entirely by the local load. Where there is no load, the membrane is taut: it can bend, carrying the influence of distant weights across the region, but it cannot bulge on its own.

Figure

The membrane picture — local curvature of V is set by the local charge

Hold on to the tension in that picture, because it is the whole subject. The equation is strictly local — curvature here is charge here, nothing else. But the value of $V$ here is determined non-locally: every charge in the universe contributes through the $1/r$ kernel of Potential of a Localized Charge Distribution. Poisson's equation is the differential statement; the Coulomb integral is its global solution; and the two are stitched together by a single identity, the fact that the Laplacian of $1/r$ is a delta function. A point source makes a $1/r$ potential, and any distribution is a sum of point sources. There is a third way to say it, worth internalizing. Because the operator is the divergence of a gradient, Poisson's equation literally says: the net outflow of $-\nabla V$ from an infinitesimal box equals the charge inside, per $\varepsilon_0$ . It is Gauss's law, run one box at a time, with the field written as slope. Nothing new has been assumed — the equation is bookkeeping. What is new is the leverage: PDEs of this type (linear, second-order, elliptic) come with a complete mathematical theory of existence, uniqueness, and numerical solution, and electrostatics inherits all of it for free.

One equation, three routes to V

The master equation of the article deserves its formal introduction. Combining the two field equations of electrostatics (the derivation below walks the two lines carefully):

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

where: $V$ — the electrostatic potential (volts); $\nabla^2=\partial_x^2+\partial_y^2+\partial_z^2$ — the Laplacian in Cartesian coordinates; $\rho$ — the volume charge density (C/m³); $\varepsilon_0 = 8.854\times10^{-12}$ F/m. Read it physically: where charge is positive the potential curves downward relative to its surroundings (a peak), where negative it curves upward (a well), and the proportionality is universal — a nanocoulomb per cubic metre bends $V$ by the same amount whether it sits in a capacitor or a thundercloud. In any region with no local charge the right side vanishes and the equation degenerates to

\nabla^2 V = 0 \qquad (2)

where: the potential is called harmonic there. Do not read "no charge in this region" as "no field": the sources may sit just outside, on plates or conductor surfaces, threading the region with strong fields whose potential is nonetheless harmonic. Equation (2) — Laplace's equation — has such rigid and beautiful solution structure that the next article is devoted to it entirely. The third pillar is the solution formula. For a localized charge distribution, with the reference at infinity,

V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,\mathrm{d}\tau' \qquad (3)

where: $\mathbf{r}$ — the field point; $\mathbf{r}'$ — the source point ranging over the distribution; $\mathrm{d}\tau'$ — the source volume element. This is exactly the superposition integral constructed in Potential of a Localized Charge Distribution — but its role has changed. There it was derived from the point-charge potential; here it is the particular solution of Poisson's equation subject to $V\to0$ at infinity. The second derivation below proves that claim, and the proof leans on one identity, established twice over in the E0 chapter (Second Derivatives and the Laplacian; The Three-Dimensional Dirac Delta Function) and recalled here as given:

\nabla^2\frac{1}{|\mathbf{r}-\mathbf{r}'|} = -4\pi\,\delta^3(\mathbf{r}-\mathbf{r}') \qquad (4)

where: $\delta^3$ — the three-dimensional Dirac delta. In words: $1/r$ is harmonic everywhere except the origin, and the singularity there carries exactly the concentrated curvature of a point source. Equation (4) is the hinge on which differential and integral electrostatics turn into each other. Because equation (1) is linear, superposition is automatic: if $V_1$ sources $\rho_1$ and $V_2$ sources $\rho_2$ , then $V_1+V_2$ sources $\rho_1+\rho_2$ . Just as useful: the difference of two solutions with the same $\rho$ satisfies Laplace's equation. That means the general solution of Poisson's equation is one particular solution — the Coulomb integral, say — plus any harmonic function, and it is boundary conditions that select which one. Stating this precisely (a solution is unique once $V$ or its normal derivative is fixed on the boundary) is the first uniqueness theorem; Laplace's Equation next door gives the proof sketch, and the E2 boundary-value chapter builds the full machinery on top of it.

Which route to V, when

situation	route	why
localized $\rho$ , no boundaries	Coulomb integral (3)	direct superposition; reference at infinity is legal
high symmetry (sphere, cylinder, plane)	Gauss's law, then $V=-\int\mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell}$	see Applications of Gauss's Law and Symmetry
conductors / prescribed boundary values	solve (1) as a boundary-value problem	integral (3) alone cannot meet the boundary data; add harmonic pieces — E2's subject
$\rho$ known only locally, complex geometry	discretize (1) numerically	elliptic solvers: relaxation, multigrid, FFT
given $V$ , want the charge	differentiate: $\rho = -\varepsilon_0\nabla^2 V$	Poisson's equation read backwards — no integral needed

The last row is easy to overlook and shows up constantly in practice (and in the first worked example): if someone hands you a potential map, two derivatives recover the charge that produced it. In one dimension the equation's anatomy is visible to the naked eye, and it is worth fixing the shape-chain in memory before the derivations: a constant charge density integrates up to a linear field and a parabolic potential. Every slab, junction, and sheet problem is a walk along this chain — and reading it backwards (potential not parabolic ⟹ charge not uniform) is a working diagnostic tool.

Figure

The 1D shape chain — constant charge makes a linear field and a parabolic potential

Derivation AFrom two field equations to one scalar equation

The goal is Poisson's equation (1), assembled from the two local laws of electrostatics: $\nabla\cdot\mathbf{E}=\rho/\varepsilon_0$ (Gauss's Law in Differential Form) and $\nabla\times\mathbf{E}=0$ (The Curl of the Electrostatic Field). The strategy is substitution: the curl-free condition is spent first, converting the unknown from a three-component field to a single scalar; Gauss's law then becomes a statement about that scalar. Watch two things: where each of the two laws is used exactly once, and how the units audit confirms the result.

Step 1. Spend the curl condition. Because $\nabla\times\mathbf{E}=0$ throughout electrostatics (a superposition of central point-charge fields), the field is the gradient of a scalar: $\mathbf{E}=-\nabla V$ . Why this is the decisive move: a vector field carries three functions of position; the curl-free condition says they are not independent, and the potential stores all three in one function. Everything that follows is Gauss's law expressed in that compressed variable.

Step 2. Substitute into Gauss's law:

\nabla\cdot\mathbf{E} = \nabla\cdot(-\nabla V) = -\nabla^2 V = \frac{\rho}{\varepsilon_0}

The divergence of a gradient is, by definition, the Laplacian — the same second-derivative combination that measures a function's deviation from the average of its neighbours.

Step 3. Rearrange:

\nabla^2 V = -\frac{\rho}{\varepsilon_0}

Why the minus sign is structural, not decorative: it descends from the convention $\mathbf{E}=-\nabla V$ (fields point downhill), so that positive charge — a source of outward field — sits at a peak of the potential, where the curvature is negative. Flip the sign and every peak becomes a well.

Step 4. Audit the dimensions. Left side: volts per metre squared, $\mathrm{V/m^2}$ . Right side: $(\mathrm{C/m^3})/(\mathrm{F/m}) = \mathrm{C\,m^{-2}\,F^{-1}} = \mathrm{V/m^2}$ , since a farad is a coulomb per volt. The equation balances. Verified: reproduces Griffiths Eq. 2.24 (§2.3.1); in Gaussian units the same two steps give $\nabla^2V=-4\pi\rho$ , confirming the constant tracks the unit system, not the physics.

Go deeper · Derivation B · 1 minThe Coulomb integral solves Poisson's equation

Derivation BThe Coulomb integral solves Poisson's equation

The goal is to prove that equation (3) — the superposition integral of Potential of a Localized Charge Distribution — satisfies $\nabla^2V=-\rho/\varepsilon_0$ , closing the circle between the integral and differential formulations. The strategy: apply the Laplacian at the field point, move it under the integral (legal because the integration variable $\mathbf{r}'$ is independent of $\mathbf{r}$ ), and let the delta identity (4) — owned by the E0 chapter, used here as given — collapse the integral. The subtle point to watch is which variable the Laplacian differentiates.

Step 1. Apply $\nabla^2$ (derivatives with respect to the field point $\mathbf{r}$ ) to the integral:

\nabla^2 V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int \rho(\mathbf{r}')\;\nabla^2\frac{1}{|\mathbf{r}-\mathbf{r}'|}\,\mathrm{d}\tau'

Why the Laplacian slides past $\rho(\mathbf{r}')$ : the source density is a function of the primed variable only — a constant as far as unprimed derivatives are concerned. The only $\mathbf{r}$ -dependence in the integrand sits in the kernel.

Step 2. Insert the identity (4):

\nabla^2 V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int \rho(\mathbf{r}')\,\big[-4\pi\,\delta^3(\mathbf{r}-\mathbf{r}')\big]\,\mathrm{d}\tau'

Why no re-derivation is offered here: the identity is the signature result of the E0 delta-function articles, proved there twice — once by the pincer argument, once by regularization. An article that re-proved it would be padding; an article that forgot to cite it would be hiding the load-bearing wall.

Step 3. Let the delta sift. The integral picks out the source density at the field point:

\nabla^2 V(\mathbf{r}) = -\frac{4\pi}{4\pi\varepsilon_0}\,\rho(\mathbf{r}) = -\frac{\rho(\mathbf{r})}{\varepsilon_0}

which is Poisson's equation. Note what the derivation quietly assumed: the integral (3) exists and $V\to0$ far away — true for localized distributions, and exactly the boundary condition under which (3) is the solution rather than a solution.

Step 4. Read the result in both directions. Forward: superposing $1/r$ kernels always manufactures a field whose local curvature matches the local charge. Backward: among all solutions of Poisson's equation, the Coulomb integral is the one with no leftover harmonic piece — no fields imposed from the boundary, nothing but the sources themselves. Verified: consistent with Griffiths §2.3.2 and Jackson §1.7; applying the same two steps to a single point charge $\rho=q\,\delta^3(\mathbf{r}')$ returns $V=q/4\pi\varepsilon_0 r$ , the potential the identity was calibrated on.

Go deeper · Derivation C · 2 minSolving Poisson directly for a uniformly charged ball

Derivation CSolving Poisson directly for a uniformly charged ball

The goal is to solve $\nabla^2V=-\rho/\varepsilon_0$ as a differential equation — no Coulomb integral, no Gauss's law shortcut — for a ball of total charge $q$ and radius $a$ with uniform $\rho=3q/4\pi a^3$ , and to see how an interior Poisson solution and an exterior Laplace solution are patched together. This is the template for every piecewise-source problem: solve in each region, then match. Watch where each integration constant dies: one to regularity at the centre, one to the reference at infinity, and the last two to matching at the surface.

Step 1. Reduce the operator. For a spherically symmetric $V(r)$ , the Laplacian keeps only its radial part, so inside the ball

\frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\!\left(r^2\frac{\mathrm{d}V}{\mathrm{d}r}\right) = -\frac{\rho}{\varepsilon_0}

Step 2. Integrate once: $r^2\,\mathrm{d}V/\mathrm{d}r = -\rho r^3/3\varepsilon_0 + C_1$ . Regularity kills $C_1$ : a nonzero $C_1$ would make $\mathrm{d}V/\mathrm{d}r\sim C_1/r^2$ — the field of a point charge at the origin, which the problem does not contain. So $C_1=0$ . Why this counts as a boundary condition: the centre of the ball is an interior point of the physics; demanding the solution stay finite and smooth there is data, just as binding as a conductor's surface value.

Step 3. Integrate again: $V_{\text{in}}(r) = -\rho r^2/6\varepsilon_0 + C_2$ , with $C_2$ pending.

Step 4. Solve outside, where $\rho=0$ and Laplace's equation holds. The same radial operator with zero right side integrates to $V_{\text{out}}=A/r+B$ ; the reference $V\to0$ at infinity kills $B$ , and the total charge fixes $A = q/4\pi\varepsilon_0$ (the exterior solution must be the point-charge potential — Applications of Gauss's Law and Symmetry established that a sphere acts from outside as if collapsed to its centre).

Step 5. Match at $r=a$ . Continuity of $V$ (no work can be extracted crossing an infinitesimally thin surface) and continuity of $\mathrm{d}V/\mathrm{d}r$ (no surface charge lives at $r=a$ ; the jump rules are Boundary Conditions on E and V's subject) give

V_{\text{in}}(r) = \frac{q}{8\pi\varepsilon_0 a}\left(3-\frac{r^2}{a^2}\right), \qquad V_{\text{out}}(r)=\frac{q}{4\pi\varepsilon_0 r}

Both matching conditions land on the same $C_2$ — a consistency check, not a coincidence: the derivative match is automatic because the enclosed charge is continuous at the surface. At the centre, $V(0)=\tfrac{3}{2}\,V(a)$ : the interior parabola adds half again the surface value. Verified: differentiating $V_{\text{in}}$ gives $E_{\text{in}} = \rho r/3\varepsilon_0 = qr/4\pi\varepsilon_0 a^3$ , matching the Gauss's-law interior field; the pair reproduces Griffiths Example 2.8 / Purcell §2.10 and the same profile Potential of a Localized Charge Distribution obtained by the integral route — two formalisms, one answer, with the $3/2$ central ratio intact.

Limiting cases

case	result	meaning
charge-free region	$\nabla^2V=0$	Laplace's equation: harmonic, mean-value property, no interior extrema — next article
point charge at origin	$\nabla^2V=-q\,\delta^3(\mathbf{r})/\varepsilon_0 \Rightarrow V=\frac{q}{4\pi\varepsilon_0 r}$	the Green-function seed from which integral (3) superposes everything
one dimension (slabs, junctions)	$\mathrm{d}^2V/\mathrm{d}x^2=-\rho/\varepsilon_0$	curvature literally visible: parabolic $V$ inside uniform $\rho$
uniform ball interior	$V\propto 3 - r^2/a^2$ , $V(0)=\tfrac{3}{2}V(a)$	Poisson solved as a PDE; parabola inside, $1/r$ outside
Gaussian units	$\nabla^2V=-4\pi\rho$	same physics; $\varepsilon_0$ absorbed, $4\pi$ surfaces
Newtonian gravity	$\nabla^2\Phi=+4\pi G\rho_m$	identical operator; sign flips because mass only attracts
time-dependent fields (E9 preview)	$\big(\nabla^2-\tfrac{1}{c^2}\partial_t^2\big)V=-\rho/\varepsilon_0$ (Lorenz gauge)	Poisson is the static limit of a wave equation — news travels at $c$

The gravity row rewards a second look. Newton's field obeys $\mathbf{g}=-\nabla\Phi$ and $\nabla\cdot\mathbf{g}=-4\pi G\rho_m$ , hence $\nabla^2\Phi=+4\pi G\rho_m$ : the same equation with the source sign flipped, because mass of one sign only attracts. Every tool in this chapter — superposition kernels, uniqueness, the ball solution — transfers to gravity by transliteration, which is why galaxy-formation codes and chip simulators can share their solvers.

Common trapThe Coulomb integral always solves the problem — no: only when infinity is the boundary

Equation (3) is the particular solution for a localized charge with $V\to0$ at infinity, and it silently fails the moment real boundaries enter. Put a grounded conductor near your charges and the integral over $\rho$ alone gives the wrong answer — it knows nothing of the induced surface charges, and it does not vanish on the conductor. The repair is structural, not cosmetic: to any particular solution you may add a harmonic function (Laplace solution), and boundary conditions decide which one. Forgetting the homogeneous piece is the single most common error in applying Poisson's equation. A second, quieter trap is the sign: it is $\nabla^2V=-\rho/\varepsilon_0$ in SI (and $-4\pi\rho$ in Gaussian units); writing $+\rho/\varepsilon_0$ turns every potential peak into a well and reverses every force downstream.

Worked examples

1 · Reading the charge off a potential map (confidence builder)

Inside a planar device region, probes map the potential as $V(x) = 500 - 2.0\times10^{4}\,x^2$ volts ( $x$ in metres, geometry uniform in $y$ and $z$ ). Find the charge density and the field.

Step 1. Poisson's equation read backwards gives the charge with two derivatives: $\mathrm{d}^2V/\mathrm{d}x^2 = -4.0\times10^{4}\ \mathrm{V/m^2}$ , so

\rho = -\varepsilon_0\,\frac{\mathrm{d}^2V}{\mathrm{d}x^2} = -\big(8.854\times10^{-12}\big)\big(-4.0\times10^{4}\big) = \boxed{+3.5\times10^{-7}\ \mathrm{C/m^3} \approx 0.35\ \mathrm{\mu C/m^3}}

The map curves downward everywhere ( $V$ has a peak at $x=0$ ), so the region holds positive charge — peaks sit on positive charge, exactly the membrane picture.

Step 2. The field is one derivative: $E_x = -\mathrm{d}V/\mathrm{d}x = +4.0\times10^{4}\,x$ V/m. At $x=5$ mm, $E_x = \boxed{200\ \mathrm{V/m}}$ , pointing away from the central peak on both sides — as it must, since the slab of positive charge pushes outward.

Step 3. Cross-check with Gauss. A symmetric slab of uniform $\rho$ gives $E(x)=\rho x/\varepsilon_0$ from a pillbox through the midplane: $\big(3.54\times10^{-7}/8.854\times10^{-12}\big)\,x = 4.0\times10^{4}\,x$ V/m ✓. Differential and integral bookkeeping agree — they are the same law.

2 · The p–n junction — Poisson's equation earning a living (exam level)

The depletion region on the n-side of a silicon junction ( $\varepsilon = 11.7\,\varepsilon_0$ ) carries exposed donor charge $\rho = +eN_D$ with $N_D = 1.0\times10^{22}\ \mathrm{m^{-3}}$ , over a width $w = 300$ nm ( $0\le x\le w$ , field vanishing at $x=w$ ). Find the maximum field and the potential drop across the region.

Step 1. Set up the one-dimensional Poisson equation in the material: $\mathrm{d}^2V/\mathrm{d}x^2 = -\rho/\varepsilon = -eN_D/\varepsilon$ , a constant. (In matter, $\varepsilon_0\to\varepsilon$ ; the honest accounting of bound charge is E3's subject — here silicon's $\varepsilon_r = 11.7$ is taken as given, with $\varepsilon = 1.036\times10^{-10}$ F/m.)

Step 2. Integrate once for the field. With $E = -\mathrm{d}V/\mathrm{d}x$ , $\mathrm{d}E/\mathrm{d}x = \rho/\varepsilon$ , and $E(w)=0$ :

E(x) = -\frac{eN_D}{\varepsilon}\,(w - x)

The magnitude grows linearly toward the junction plane and peaks at $x=0$ :

|E|_{\max} = \frac{eN_D w}{\varepsilon} = \frac{(1.6\times10^{-19})(1.0\times10^{22})(3.0\times10^{-7})}{1.036\times10^{-10}} = \boxed{4.6\times10^{6}\ \mathrm{V/m}}

Step 3. Integrate again for the potential drop — the area under the triangular field profile:

\Delta V = \frac{eN_D w^2}{2\varepsilon} = \frac{(1.6\times10^{-19})(1.0\times10^{22})(3.0\times10^{-7})^2}{2\times1.036\times10^{-10}} = \boxed{0.70\ \mathrm{V}}

equivalently $\tfrac12\,|E|_{\max}\,w = \tfrac12(4.63\times10^6)(3.0\times10^{-7}) = 0.70$ V ✓ — a realistic built-in voltage for silicon, which is why this piecewise-parabola calculation (the depletion approximation) is the first thing every device engineer learns.

Step 4. Read the shape: constant $\rho$ ⟹ linear $E$ ⟹ parabolic $V$ . One glance at a measured $V(x)$ that is not parabolic tells the engineer the charge is not uniform — Poisson's equation as a diagnostic instrument, again read backwards.

Origins · 1 minSiméon Denis Poisson and how this was discovered

Siméon Denis Poisson

1781–1840 · Paris, École Polytechnique

Poisson was the model of the French mathematical physicist manufactured by the Revolution's new schools: a star student of Lagrange and Laplace at the École Polytechnique, professor there by 21, and a machine of output — over 300 works spanning celestial mechanics, elasticity, probability (the distribution bearing his name), and electricity. His 1813 result was characteristically sharp: Laplace's celebrated equation $\nabla^2V=0$ , everyone's tool for the potential of gravitating bodies, simply fails inside the body — the correct statement carries the local density as a source. Legend has it the observation was almost a footnote in a paper on spheroids; the footnote now underwrites every electrostatics solver on Earth. Poisson himself distrusted Fresnel's wave optics and predicted, as a refutation, a bright spot behind an illuminated disc — Arago promptly found the spot, and "Poisson's spot" became the wave theory's trophy. Being productively wrong twice over: the source term he added outlived his objections.

History

1782

— Laplace shows the gravitational potential of a body obeys

\nabla^2V=0

in the empty space around it; potential theory becomes a subject.

1813

— Poisson extends the equation into the interior of matter: the potential's curvature equals the local source density — Poisson's equation.

1828

— Green's self-published Essay supplies the identities and Green functions that turn Poisson's equation into a solution machine.

1845

— Thomson (Kelvin) rediscovers and republishes Green's Essay; uniqueness and boundary-value methods enter mainstream physics.

1928

— Courant, Friedrichs and Lewy analyze finite-difference schemes for elliptic equations, opening the numerical era that now dominates practice.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §2.3.1 (Eq. 2.24) and §3.1 · Jackson, Classical Electrodynamics, 3rd ed., §1.7 and §1.10 · Purcell & Morin, Electricity and Magnetism, 3rd ed., §2.9-2.12 · Zangwill, Modern Electrodynamics, §7.2-7.3

Provenance. Equation forms and signs verified against Griffiths §2.3.1, §3.1 (Poisson Eq. 2.24, uniqueness theorems), §1.5.3 (Laplacian of 1/r, Eq. 1.102); Jackson §1.7-1.10 (Green's functions, uniqueness); Purcell §2.9-2.12; Zangwill §7.2-7.3. Constant $\varepsilon_0=8.854\times10^{-12}\,\mathrm{F/m}$ is the CODATA standard value. A verifier should: (1) rerun the two-line assembly and the dimensional audit $(\mathrm{C/m^3})/(\mathrm{F/m})=\mathrm{V/m^2}$ ; (2) recompute example 1: $\rho=-\varepsilon_0(-4.0\times10^4)=3.54\times10^{-7}$ C/m³ and the Gauss cross-check $\rho/\varepsilon_0=4.0\times10^4$ ; (3) recompute the junction numbers $|E|_{\max}=eN_Dw/\varepsilon=4.63\times10^6$ V/m and $\Delta V=eN_Dw^2/2\varepsilon=0.695$ V with $\varepsilon=11.7\times8.854\times10^{-12}$ F/m; (4) verify the ball solution by differentiating $V_{\text{in}}$ to $E=\rho r/3\varepsilon_0$ and checking $V(0)=\tfrac32V(a)$ ; the delta identity (4) is proved in the E0 articles cited, not here.

Builds on: Gauss's Law in Differential Form · The Electrostatic Potential · The Curl of the Electrostatic Field · Continuous Charge Distributions Leads to: Potential of a Localized Charge Distribution · Applications of Gauss's Law and Symmetry · Coulomb's Law

Why is there a minus sign in Poisson's equation, and how do I stop dropping it?

The sign descends directly from $\mathbf{E}=-\nabla V$ — fields point downhill — combined with Gauss's law, in which positive charge is a source of outward field. Chase the two through: outward field around a point means $V$ falls off in every direction, i.e. curves downward, i.e. $\nabla^2V<0$ where $\rho>0$ . Hence the minus. The reliable habit is a one-second limit check after writing the equation: a positive point charge must come out as a potential peak. If your signs make it a well, you dropped the minus (or you are secretly in the gravitational twin, where the flip is physical: mass only attracts, so mass sits at the bottom of wells and $\nabla^2\Phi=+4\pi G\rho_m$ ).

Is Poisson's equation more fundamental than Coulomb's law, or just a rewrite?

Within electrostatics they are exactly equivalent — the second derivation walks the bridge in one direction, and integrating the point-source solution walks back. But the two package the physics differently, and the packaging matters when you generalize. Coulomb's law hard-codes action at a distance and the specific $1/r^2$ kernel; Poisson's equation states a local relation between field curvature and source that survives almost unchanged into harder theories. Add time dependence and Poisson upgrades to a wave equation (news about charges travels at $c$ — E9's retarded potentials); Coulomb's law, taken literally, becomes wrong. Add a medium and the equation absorbs it gracefully. Physicists' strong preference for local field equations over force laws starts precisely here.

If the equation is local, how does the potential at my point know about far-away charges?

Through the boundary — and this is worth getting straight, because "local" and "long-ranged" sound contradictory. Poisson's equation constrains only curvature: at each point, $V$ bends according to the charge there. But a constraint on curvature everywhere, plus data on the boundary of the region, pins the function globally — that is what solving an elliptic PDE means. A distant charge changes the values $V$ takes on your region's boundary, and the harmonic interior response transmits that change smoothly across the whole region. In the membrane picture: a weight outside your patch of membrane still tilts your patch, not by loading it but by pulling its rim.

When do I actually need the PDE instead of just integrating over the charge?

Whenever you do not know all the charge in advance — which is nearly always, once conductors appear. A grounded electrode carries induced surface charge that arranges itself; you cannot integrate over it because you do not know it yet. The boundary-value formulation cuts the knot: solve $\nabla^2V=-\rho/\varepsilon_0$ using only the known charges, imposing $V=0$ on the electrode as a boundary condition, and the induced charge comes out at the end (from the normal derivative at the surface) instead of going in at the start. Numerics is the other driver: a grid solver wants a local update rule, and the discretized Laplacian is exactly that, whereas the integral couples every pair of points.

Does Poisson's equation hold inside materials like glass or silicon?

Yes, with an honest reading of $\rho$ : it must be the total charge density — free charges you placed plus the bound charge from stretched and rotated molecules. That bound contribution is awkward to know in advance, which is why E3 introduces the displacement field and the practical variant $\nabla\cdot(\varepsilon\nabla V)=-\rho_{\text{free}}$ ; for a uniform linear dielectric this collapses to Poisson's equation with $\varepsilon_0\to\varepsilon$ , exactly the move made in the junction example. The vacuum equation is never wrong — it is just bookkeeping-heavy in matter, and the dielectric constant is the tidy summary of the bound charge's response.

What does it mean that the equation is elliptic, and why should I care?

Second-order PDEs come in three characters, and the classification decides everything about how solutions behave. Hyperbolic equations (waves) propagate signals at finite speed with memory of initial data; parabolic equations (diffusion) smooth initial data forward in time; elliptic equations like Poisson's have no time direction at all — they pose equilibrium problems, where the solution everywhere adjusts simultaneously to the sources and the boundary. Practical consequences: no ripples or shocks (solutions are as smooth as the sources allow), boundary data on a closed surface (not initial conditions) is the right input, and uniqueness theorems are strong. When E9 restores time, the operator gains a $-\partial_t^2/c^2$ and turns hyperbolic — statics is exactly the limit in which the field stops being news.

Is there a Poisson equation in quantum mechanics too?

The operator is everywhere; only the interpretation shifts. In atomic and condensed-matter physics one solves Poisson's equation for the electrostatic potential generated by the quantum charge density $\rho=-e|\psi|^2$ — that is the Hartree term of density-functional theory, computed by exactly the solvers this article describes, inside every materials-prediction code. The Schrödinger equation itself carries the same Laplacian as its kinetic term, so the numerical machinery (finite differences, spectral methods, multigrid) transfers wholesale. And in semiclassical devices, Poisson and Schrödinger are solved together self-consistently: the wavefunction sets the charge, the charge sets the potential, the potential feeds back into the Hamiltonian — a loop of two Laplacians until they agree.