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Derivation

Scalar Potential, Poisson and Laplace Equations

Statement

Because the electrostatic field is irrotational, \(\nabla\times\mathbf{E}=\mathbf{0}\), it can be written as the gradient of a scalar potential, \(\mathbf{E}=-\nabla\varphi\). Substituting this into the differential form of Gauss's law, \(\nabla\cdot\mathbf{E}=\rho/\varepsilon_0\), yields Poisson's equation \(\nabla^2\varphi=-\rho/\varepsilon_0\), whose source-free limit \(\rho=0\) is Laplace's equation \(\nabla^2\varphi=0\).

Why it matters

The scalar potential replaces the three coupled components of \(\mathbf{E}\) with a single scalar field, collapsing an over-determined first-order vector problem into one well-posed second-order scalar equation. Almost every practical electrostatics calculation — capacitors, image charges, conductors, dielectric boundary-value problems — is solved by finding \(\varphi\) first and differentiating afterwards.

Poisson's equation is the archetype of an elliptic partial differential equation, and its structure recurs verbatim in Newtonian gravitation and steady-state heat and diffusion problems. Mastering its derivation and its Laplace limit gives the analytic backbone for potential theory across physics.

Assumptions
Electrostatics (time-independent).If \(\partial\mathbf{B}/\partial t\neq 0\), Faraday's law gives \(\nabla\times\mathbf{E}=-\partial\mathbf{B}/\partial t\neq\mathbf{0}\), the field is no longer a pure gradient, and \(\varphi\) alone cannot describe \(\mathbf{E}\); a vector-potential term is required.
Linear, homogeneous medium with permittivity \(\varepsilon_0\) (or constant \(\varepsilon\)).If \(\varepsilon=\varepsilon(\mathbf{r})\) varies in space, the correct equation is \(\nabla\cdot(\varepsilon\nabla\varphi)=-\rho_{\text{free}}\), which does not reduce to \(\nabla^2\varphi=-\rho/\varepsilon\); extra \(\nabla\varepsilon\cdot\nabla\varphi\) terms appear.
Simply connected domain, so curl-free implies gradient.On a multiply connected region a curl-free field need not be a single-valued gradient; \(\varphi\) can become multivalued and the potential representation fails globally even though it holds locally.
Fields and sources smooth enough that \(\nabla\cdot(\nabla\varphi)=\nabla^2\varphi\) holds classically.At a point or surface charge, \(\rho\) is singular and derivatives exist only in the distributional sense; the equation then holds as a statement about distributions, not classical functions.
Derivation
1
\[ \nabla\times\mathbf{E}=\mathbf{0} \]
Prior result (curl-free electrostatic field): in the static regime Faraday's law has zero right-hand side. A
2
\[ \nabla\times\mathbf{E}=\mathbf{0}\;\Longrightarrow\;\mathbf{E}=-\nabla\varphi \]
Any irrotational field on a simply connected domain is a gradient, since \(\nabla\times(\nabla\varphi)=\mathbf{0}\) identically for twice-differentiable \(\varphi\). The minus sign is a convention making \(\mathbf{E}\) point from high to low potential. B
3
\[ \nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_0} \]
Prior result (Gauss's law, differential form): the divergence of \(\mathbf{E}\) is set by the local charge density. A
4
\[ \nabla\cdot\left(-\nabla\varphi\right)=\frac{\rho}{\varepsilon_0} \]
Substitute the potential representation of step 2 into Gauss's law of step 3; a legitimate replacement of equal quantities. A
5
\[ \nabla\cdot\left(\nabla\varphi\right)=\nabla^2\varphi=\frac{\partial^2\varphi}{\partial x^2}+\frac{\partial^2\varphi}{\partial y^2}+\frac{\partial^2\varphi}{\partial z^2} \]
The divergence of a gradient defines the Laplacian; valid wherever the second partials exist and commute (Clairaut's theorem). B
6
\[ -\nabla^2\varphi=\frac{\rho}{\varepsilon_0}\;\Longrightarrow\;\nabla^2\varphi=-\frac{\rho}{\varepsilon_0} \]
Insert the Laplacian identity and multiply through by \(-1\); pure algebra. This is Poisson's equation. A
7
\[ \rho=0\;\Longrightarrow\;\nabla^2\varphi=0 \]
In any region free of charge the source term vanishes, giving Laplace's equation as the homogeneous special case. A
Result
\[ \mathbf{E}=-\nabla\varphi,\qquad \nabla^2\varphi=-\frac{\rho}{\varepsilon_0},\qquad \left.\nabla^2\varphi\right|_{\rho=0}=0 \]

Reading. The electric field is the negative slope of a scalar potential landscape. The curvature of that landscape, measured by the Laplacian, is fixed pointwise by the local charge density: positive charge makes \(\varphi\) locally peaked (\(\nabla^2\varphi<0\)), and where there is no charge the potential can have no interior maxima or minima — it is the smooth interpolation set entirely by the boundaries.

Units check. \(\varphi\) is measured in volts, \(\text{V}=\text{J/C}\), so \(\nabla^2\varphi\) has units \(\text{V}\,\text{m}^{-2}\). On the right, \(\dfrac{\rho}{\varepsilon_0}\) has units \(\dfrac{\text{C}\,\text{m}^{-3}}{\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}}=\dfrac{\text{N}\,\text{m}^{-1}}{\text{C}}=\dfrac{\text{J}\,\text{m}^{-2}}{\text{C}}=\text{V}\,\text{m}^{-2}.\) Both sides are \(\text{V}\,\text{m}^{-2}\); consistent.

Limiting cases
  • Charge-free region (\(\rho=0\)): Poisson collapses to Laplace, \(\nabla^2\varphi=0\); the solution is harmonic and fixed only by boundary data.
  • Uniform field: \(\varphi=-E_0\,x\) gives \(\mathbf{E}=E_0\,\hat{\mathbf{x}}\) and \(\nabla^2\varphi=0\), consistent with zero charge in the bulk.
  • Point charge: \(\varphi=\dfrac{q}{4\pi\varepsilon_0 r}\) satisfies \(\nabla^2\varphi=0\) for \(r>0\) and reproduces \(\nabla^2\varphi=-\dfrac{q}{\varepsilon_0}\,\delta^3(\mathbf{r})\) at the origin.
  • Constant potential (\(\varphi=\text{const}\)): \(\mathbf{E}=\mathbf{0}\) and \(\rho=0\); the interior of a conductor in equilibrium.
  • Gravitational analogue: replacing \(\dfrac{\rho}{\varepsilon_0}\to -4\pi G\rho_m\) gives \(\nabla^2\Phi=4\pi G\rho_m\), the Newtonian gravitational Poisson equation.
Breaks when
  • Time-varying magnetic fields. With \(\partial\mathbf{B}/\partial t\neq\mathbf{0}\), \(\mathbf{E}\) is no longer curl-free, so \(\mathbf{E}=-\nabla\varphi\) is incomplete; one must write \(\mathbf{E}=-\nabla\varphi-\partial\mathbf{A}/\partial t\), and in a chosen gauge \(\varphi\) obeys a wave equation, not Poisson's.
  • Spatially varying permittivity or nonlinear/anisotropic media. When \(\varepsilon=\varepsilon(\mathbf{r})\) the governing law is \(\nabla\cdot(\varepsilon\nabla\varphi)=-\rho_{\text{free}}\); pulling \(\varepsilon\) outside the divergence is illegal and \(\nabla^2\varphi=-\rho/\varepsilon\) is wrong.
  • Singular or non-smooth sources treated classically. At point, line, or surface charges \(\rho\) is a distribution; the classical Laplacian does not exist there, and the equation holds only weakly, with boundary jump conditions replacing the local PDE across charged surfaces.
  • Multiply connected domains. A curl-free field around a topological hole can have nonzero circulation, so \(\varphi\) becomes multivalued and the single-valued gradient representation fails globally.
Failure modes
  • Dropping the minus sign: writing \(\mathbf{E}=+\nabla\varphi\) flips the field direction and the sign of Poisson's equation to \(\nabla^2\varphi=+\rho/\varepsilon_0\), inverting the physics of energy hills.
  • Scalar vs. vector Laplacian: \(\nabla^2\varphi\) is a scalar Laplacian; students wrongly apply Poisson to \(\mathbf{E}\) component-wise without the identity \(\nabla^2\mathbf{E}=\nabla(\nabla\cdot\mathbf{E})-\nabla\times(\nabla\times\mathbf{E})\).
  • Assuming \(\varphi=0\) means \(\mathbf{E}=\mathbf{0}\): only \(\nabla\varphi\) is physical; a point with \(\varphi=0\) can carry a large field, and adding a constant to \(\varphi\) changes nothing.
  • Believing a charge-free region forces \(\mathbf{E}=\mathbf{0}\): Laplace's equation permits strong fields (e.g. inside a capacitor) driven entirely by boundary values.
  • Expecting interior extrema of \(\varphi\) in empty space: harmonic functions obey the maximum principle, so extrema live only on boundaries; predicting a potential well in source-free bulk is impossible.
  • Using \(\nabla^2\varphi=-\rho/\varepsilon_0\) inside a dielectric: the free-charge form uses \(\varepsilon\) and \(\rho_{\text{free}}\); mixing bound and free charge with \(\varepsilon_0\) double-counts the polarization.
Discussion

The derivation shows why electrostatics is so much simpler than the full Maxwell theory. The two Maxwell laws for \(\mathbf{E}\) map onto the two halves of the potential formulation: the curl equation is spent to guarantee that a potential exists at all, and the divergence equation then fixes what that potential must be. Poisson's equation is therefore not a new postulate but the divergence law rewritten in the natural variable \(\varphi\).

Elliptic equations like Poisson's are boundary-value problems, not initial-value problems: \(\varphi\) at any interior point depends on the sources everywhere plus the values (Dirichlet data) or normal derivatives (Neumann data) on the boundary. The uniqueness theorem guarantees that specifying \(\varphi\) or \(\partial\varphi/\partial n\) on a closed surface, together with \(\rho\) inside, determines \(\varphi\) uniquely up to an additive constant. This licenses powerful shortcuts such as the method of images, where a fictitious charge reproduces the correct boundary conditions and hence, by uniqueness, the correct field.

The formal solution is the convolution of the source with the Green's function of the Laplacian, \(\varphi(\mathbf{r})=\dfrac{1}{4\pi\varepsilon_0}\displaystyle\int\dfrac{\rho(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'\), because \(G(\mathbf{r},\mathbf{r}')=\dfrac{1}{4\pi\left|\mathbf{r}-\mathbf{r}'\right|}\) satisfies \(\nabla^2 G=-\delta^3(\mathbf{r}-\mathbf{r}')\). Thus the Coulomb potential and superposition are not independent inputs; they are the integral inversion of Poisson's differential equation. The same Green's-function machinery, with the operator \(\nabla^2-\kappa^2\), yields the screened (Yukawa) potential, and with \(\nabla^2\) alone underlies gravitation and steady diffusion.

Common misconceptions. The potential is a computational scaffold, not a directly measured quantity — only potential differences and fields are physical, reflecting the gauge freedom \(\varphi\to\varphi+\text{const}\). Laplace's equation does not mean "no field"; it means "no local sources," and empty space between charged plates is its natural home. Finally, Poisson's equation is local: the Laplacian at a point sees only the charge density at that point, even though \(\varphi\) itself is nonlocal through the boundary conditions.

Worked examples
1
Parallel-plate gap: find \(\varphi(x)\) between plates at \(x=0\) (\(\varphi=0\)) and \(x=d\) (\(\varphi=V_0\)), no charge in the gap.
Symbols first. Charge-free gap \(\Rightarrow\) 1D Laplace equation. A
2
\[ \nabla^2\varphi=\frac{d^2\varphi}{dx^2}=0\;\Rightarrow\;\varphi(x)=Ax+B \]
Integrate the 1D Laplace equation twice; two constants for two boundary conditions. A
3
\[ \varphi(0)=0\Rightarrow B=0,\qquad \varphi(d)=V_0\Rightarrow A=\frac{V_0}{d} \]
Apply the Dirichlet boundary conditions. A
4
\[ \varphi(x)=\frac{V_0}{d}\,x,\qquad \mathbf{E}=-\frac{d\varphi}{dx}\,\hat{\mathbf{x}}=-\frac{V_0}{d}\,\hat{\mathbf{x}} \]
Differentiate to recover the uniform field, pointing from high to low potential. A
5
Numbers: \(V_0=12\ \text{V}\), \(d=2.0\ \text{mm}=2.0\times10^{-3}\ \text{m}\).
Insert values only after the symbolic result. A
\[ E=\frac{V_0}{d}=\frac{12}{2.0\times10^{-3}}=6.0\times10^{3}\ \text{V/m} \]

Reading. A 12 V bias across a 2 mm gap gives a uniform 6 kV/m field directed from the high plate to the low. The linear profile is the unique harmonic function matching the plate voltages.

1
Uniformly charged ball: find \(\varphi\) at the centre of a ball of radius \(R\), total charge \(Q\), uniform \(\rho\), using Poisson inside.
Symbols first. Spherical symmetry \(\Rightarrow\) \(\varphi=\varphi(r)\). B
2
\[ \frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{d\varphi}{dr}\right)=-\frac{\rho}{\varepsilon_0},\qquad \rho=\frac{Q}{\tfrac{4}{3}\pi R^3} \]
Radial Laplacian in Poisson's equation; density from total charge over volume. B
3
\[ \varphi_{\text{in}}(r)=\frac{Q}{4\pi\varepsilon_0}\,\frac{3R^2-r^2}{2R^3} \]
Integrate twice, discard the singular \(1/r\) part by regularity at \(r=0\), then match \(\varphi\) and \(d\varphi/dr\) to the exterior \(Q/4\pi\varepsilon_0 r\) at \(r=R\). C
4
\[ \varphi(0)=\frac{Q}{4\pi\varepsilon_0}\,\frac{3R^2}{2R^3}=\frac{3}{2}\,\frac{Q}{4\pi\varepsilon_0 R} \]
Evaluate at the centre; it is \(3/2\) times the surface potential. B
5
Numbers: \(Q=1.0\ \text{nC}=1.0\times10^{-9}\ \text{C}\), \(R=5.0\ \text{cm}=0.050\ \text{m}\), \(\dfrac{1}{4\pi\varepsilon_0}=8.99\times10^{9}\ \text{N}\,\text{m}^2\,\text{C}^{-2}\).
Insert values. A
\[ \varphi(0)=\frac{3}{2}\left(8.99\times10^{9}\right)\frac{1.0\times10^{-9}}{0.050}\approx 2.7\times10^{2}\ \text{V} \]

Reading. The potential peaks at the centre near 270 V, one and a half times the 180 V surface value, because building charge uniformly through the interior stacks contributions from all shells. The interior obeys Poisson (source present); the exterior obeys Laplace.

Problems
  1. Sign and structure. Starting from \(\mathbf{E}=-\nabla\varphi\) and \(\nabla\cdot\mathbf{E}=\rho/\varepsilon_0\), derive Poisson's equation and state which prior result supplies the potential representation.
    Solution Take the divergence: \(\nabla\cdot\mathbf{E}=\nabla\cdot(-\nabla\varphi)=-\nabla^2\varphi\). Set equal to \(\rho/\varepsilon_0\): \(-\nabla^2\varphi=\rho/\varepsilon_0\), i.e. \(\nabla^2\varphi=-\rho/\varepsilon_0\). The representation \(\mathbf{E}=-\nabla\varphi\) is licensed by the curl-free result \(\nabla\times\mathbf{E}=\mathbf{0}\) on a simply connected domain, since \(\nabla\times(\nabla\varphi)\equiv\mathbf{0}\).
  2. 1D Poisson slab. An infinite slab \(0\le x\le a\) carries uniform \(\rho_0\); outside it \(\rho=0\). With \(\varphi(0)=0\) and \(\left.\dfrac{d\varphi}{dx}\right|_{0}=0\), find \(\varphi(x)\) inside the slab and the field \(E_x\).
    Solution \(\dfrac{d^2\varphi}{dx^2}=-\dfrac{\rho_0}{\varepsilon_0}\). Integrate once: \(\dfrac{d\varphi}{dx}=-\dfrac{\rho_0}{\varepsilon_0}x+C_1\); the condition \(\varphi'(0)=0\) gives \(C_1=0\). Integrate again: \(\varphi=-\dfrac{\rho_0}{2\varepsilon_0}x^2+C_2\); \(\varphi(0)=0\) gives \(C_2=0\). So \(\varphi(x)=-\dfrac{\rho_0}{2\varepsilon_0}x^2\), and \(E_x=-\dfrac{d\varphi}{dx}=\dfrac{\rho_0}{\varepsilon_0}x\), linear in \(x\).
  3. Harmonic check. Show that \(\varphi(x,y)=x^2-y^2\) satisfies Laplace's equation, find the field, and say whether such a \(\varphi\) can exist in a charge-free region.
    Solution \(\dfrac{\partial^2\varphi}{\partial x^2}=2\), \(\dfrac{\partial^2\varphi}{\partial y^2}=-2\), so \(\nabla^2\varphi=2-2=0\): harmonic. Field \(\mathbf{E}=-\nabla\varphi=(-2x,\,2y,\,0)\). Yes — since \(\nabla^2\varphi=0\) it corresponds to \(\rho=0\). The critical point at the origin is a saddle, not an extremum, illustrating the maximum principle: harmonic functions have no interior extrema.
  4. Density from potential. In a region \(\varphi(x)=\alpha x^2\) with \(\alpha=50\ \text{V/m}^2\). Find \(\rho\) and evaluate \(E\) at \(x=0.10\ \text{m}\). Use \(\varepsilon_0=8.85\times10^{-12}\ \text{F/m}\).
    Solution \(\nabla^2\varphi=\dfrac{d^2\varphi}{dx^2}=2\alpha=100\ \text{V/m}^2\). Poisson: \(\rho=-\varepsilon_0\nabla^2\varphi=-(8.85\times10^{-12})(100)=-8.85\times10^{-10}\ \text{C/m}^3\). Field \(E_x=-\dfrac{d\varphi}{dx}=-2\alpha x=-2(50)(0.10)=-10\ \text{V/m}\); magnitude \(10\ \text{V/m}\) along \(-x\).
  5. Spherical Poisson, centre-to-surface. A ball of radius \(R=0.10\ \text{m}\) carries uniform \(\rho=2.0\times10^{-8}\ \text{C/m}^3\). Compute the potential difference \(\varphi(0)-\varphi(R)\).
    Solution Inside, \(\varphi(r)=\dfrac{Q}{4\pi\varepsilon_0}\dfrac{3R^2-r^2}{2R^3}\), so \(\varphi(0)=\dfrac{Q}{4\pi\varepsilon_0}\dfrac{3}{2R}\) and \(\varphi(R)=\dfrac{Q}{4\pi\varepsilon_0}\dfrac{1}{R}\). Hence \(\varphi(0)-\varphi(R)=\dfrac{Q}{4\pi\varepsilon_0}\left(\dfrac{3}{2R}-\dfrac{1}{R}\right)=\dfrac{Q}{8\pi\varepsilon_0 R}\). Writing \(Q=\tfrac{4}{3}\pi R^3\rho\) gives the compact form \(\Delta\varphi=\dfrac{\rho R^2}{6\varepsilon_0}\). Numerically \(\Delta\varphi=\dfrac{(2.0\times10^{-8})(0.10)^2}{6(8.85\times10^{-12})}=\dfrac{2.0\times10^{-10}}{5.31\times10^{-11}}\approx 3.8\ \text{V}\). The centre is higher by about 3.8 V.