Electromagnetism · Upper Division

Electrostatics: Boundary Problems

Laplace's and Poisson's equations govern the electric potential. Their solutions via separation of variables and the method of images are the central technical tools of classical electrostatics.

Prerequisites

Electric fields (Ch. 13) \cdot Maxwell's equations (Ch. M) \cdot Partial differential equations \cdot

Fourier series

Learning Goals

$Derive Poisson's and Laplace's equations from Gauss's law and E = -\nablaV$ .
Apply separation of variables in Cartesian coordinates to solve Laplace's equation with given boundary conditions.
Expand the solution in spherical coordinates using Legendre polynomials and match boundary conditions.
Use the method of images to find the potential and surface charge for a charge above a grounded conductor.
Construct the multipole expansion and identify the leading term for a neutral charge distribution.

ES.1 Poisson's and Laplace's Equations

From Gauss's law (∇·E = ρ/ε₀) and E = −∇V, the electric potential V satisfies Poisson's equation:

\nabla^{2}V = -\rho/\varepsilon_{0} \qquad (Poisson)

(ES.1)

In charge-free regions (ρ = 0), this reduces to Laplace's equation:

\nabla^{2}V = 0 \qquad (Laplace)

(ES.2)

Solutions to Laplace's equation are called harmonic functions. They satisfy the mean value theorem: the value at any point equals the average over any sphere centered there. Consequence: harmonic functions have no local maxima or minima in a charge-free region — the potential maximum is always on the boundary.

Theorem ES.1 — Uniqueness Theorem

A solution to Laplace'

s equation with specified boundary conditions is unique. Specifically: if two solutions V_{1}

V = 0 in a region and agree on all boundaries, then V_{1} = V_{2} everywhere inside. Proof: let

U = V_{1} - V_{2}, which satisfies \nabla^{2}U = 0 and U = 0 on boundaries. By the mean value theorem, U

has no interior extremum \to U = 0 everywhere. \blacksquare This theorem justifies the method of images

if you find any solution satisfying both Laplace's equation and the boundary conditions — by any method, however clever — it must be the correct solution.

ES.2 Separation of Variables

In Cartesian coordinates, try V(x,y,z) = X(x)Y(y)Z(z). Laplace's equation becomes:

X''/X + Y''/Y + Z''/Z = 0

(ES.3)

Each term must be a constant (separation constants). For boundary conditions that demand oscillatory behavior in x and y but exponential in z, choose: X'' = −k²X, Y'' = −l²Y, Z'' = (k²+l²)Z. The general solution is built from products of sines/cosines and exponentials. Boundary conditions then fix the allowed k, l and the coefficients (via Fourier series on the boundary).

Example ES.1 — Infinite Rectangular Pipe

$A rectangular pipe (0 \le x \le a, 0 \le y \le b) extends in z. Three walls are grounded (V=0); th$ $e top wall (y=b) is held at V_{0}(x). Find V(x,y$ .

Separated solutions:

V(x,y) = X(x)Y(y). Boundary conditions: V=0 at x=0,a \to X = sin(n\pix/a). V=0 at y=0 \to Y(0)=0

\to Y = sinh(n\piy/a

General solution:

V(x,y) = \sum(n) C_{n} sin(n\pix/a) sinh(n\piy/a

Top boundary:

At y=b: V_{0}(x) = \sum C_{n} sin(n\pix/a) sinh(n\pib/a). This is a Fourier sine series

Coefficients:

C_{n} = [2/(a sinh(n\pib/a))] \int(0 to a) V_{0}(x) sin(n\pix/a) dx

Special case:

For V_{0}(x) = V_{0} (constant): C_{n} = 4V_{0}/(n\pi sinh(n\pib/a)) for odd n, 0 for even n

ES.3 Separation in Spherical Coordinates

In spherical coordinates, Laplace's equation becomes:

(1/r^{2}) d/dr(r^{2} dV/dr) + (1/r^{2} sin \theta) \partial/\partial\theta(sin \theta \partialV/\partial\theta) + \cdots = 0

(ES.4)

For azimuthal symmetry (V independent of φ), the separated solutions are:

V(r,\theta) = \sum(l=0 to \infty) (A_{l} rˡ + B_{l} r^{-l-1}) P_{l}(cos \theta)

(ES.5)

where Pₗ(cos θ) are Legendre polynomials: P₀ = 1, P₁ = cos θ, P₂ = ½(3cos²θ−1), P₃ = ½(5cos³θ−3cosθ), ... The rˡ term is regular at the origin; r^(−l−1) is regular at infinity (multipole fields). A pure r^(−l−1) term with Pₗ is called a 2ˡ-pole: l=0 monopole, l=1 dipole, l=2 quadrupole.

Example ES.2 — Conducting Sphere in a Uniform Field

$A grounded conducting sphere of radius R is placed in a uniform external field E_{0}ẑ. Find t$ he potential outside.

Far-field:

V \to -E_{0}r cos \theta = -E_{0}r P_{1}(cos \theta) as r \to \infty

General l=1 solution:

V = (Ar + B/r^{2}) cos \theta. Far field: A = -E_{0}

Boundary condition:

V(R, \theta) = 0 \to (-E_{0}R + B/R^{2}) cos \theta = 0 \to B = E_{0}R^{3}

Result:

V(r,\theta) = -E_{0}r cos \theta + E_{0}R^{3} cos \theta / r^{2} = -E_{0}(r - R^{3}/r^{2}) cos \theta

Interpretation:

The r term is the applied field; the R^{3}/r^{2} term is an induced dipole with moment p = 4\pi\varepsilon_{0}E

_{0}R^{3}. Near the sphere (at r=R): E_{r} = -\partialV/\partialr = 3E_{0} cos \theta — the field is three times the app

lied value at the poles.

ES.4 The Method of Images

The method of images uses the uniqueness theorem to find potentials near conductors: replace the conductor with fictitious image charges placed so that the boundary condition (V = 0 on conductor surface) is satisfied.

Example ES.3 — Point Charge Above a Grounded Plane

$Charge +q at height d above an infinite grounded conducting plane (z = 0). Find V and the$ surface charge density.

Image charge:

Place -q at position z = -d (mirror image below the plane

Verify BC:

On the plane (z=0), the potential from +q and -q cancels by symmetry: V=0 ✓

Potential (z > 0):

V = q/(4\pi\varepsilon_{0}) [1/r_{+} - 1/r_{-}], where r\pm = \sqrt(x^{2}+y^{2}+(z∓d)^{2}

Surface charge:

\sigma = -\varepsilon_{0} \partialV/\partialz|_{z=0} = -qd/(2\pi(x^{2}+y^{2}+d^{2})^(3/2

Total charge:

\int\sigma dA = -q (the plane carries total charge -q, as required by Gauss's law

Force:

F = q \times (field from -q) = -q^{2}/(16\pi\varepsilon_{0}d^{2}) ẑ — attractive, as expected

Definition ES.2 — Common Traps

Uniqueness depends on boundary conditions: satisfying the PDE alone is not enough.
Image charges are mathematical devices: they replace conductors only in the allowed physical region.
Regularity chooses radial terms: discard terms that diverge at the origin or infinity when the physical region requires it.
Surface charge comes from the normal field: use the conductor boundary field, not the full image-space fiction.

Exercises — ES.1–ES.4 Electrostatics Boundary Problems

A point charge q = 2 \muC is held at d = 0.3 m above a grounded conducting plane. Using the

method of images, find the attractive force on the charge.

Straightforward

1.State Green's reciprocity theorem for electrostatics and prove it using integration by parts on Poisson's equation.

Straightforward

Solve for the potential of a dielectric sphere (relative permittivity \varepsilon_{r}) in a uniform app

lied field E_{0}. Find the field inside and the polarization

Intermediate

3.A charge q is placed between two parallel grounded conducting planes separated by distance d. Set up the method of images (infinite series). How does this connect to electrostatic energy in ionic crystals?

Intermediate

4.Derive the multipole expansion for the potential of a localized charge distribution. Find the leading terms for a neutral system and an electric dipole. What is the practical importance for intermolecular forces?

Challenging

Key Takeaways

$Poisson: \nabla^{2}V = -\rho/\varepsilon_{0}. Laplace: \nabla^{2}V = 0 in charge-free regions$ .
Uniqueness: BCs uniquely determine the solution — justifies any method that satisfies both PDE and BCs.
$Separation of variables: Cartesian \to sines/exponentials; spherical \to rˡ and P_{l}(cos \theta$ .
$Legendre polynomials P_{l}(cos \theta): monopole (l=0), dipole (l=1), quadrupole (l=2$ ...
Method of images: replace conductors with image charges; uniqueness guarantees correctness.
$Multipole expansion: V = Q/r + p\cdotr̂/r^{2} + \cdots Neutral systems \to dipole falls as 1/r^{2}$ .