Laplace's and Poisson's Equations: Uniqueness, Images and Separation of Variables

The previous chapter reduced electrostatics to a single partial differential equation, $\nabla^2 V = -\rho/\varepsilon_0$, with $\vec{E} = -\nabla V$. This chapter is about solving it when the sources are not all known in advance — the usual situation, because conductors carry induced charge that is itself part of the answer. The tools are three: the uniqueness theorems, which tell you that any solution meeting the boundary data is the solution; the method of images, which replaces a conductor by fictitious charges that reproduce its boundary condition; and separation of variables, which builds the solution as an infinite series of elementary product solutions matched to the boundary term by term.

The spirit here is Griffiths' Chapter 3. Laplace's equation $\nabla^2 V = 0$ is deceptively simple and enormously constrained: its solutions have no interior maxima or minima, they equal their own spherical averages, and they are pinned down completely by their boundary values. Every technique in this chapter is really a way of exploiting those properties. Once you believe the uniqueness theorem, solving a boundary value problem becomes a licensed guessing game: produce a potential that satisfies Laplace's (or Poisson's) equation in the region and matches the boundary data, and you are done — it must be the right one.

How to read this chapter. The uniqueness theorems (Section 2) are the logical foundation; everything after them is a construction method that uniqueness then certifies. For images, always check two things: the image charges lie outside the region where you want the field, and the boundary condition on the conductor is exactly reproduced. For separation of variables, the workflow is fixed — separate, impose the homogeneous boundary conditions to fix the mode structure, then Fourier-expand the remaining inhomogeneous condition.

1. The Two Equations and the Boundary Data

In a region $\mathcal{V}$ the potential obeys

To single out one solution you must also specify data on the boundary $\partial\mathcal{V}$. Two standard choices:

• Dirichlet: $V$ is given at every point of $\partial\mathcal{V}$ (e.g. conductors held at fixed potentials by batteries).
• Neumann: the normal derivative $\partial V/\partial n$ is given on $\partial\mathcal{V}$ (equivalently the surface charge, since $\partial V/\partial n = -\sigma/\varepsilon_0$ just outside a conductor).

Physically reasonable problems supply one or the other (or a mix, on different parts of the boundary). The uniqueness theorems say that is always exactly enough.

2. Uniqueness Theorems

First uniqueness theorem. The solution of Poisson's equation in a volume $\mathcal{V}$ is uniquely determined by $\rho$ in $\mathcal{V}$ and the value of $V$ on the boundary $\partial\mathcal{V}$.

Proof. Suppose $V_1$ and $V_2$ both solve $\nabla^2 V = -\rho/\varepsilon_0$ in $\mathcal{V}$ and agree on $\partial\mathcal{V}$. Their difference $W = V_1 - V_2$ satisfies $\nabla^2 W = 0$ in $\mathcal{V}$ and $W = 0$ on $\partial\mathcal{V}$. A harmonic function takes its maximum and minimum on the boundary (the mean-value property forbids interior extrema), so $W$ is squeezed between $0$ and $0$: $W \equiv 0$, hence $V_1 = V_2$. $\;\blacksquare$

An equivalent energy argument: $\int_{\mathcal{V}} |\nabla W|^2\,d\tau = \oint_{\partial\mathcal{V}} W\,\nabla W\cdot d\vec{a} - \int_{\mathcal{V}} W\,\nabla^2 W\,d\tau = 0$, and a non-negative integrand with zero integral must vanish, so $\nabla W = 0$ and $W$ is constant — zero, by the boundary condition.

Second uniqueness theorem. In a region containing conductors, the field is uniquely determined by $\rho$ in the region between the conductors and the total charge on each conductor.

This is the one that licenses the method of images: you do not need to know how the induced charge is distributed, only its total on each conductor (often zero net, or "grounded" meaning whatever charge is needed to hold $V = 0$).

3. Properties of Harmonic Functions

Solutions of $\nabla^2 V = 0$ satisfy:

1. Mean-value property. $V$ at the centre of any sphere (lying in the charge-free region) equals the average of $V$ over that sphere's surface.
2. No interior extrema. $V$ has no local maximum or minimum inside the region; extreme values occur only on the boundary. (Consequence: Earnshaw's theorem — no electrostatic trap.)
3. Smoothness. A harmonic function is infinitely differentiable in the interior, however rough its boundary values.

Worked example — the mean-value property in one line. Place a point charge $q$ far away and surround it by an imaginary sphere of radius $R$ centred at a field point $\vec{r}_0$ where there is no charge. The average of the point-charge potential over the sphere is $\dfrac{1}{4\pi R^2}\displaystyle\oint \dfrac{q}{4\pi\varepsilon_0\,|\vec{r} - \vec{r}_q|}\,da$, and a direct integration gives exactly $\dfrac{q}{4\pi\varepsilon_0\,|\vec{r}_0 - \vec{r}_q|} = V(\vec{r}_0)$. Superposition extends it to any charge distribution outside the sphere.

4. The Method of Images: Point Charge and Grounded Plane

A point charge $q$ sits a distance $d$ above an infinite grounded conducting plane at $z = 0$. In the region $z > 0$ we must solve $\nabla^2 V = -\dfrac{q}{\varepsilon_0}\,\delta^3(\vec{r} - d\hat{z})$ with $V = 0$ on $z = 0$ and $V \to 0$ at infinity.

Claim: in $z > 0$, the potential is the same as that of $q$ at $(0,0,d)$ together with an image charge $-q$ at $(0,0,-d)$:

This satisfies Poisson's equation in $z > 0$ (the image charge is at $z < 0$, outside the region, so it adds nothing to $\rho$ there), it vanishes on $z = 0$ by symmetry, and it dies at infinity. By the first uniqueness theorem it is the solution.

Consequences. The induced surface charge is $\sigma(s) = \varepsilon_0 E_z|_{z=0^+} = -\dfrac{qd}{2\pi(s^2+d^2)^{3/2}}$, whose integral over the plane is exactly $-q$. The force on $q$ is the Coulomb attraction to its image, $F = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{(2d)^2}$, directed toward the plane. The energy of the configuration is

— half the interaction energy of a real $q,-q$ pair at separation $2d$, because the true field fills only the half-space $z > 0$, so $\tfrac{\varepsilon_0}{2}\int E^2\,d\tau$ collects only half of what the two-charge integral would.

5. The Method of Images: Point Charge and Grounded Sphere

A charge $q$ is a distance $a$ from the centre of a grounded sphere of radius $R$ ($a > R$). Try a single image charge $q'$ on the line from the centre to $q$, at distance $b$ from the centre. Requiring $V = 0$ at the two points where that line pierces the sphere gives

One then checks that $V = 0$ on the entire sphere, not just those two points — it does. The force on $q$ is the Coulomb attraction to $q'$ at separation $a - b$:

always attractive. If instead the sphere is isolated and carries total charge $Q$, superpose a second image $Q - q'$ at the centre, chosen so the sphere's total image charge is $Q$; the second uniqueness theorem guarantees this is right.

Worked example — a charge near an isolated neutral sphere. Set $Q = 0$. The images are $q' = -qR/a$ at $b = R^2/a$ and $-q' = +qR/a$ at the centre. Far away the sphere looks neutral (the two images sum to zero), but at finite $a$ the charge $q$ still polarises it and is attracted: the near image dominates. This is the mechanism by which a charged rod attracts an uncharged conductor.

6. Separation of Variables in Cartesian Coordinates

For Laplace's equation in a region bounded by planes, seek product solutions $V(x,y,z) = X(x)\,Y(y)\,Z(z)$. Substituting and dividing by $XYZ$:

Each term depends on a different variable, so each must be a constant, with the three constants summing to zero. The signs are dictated by the boundary conditions: a direction with two grounded walls needs an oscillatory factor ($X'' / X = -k^2$, giving $\sin$/$\cos$); a direction that must decay needs a real exponential ($Z''/Z = +k^2$, giving $e^{\pm kz}$).

Worked example — the grounded slot. Two grounded planes at $x = 0$ and $x = a$, a strip at $y = 0$ held at $V_0(x)$, and $V \to 0$ as $y \to \infty$. The $x$-factor must vanish at $0$ and $a$: $X_n = \sin(n\pi x/a)$. Then $Y_n'' = (n\pi/a)^2 Y_n$, and decay picks $Y_n = e^{-n\pi y/a}$. The general solution is

the coefficients being the Fourier sine series of the strip potential. For a constant strip $V_0(x) = V_0$, only odd $n$ survive and $C_n = 4V_0/n\pi$.

7. Separation of Variables in Spherical Coordinates

With azimuthal symmetry ($V$ independent of $\phi$), Laplace's equation separates into a radial equation and the Legendre equation. The general solution is

The radial building blocks are $r^{\ell}$ (regular at the origin, blows up at infinity) and $r^{-(\ell+1)}$ (the reverse). Keep only $r^{\ell}$ in a region that includes the origin; only $r^{-(\ell+1)}$ in a region extending to infinity; both in a shell. The $P_\ell(\cos\theta)$ are orthogonal, $\int_{-1}^{1}P_\ell P_{\ell'}\,dx = \tfrac{2}{2\ell+1}\delta_{\ell\ell'}$, which is what lets you project a boundary condition onto individual $\ell$.

Worked example — conducting sphere in a uniform field. A grounded sphere of radius $R$ sits in an external field $\vec{E}_0 = E_0\hat{z}$, whose potential is $-E_0 r\cos\theta = -E_0 r P_1$. Outside the sphere write $V = -E_0 r P_1 + \sum_\ell B_\ell r^{-(\ell+1)} P_\ell$. The condition $V(R,\theta) = 0$ for all $\theta$ forces every $B_\ell = 0$ except $B_1 = E_0 R^3$:

The $R^3/r^2$ term is a pure dipole: the induced dipole moment is $p = 4\pi\varepsilon_0 R^3 E_0$. The induced surface charge is $\sigma(\theta) = -\varepsilon_0\,\partial V/\partial r|_{r=R} = 3\varepsilon_0 E_0\cos\theta$ — positive on the hemisphere toward which $\vec{E}_0$ points ($z>0$), negative on the other.

8. Common Pitfalls

• Putting an image charge inside the region of interest. The image must lie in the excluded region (behind the conductor). An image charge in the physical region would change $\rho$ there and give the wrong Poisson equation.
• Checking the boundary condition at too few points. For the grounded sphere, matching $V = 0$ at two points fixes $q'$ and $b$; you must then verify $V = 0$ over the whole sphere. It works, but the logic requires the check.
• Forgetting the half-space factor in image energies. The interaction energy of a charge with its image is not the physical energy; integrate the field energy over the physical region, which halves it for the plane.
• Keeping $r^\ell$ terms where they diverge. In an exterior region ($r \to \infty$), all $A_\ell$ with $\ell \ge 1$ must vanish (except any prescribed applied-field term); in an interior region, all $B_\ell$ must vanish.
• Using the wrong sign of separation constant. Two grounded walls in a direction require sines (negative constant); a decaying direction requires real exponentials (positive constant). Getting this backwards makes the boundary conditions impossible to satisfy.
• Truncating the series before Fourier-matching. The single term that "looks right" is usually not enough; the inhomogeneous boundary condition generally excites infinitely many modes.
• Confusing grounded with isolated. Grounded means $V = 0$ (charge free to flow); isolated means fixed total charge (potential floats). The image constructions differ.

Summary

• Uniqueness (first): $\rho$ in $\mathcal{V}$ plus $V$ on $\partial\mathcal{V}$ determine $V$ uniquely; proof via the maximum principle for the harmonic difference $W = V_1 - V_2$.
• Uniqueness (second): $\rho$ between conductors plus the total charge on each conductor determine $\vec{E}$ uniquely — the basis of the method of images.
• Harmonic functions: mean-value property, no interior extrema, infinitely smooth interior.
• Image — grounded plane: image $-q$ at the mirror point; $\sigma(s) = -qd/2\pi(s^2+d^2)^{3/2}$, force $-\tfrac{1}{4\pi\varepsilon_0}q^2/(2d)^2$, energy $-\tfrac{1}{4\pi\varepsilon_0}q^2/4d$ (half the pair value).
• Image — grounded sphere: $q' = -qR/a$ at $b = R^2/a$; force $-\tfrac{1}{4\pi\varepsilon_0}q^2 Ra/(a^2-R^2)^2$. Isolated sphere: add a central image to fix the total charge.
• Separation (Cartesian): $V = \sum_n C_n \sin(n\pi x/a)\,e^{-n\pi y/a}$ for the grounded slot; $C_n$ from a Fourier sine series of the driven boundary.
• Separation (spherical, azimuthal symmetry): $V = \sum_\ell (A_\ell r^\ell + B_\ell r^{-\ell-1})P_\ell(\cos\theta)$; conducting sphere in $E_0\hat{z}$ gives $V = -E_0(r - R^3/r^2)\cos\theta$, induced $p = 4\pi\varepsilon_0 R^3 E_0$, $\sigma = 3\varepsilon_0 E_0\cos\theta$.