Review / Chapter 3 — Conductors

The Uniqueness Theorem

Chapter 3 — Conductors

When you solve for the field in a region containing conductors, you often find yourself guessing — assuming a symmetric charge distribution, placing an image charge, picking a potential that looks right. The uniqueness theorem is your license to do that: if your guess satisfies the equations and the boundary conditions, it's the answer.

Statement

In a region of space where charge density \(\rho(\vec r)\) is specified, the potential \(V(\vec r)\) is uniquely determined by any one of the following on the boundary:

In particular: if \(N\) conductors are held at fixed potentials \(V_1,\dots,V_N\) and the charge density outside them is given, then the field everywhere is determined.

Sketch of proof (Dirichlet version)

Suppose two candidate potentials \(\phi\) and \(\psi\) both solve Poisson's equation \(\nabla^2\phi = -\rho/\epsilon_0\) with the same boundary values. Let \(W = \phi - \psi\). Then:

  1. \(\nabla^2 W = \nabla^2\phi - \nabla^2\psi = 0\) — so \(W\) satisfies Laplace's equation.
  2. \(W = 0\) on every boundary (both candidates match the same BCs).
  3. Solutions of Laplace's equation have no interior maxima or minima (mean-value property). So \(W\)'s extremes occur on the boundary — where \(W = 0\). Therefore \(W \equiv 0\) everywhere.

Conclusion: \(\phi = \psi\). The solution is unique.

Why this is so useful

The theorem legitimizes the "guess and check" style of conductor problem-solving. In particular, it's the foundation of the method of images: replace a conductor with fictitious point charges that reproduce the same boundary condition, and by uniqueness you've solved the original problem in the physical region.

Reverse method of images

Going the other direction: pick any nice charge distribution, compute its equipotentials, then "replace" one of those equipotential surfaces with a conductor at that potential. Outside the conductor, the field is unchanged. This is a useful trick for designing electrode shapes.

A corollary: Earnshaw revisited

You can't stably trap a point charge using only other static charges and conductors. A local potential minimum would require \(\nabla^2 V > 0\), impossible in a charge-free region. The uniqueness + maximum-principle structure rules it out.

Practice Problems

Problem 1easy
The inside of a closed, uncharged conducting shell is empty of charge. Argue that the field inside is zero.
Hint
Find any one solution that satisfies the boundary conditions.
Solution

Inside the cavity, Laplace's equation \(\nabla^2 V = 0\) holds, and the boundary (the inner wall of the conductor) is at some constant potential \(V_0\). The function \(V(\vec r) \equiv V_0\) satisfies both. By uniqueness it's the only solution, so \(\vec E = -\nabla V = 0\) throughout the cavity.

Answer: \(\vec E = 0\) inside, regardless of external field.

Problem 2medium
A region is bounded by two concentric conducting shells. The inner shell (radius \(R_1\)) is held at \(V_1\), the outer (radius \(R_2\)) at \(V_2\). Use the uniqueness theorem plus spherical symmetry to guess the solution.
Hint
Try \(V(r) = A + B/r\).
Solution

Between the shells, \(\nabla^2 V = 0\) and by symmetry \(V\) depends only on \(r\). The radial Laplacian gives \(\frac{1}{r^2}\frac{d}{dr}(r^2 V') = 0\), with general solution \(V(r) = A + B/r\). Applying \(V(R_1) = V_1\) and \(V(R_2) = V_2\) fixes \(A, B\). This satisfies both the PDE and the BCs, so by uniqueness it's the answer.

Answer: \(V(r) = \dfrac{V_1(R_1^{-1}-R_2^{-1})^{-1}(1/r - 1/R_2) + V_2\cdots}{\cdots}\) — any guess of the form \(A + B/r\) that matches the two endpoints works.

Problem 3medium
Two solutions \(\phi_1\) and \(\phi_2\) to a Neumann problem agree on \(E_\perp\) at every boundary. Show they differ by at most a constant.
Hint
Same \(W = \phi_1 - \phi_2\) argument, but now \(\partial W/\partial n = 0\).
Solution

\(W = \phi_1-\phi_2\) satisfies \(\nabla^2 W = 0\) and \(\partial W/\partial n = 0\) on the boundary. The identity \(\int (W\nabla W)\cdot d\vec A = \int (|\nabla W|^2 + W\nabla^2 W)\,dV\) becomes \(0 = \int |\nabla W|^2\,dV\), forcing \(\nabla W = 0\). So \(W\) is constant.

Answer: \(\phi_1 - \phi_2 = \text{const}\).

Problem 4hard
A point charge \(q\) sits at the center of a spherical cavity inside a grounded conductor. Guess the potential inside the cavity using uniqueness — no integral calculation.
Hint
The cavity wall is at \(V = 0\). Try \(V = kq/r + C\).
Solution

Inside the cavity, Poisson's equation reads \(\nabla^2 V = -q\delta(\vec r)/\epsilon_0\) (just a point source). A function of the form \(V(r) = \frac{1}{4\pi\epsilon_0}\frac{q}{r} + C\) solves this. To hit \(V=0\) at \(r=R\) (cavity radius), pick \(C = -\frac{q}{4\pi\epsilon_0 R}\). Uniqueness guarantees no other function does the job.

Answer: \(V(r) = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r} - \frac{1}{R}\right)\) for \(r < R\).