Questions: Laplace's and Poisson's Equations

Question 1

A physicist needs the electric potential inside a charge-free cavity in a conductor. She guesses φ = A + B·r² and verifies it satisfies ∇²φ = 0 inside the cavity and φ = 0 on the conductor surface. Can she conclude she has found the correct potential?

• A. No — guessing solutions is not rigorous; she must derive φ using systematic separation of variables
• B. No — the uniqueness theorem applies only to Dirichlet conditions on flat boundaries, not curved surfaces
• C. Yes — if a function satisfies ∇²φ = 0 inside and the boundary conditions on all surfaces, the uniqueness theorem guarantees it is the only solution
• D. Yes, but only provisionally — she must also verify Green's function compatibility

Answer: (C)

The uniqueness theorem for Laplace's equation is what makes this approach rigorous. Given Dirichlet boundary conditions (φ specified on all bounding surfaces), there is exactly one function satisfying ∇²φ = 0 inside and matching those conditions. The method of finding that function is irrelevant — guessing, separation of variables, conformal mapping, physical intuition. Once you have verified the PDE and boundary conditions are satisfied, you have the unique solution. This is what makes techniques like the method of images mathematically legitimate rather than merely a useful fiction.

Question 2

What is the fundamental advantage of reformulating an electrostatics problem using Poisson's equation (∇²φ = −ρ/ε₀) rather than applying Coulomb's law directly to every charge element?

• A. The potential φ is a scalar, and uniqueness theorems guarantee that any solution satisfying the PDE and boundary conditions is the correct and only solution
• B. Poisson's equation eliminates boundary conditions, so the solution region need not be specified
• C. Poisson's equation applies only inside conductors where Coulomb's law breaks down
• D. Coulomb's law gives the field directly while Poisson's equation only gives the potential, requiring an extra differentiation step that reduces accuracy

Answer: (A)

The Poisson/Laplace reformulation has two key advantages over Coulomb superposition: (1) the potential φ is a scalar, avoiding the vector addition of field contributions from every charge element — a dramatic simplification for complex geometries where the vector integral is intractable; (2) uniqueness theorems guarantee that any function satisfying the PDE and boundary conditions is the solution, licensing any method of solution. Option D correctly notes the differentiation step (E = −∇φ) but mischaracterizes it as a disadvantage — this trivial step is far simpler than performing vector integrals over distributed charge.

Question 3

A harmonic function (solution to Laplace's equation) can attain a local maximum value at an interior point of a charge-free region, provided the boundary values are arranged to create a sufficiently steep potential hill.

• T. True
• F. False

Answer: False

The mean value theorem for harmonic functions prohibits interior extrema. The theorem states that the value of a harmonic function at any interior point equals the average of its values over any sphere centered on that point. If φ had a local maximum at an interior point P, then φ(P) would exceed the values on a small enclosing sphere, violating the mean-value property (an average cannot exceed all of its terms). Therefore all maxima and minima of φ must lie on the boundary — a key physical result: electric potential has no peaks or valleys in empty space.

Question 4

Poisson's equation reduces to Laplace's equation in any region where the free charge density is zero.

• T. True
• F. False

Answer: True

Poisson's equation is ∇²φ = −ρ/ε₀. When ρ = 0 (no free charges in the region), the source term vanishes and the equation becomes ∇²φ = 0, which is Laplace's equation. This is why Laplace's equation governs the potential between conductors, outside charge distributions, and inside charge-free cavities — those regions have ρ = 0. Poisson's equation with ρ ≠ 0 applies where charges are present (inside a charged sphere, in a plasma, within a dielectric with polarization charge).

Question 5

What does the uniqueness theorem for Laplace's equation imply about how you are allowed to solve a boundary value problem in electrostatics?

Model answer: The uniqueness theorem states that specifying the potential φ on all boundaries of a charge-free region (Dirichlet conditions) uniquely determines the solution inside. This means you may find the solution by any method — guessing, separation of variables, the method of images, conformal mapping, physical symmetry — and if your candidate satisfies ∇²φ = 0 inside and matches the boundary values, it is provably the only solution that exists. The method of discovery is mathematically irrelevant; only verification against the PDE and boundary conditions matters.

The method of images exploits this directly: to find the potential above a grounded conducting plane with a point charge above it, you place an image charge of opposite sign below the plane (in the excluded region), construct the combined Coulomb potential, and verify φ = 0 on the plane. No image charge actually exists below the plane — it is a mathematical fiction — but uniqueness guarantees that the resulting potential, which satisfies ∇²φ = 0 in the upper half-space and φ = 0 on the boundary, must be the actual physical potential. Uniqueness is what licenses such 'tricks' as rigorous.