Questions: Boundary Value Problems in Electrostatics

Question 1

A physicist uses the method of images — replacing a grounded conductor with a fictitious charge placed below the surface — to find the potential above the conductor. The resulting potential satisfies Laplace's equation above the conductor and equals zero on its surface. Is this the correct physical solution?

• A. No — the image charge is not physically real, so the potential it produces cannot represent the actual field
• B. Yes — the uniqueness theorem guarantees that any function satisfying Laplace's equation and the boundary conditions is the unique correct solution
• C. Only approximately — the method of images gives the right answer far from the surface but fails near it
• D. Only if the image charge was derived by solving the full boundary value problem, not guessed

Answer: (B)

The uniqueness theorem is the justification: given Poisson's (or Laplace's) equation in a region and appropriate boundary conditions on its surface, the solution is unique. If you can find any function — by whatever means, including inspired guessing — that satisfies both the differential equation and the boundary conditions, that function must be the correct answer. The method of images exploits this: the image charge is unphysical inside the conductor, but above the surface it reproduces the correct potential, so uniqueness guarantees it is the solution.

Question 2

What information is required to uniquely determine the electrostatic potential in a bounded region?

• A. Only the charge distribution inside the region — boundary conditions are derivable from the charges
• B. Only the boundary conditions — the charge distribution inside is determined by the field equations
• C. Poisson's equation in the region (specifying charge distribution ρ) and appropriate boundary conditions on the enclosing surface
• D. The full charge distribution throughout all of space, not just the region of interest

Answer: (C)

Poisson's equation ∇²Φ = −ρ/ε₀ constrains the potential inside the region based on charge distribution, but by itself has infinitely many solutions — all harmonic functions satisfying it. The boundary conditions (Dirichlet, specifying Φ on the surface; Neumann, specifying ∂Φ/∂n; or mixed) select the unique physical solution. Neither piece alone is sufficient: you need both the PDE and the boundary data.

Question 3

Dirichlet boundary conditions specify the value of the electrostatic potential on a bounding surface, while Neumann boundary conditions specify the normal derivative of the potential on the surface.

• T. True
• F. False

Answer: True

Correct. A grounded conductor enforces a Dirichlet condition (Φ = 0 on the surface). Specifying the surface charge density σ enforces a Neumann condition, since E_n = σ/ε₀ at a conductor surface and E = −∇Φ, so ∂Φ/∂n = −σ/ε₀. Mixed boundary conditions can specify Dirichlet on part of the boundary and Neumann on the rest. The uniqueness theorem holds for all three types.

Question 4

Because Laplace's equation has infinitely many harmonic solutions, knowing the potential on the boundary of a region is not sufficient to uniquely determine the potential inside.

• T. True
• F. False

Answer: False

The uniqueness theorem says the opposite: Dirichlet boundary conditions (specifying Φ everywhere on the bounding surface) uniquely determine the potential inside. The infinite family of harmonic functions is large, but the constraint of matching prescribed values on the entire boundary singles out exactly one member. This is the theorem's entire point — and it is what justifies creative solution methods. Neumann conditions (specifying the normal derivative) also uniquely determine the solution up to an additive constant.

Question 5

Explain the significance of the uniqueness theorem for boundary value problems in electrostatics, and describe how it justifies solution methods like the method of images.

Model answer: The uniqueness theorem states that given Poisson's equation in a region and appropriate boundary conditions on its surface (Dirichlet, Neumann, or mixed), the electrostatic potential is uniquely determined. This matters because it converts the problem of 'finding the solution' into the problem of 'verifying a candidate solution.' If you can produce any function — by symmetry arguments, physical intuition, or outright guessing — that satisfies both the differential equation and the boundary conditions, uniqueness guarantees it is the correct physical answer. The method of images exploits this: place fictitious image charges to reproduce the correct boundary conditions, verify Laplace's equation is satisfied, and uniqueness does the rest.

Without uniqueness, you could never trust a solution obtained by non-systematic means. With it, the mathematician's instinct — 'find something that works and prove it satisfies the conditions' — becomes a rigorous strategy. This is why Griffiths calls the method of images 'a legitimate short-cut' rather than mere guessing.