Questions: Green Function Method for Electrostatics

Question 1

What is the key practical advantage of the Green function method over expanding in eigenfunctions (Fourier series, spherical harmonics) for electrostatic boundary value problems?

• A. It works only for symmetric charge distributions, making it faster for common cases
• B. Once G is found for a geometry, any charge distribution's potential follows by a single integration — the hard work of satisfying boundary conditions is done once and encoded in G
• C. It replaces the Poisson equation with an algebraic equation, eliminating differential equations
• D. It always yields closed-form analytic solutions, unlike eigenfunction expansions which require infinite sums

Answer: (B)

The Green function encodes, once, how a given geometry responds to a unit point source at every location — including all boundary condition effects. After that, computing the potential for any charge distribution ρ(r') is just integration: φ(r) = ∫ G(r,r') ρ(r')/ε₀ d³r'. With eigenfunction methods, each new charge distribution requires re-solving the matching conditions at boundaries. The Green function moves the boundary-condition labor from 'per problem' to 'per geometry,' making it especially powerful for geometries that host many different charge configurations.

Question 2

The free-space Green function G₀(r,r') = 1/(4π|r−r'|) satisfies ∇²G₀ = −δ³(r−r'). When a grounded conducting sphere is present, why must G be modified?

• A. The Poisson equation changes form inside conductors, requiring a different differential equation
• B. The boundary condition requires G = 0 on the conducting surface, which G₀ does not satisfy
• C. The delta function source must be repositioned to lie on the conducting surface
• D. Conductors attenuate fields, so G₀ must include an exponential decay factor

Answer: (B)

For a Dirichlet problem (grounded conductor), the Green function must vanish on the boundary: G(r,r') = 0 whenever r is on the conductor surface. G₀ = 1/(4π|r−r'|) is nonzero everywhere except at r = r', so it fails to satisfy this condition. The Green function must be constructed specifically for the geometry — which is exactly where the method of images enters: adding an image charge outside the domain adjusts the potential so that G = 0 on the conductor surface.

Question 3

The Green function satisfies reciprocity: G(r,r') = G(r',r), meaning the potential at r due to a unit source at r' equals the potential at r' due to a unit source at r.

• T. True
• F. False

Answer: True

Reciprocity follows from the self-adjoint nature of the Laplacian operator. Applying Green's second identity to G(r,r') and G(r,r'') and using their defining equations yields G(r',r'') = G(r'',r'). Physically, this is the statement that mutual capacitance coefficients are symmetric: how much conductor A's potential rises per unit charge on B equals how much B's potential rises per unit charge on A. Reciprocity provides an important consistency check when computing Green functions numerically or analytically.

Question 4

The free-space Green function G₀(r,r') = 1/(4π|r−r'|) is the correct Green function to use for electrostatics problems involving grounded conducting boundaries.

• T. True
• F. False

Answer: False

G₀ is the Green function for free space only — it satisfies the delta function equation but ignores boundary conditions. A problem with a grounded conductor requires a Green function that satisfies both ∇²G = −δ³(r−r') and G = 0 on the conducting surface. This must be constructed separately for each geometry. The method of images is one technique for building the corrected G; the key physical content is that the induced surface charge on the conductor is implicitly encoded in the boundary-satisfying G.

Question 5

Explain why the Green function method is analogous to an impulse response in signal processing, and what this analogy reveals about solving the Poisson equation.

Model answer: In linear signal processing, a system is fully characterized by its impulse response — the output when the input is a delta function. Any other input is a superposition of scaled and shifted delta functions, so the output is the convolution of the input with the impulse response. The Poisson equation ∇²φ = −ρ/ε₀ is a linear PDE: the 'input' is the charge distribution ρ and the 'output' is the potential φ. The Green function G(r,r') is the response to a unit point source (delta function input) at r', satisfying ∇²G = −δ³(r−r'). Because the equation is linear, the potential for any ρ is the integral (convolution) of G with ρ. The boundary conditions are encoded in G, so this integration gives the correct potential automatically.

This analogy is exact, not just metaphorical. The Poisson operator is self-adjoint and linear, so the theory of linear operators guarantees that G exists and that the convolution integral gives the full solution. The same framework appears in heat conduction, quantum mechanics (propagators), and acoustics.