Question

A scalar potential \( \psi \) of a vector field \( \vec{F} \) satisfies the Laplace equation \( \nabla^2 \psi = 0 \) in free space. \( \psi \) can be uniquely determined at any point inside the closed surface \( S \) using:

• A. \( \nabla \cdot \vec{F} = 0 \)
• B. \( \nabla \times \vec{F} = 0 \)
• C. \( \psi(\vec{x}) = {constant}, \, \vec{x} \in S \)
• D. \( \nabla \cdot \vec{F} \ne 0 \)

Hint:

If a scalar potential \( \psi \) satisfies \( \nabla^2 \psi = 0 \), then it’s a harmonic function. Specifying \( \psi \) on the boundary (Dirichlet condition) uniquely determines \( \psi \) inside the region.

Answer 1

The Correct Option is C

Step 1: Understanding the Laplace equation. 
The Laplace equation \( \nabla^2 \psi = 0 \) implies that \( \psi \) is a harmonic function in the region. 
Step 2: Applying uniqueness theorem. 
According to the uniqueness theorem in potential theory, a solution \( \psi \) to the Laplace equation within a volume is uniquely determined if either:
The value of \( \psi \) is specified on the boundary (Dirichlet boundary condition), or
The normal derivative \( \frac{\partial \psi}{\partial n} \) is specified on the boundary (Neumann boundary condition).
Step 3: Analyzing the given options. 
Option (C) provides the Dirichlet condition — that \( \psi \) is known (even constant) on the boundary \( S \), which is sufficient to determine the solution uniquely inside \( S \).