- Asked 2/13
- 2075 Chaitra · 2 marks
- 2074 Asoj · 1 mark
Derive Poisson's and Laplace's equations
Short answer
Poisson's equation: ∇²V = −ρ_v/ε, derived from Gauss's law ∇·D = ρ_v, D = εE, and E = −∇V for homogeneous medium. Laplace's equation: ∇²V = 0, obtained when ρ_v = 0.
Derive Poisson's and Laplace's equations.
Derivation of Poisson's equation
Start with three basic relations of electrostatics:
- Gauss's law, point form: ∇ · D = ρ_v
- Constitutive relation: D = εE
- Relation between field and potential: E = −∇V
Substituting (2) and (3) into (1):
∇ · (ε E) = ρ_v
∇ · (−ε ∇V) = ρ_v
For a homogeneous medium (ε constant):
∇ · ∇V = ∇²V = −ρ_v / ε ← Poisson's equation
Laplace's equation
In a region with no free charge, ρ_v = 0, and Poisson's equation reduces to
∇²V = 0 ← Laplace's equation
Forms of ∇²V
Cartesian: ∇²V = ∂²V/∂x² + ∂²V/∂y² + ∂²V/∂z²
Cylindrical: ∇²V = (1/ρ) ∂/∂ρ (ρ ∂V/∂ρ) + (1/ρ²) ∂²V/∂φ² + ∂²V/∂z²
Spherical: ∇²V = (1/r²) ∂/∂r (r² ∂V/∂r) + (1/(r² sinθ)) ∂/∂θ (sinθ ∂V/∂θ)
+ (1/(r² sin²θ)) ∂²V/∂φ²
Poisson's and Laplace's equations are used to find the potential in a region when boundary conditions (or charge distribution) are known, for example in capacitors.
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- State and prove uniqueness theorem · Electromagnetism, chapter 2 · asked 4×
- Capacitance of coaxial capacitor · Electromagnetism, chapter 2 · asked 2×
- Derive slope deflection equations · Theory of Structures II, chapter 3 · asked 4×
- Derive Cauchy-Riemann equations (Cartesian) · Applied Mathematics, chapter 1 · asked 2×
- Derive principal and shear stress equations at toe/heel · Hydropower Engineering, chapter 4
- Relation between E, G and Poisson's ratio · Strength of Materials, chapter 3 · asked 3×