ECE
class_14(E_boundary_value)

# class_14(E_boundary_value) - Electrostatic boundary value...

• Notes
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Electrostatic boundary value problem v D E E V ∇⋅ = ∇⋅ε = ρ = -∇ r r Deriving Poisson’s equation: Gauss’s law 2 v V / = -ρ ε Poisson’s equation for electrostatics: Laplace’s equation: 2 V 0 = v 0 ρ =

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Laplace operator 2 2 2 2 2 2 2 V V V V x y z = + + Cartesian: Cylindrical: 2 2 2 2 2 2 1 V 1 V V V ( ) z = ⋅ ρ + + ρ ∂ρ ∂ρ ρ ∂φ Spherical: 2 2 2 2 2 2 2 2 1 V 1 V 1 V V (r ) (sin ) r r r r sin r sin = + θ + θ ∂θ ∂θ θ ∂φ
Uniqueness Theorem If a solution to Laplace’ equation can be found that satisfies the boundary conditions, the solution is unique

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General approach for solving electrostatic field using Posson’s or Laplace’s equation Step 1. Determine the boundary conditions Step 2. Solve potential using Posson’s or Laplace’s equation Step 3. From the potential, find the electric field V E - = r r
A simple example π /4 V=0 V=1Volt Cylindrical: 2 2 2 2 2 2 1 V 1 V V V ( ) 0 z = ⋅ ρ + + = ρ ∂ρ ∂ρ ρ ∂φ V only depends on φ 2 2 2 1 V 0 = ρ ∂φ V( ) a b φ = φ + V( 0) 0 φ = = V( / 4) 1 φ = π = Boundary condition for the ODE: a 4/ ,b 0 = π = V( ) (4/ ) φ = π φ

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• Fall '08
• YifeiLi
• Electromagnet, Boundary value problem, shaded region, Boundary conditions, Green's function

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