This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 Poisson and Laplace's Equations Recall Gauss's law and Faraday's law (under electrostatic conditions) in differential form: ∇⋅ ⇒ ∇⋅ D E = = v v ρ ε ρ ∇ × ⇒ ∇Φ E = E =  Substituting E =  ∇Φ into Gauss's law yields ∇⋅ − ∇Φ = ε ρ b g v . Assuming that the permittivity is a constant throughout the given region (i.e. the medium is homogeneous), then ε can be taken out of the divergence operation. Thus, ∇⋅ −∇Φ = b g ρ ε v . By substituting in for the del operator, ∇⋅∇Φ = ∂ ∂ + ∂ ∂ + ∂ ∂ F H G I K J ⋅ ∂Φ ∂ + ∂Φ ∂ + ∂Φ ∂ F H G I K J = ∂ ∂ + ∂ ∂ + ∂ ∂ ≡ ∇ ¡ ¡ ¡ ¡ ¡ ¡ , x y z x y z x y z x y z x y z 2 2 2 2 2 2 2 Φ Φ Φ Φ Poisson's equation is obtained: ∇ = − 2 Φ ρ ε v . The ∇ 2 operator is referred to as the Laplacian operator and is defined (as above in rectangular coordinates) as This set of notes is related to the material in Section 3.11 in the text by D.K. Cheng. 7/23/99 Poisson's Equation 2 ∇ = ∂ ∂ ∂Φ ∂ F H G I K J + ∂ ∂ + ∂ ∂ 2 2 2 2 2 2 1 1 Φ Φ Φ ρ ρ ρ ρ ρ φ z in cylindrical coordinates, and as ∇ = ∂ ∂ ∂Φ ∂ F H G I K J + ∂ ∂ ∂Φ ∂ F H G I K J + ∂ ∂ 2 2 2 2 2 2 2 2 1 1 1 Φ Φ r r r r r r sin sin sin θ θ θ θ θ φ in spherical coordinates. The equation reduces to Laplace's equation in a sourcefree region. That is, letting ρ v = 0 in Poisson's equation yields ∇ = 2 Φ It is important to recognize that the solution for the potential is obtained by solving these equations with the correct sources and subject to the appropriate boundary conditions. equations with the correct sources and subject to the appropriate boundary conditions....
View
Full Document
 Fall '08
 Staff
 Electrostatics, Laplace, Electric charge, potential function, Boundary conditions

Click to edit the document details