Notes6 - Poisson and Laplace's Equations Recall Gauss's law...

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11/3/03 Poisson and Laplace's Equations Recall Gauss's law and Faraday's law (under electrostatic conditions) in differential form: free charge! ( ) v == ε ρ ⋅⋅ DE ∇∇ =0 = × ⇒− EE Φ Substituting E into Gauss's law yields =− Φ ( ) v −Φ = . Assuming that the permittivity is a constant throughout the given region (i.e. the medium is homogeneous), then can be taken out of the divergence operation. Thus, free charge! () v ⋅Φ = . Substituting for the del operator in rectangular coordinates, 222 2 ˆˆˆ ˆ ˆ ˆ , x yz x y z xyz ⎛⎞ ∂∂∂ ∂ Φ∂ Φ Φ = + + + + ⎜⎟ ∂∂∂ ∂ ∂ ∂ ⎝⎠ ∂Φ =++ ≡∇ Φ xyz x y z we obtain Poisson's equation : ∇= 2 Φ v . Poisson's Equation The operator 22 2 2 2 x + + in rectangular coordinates and is referred to as the Laplacian operator . In cylindrical coordinates, it is
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22 2 2 11 z ρ ρρ φ ⎛⎞ ∂Φ ∇Φ= + + ⎜⎟ ∂∂ ⎝⎠ and in spherical coordinates is 2 2 2 1 sin sin sin r rr r r r θ 2 θθ Φ Φ + + Φ . Poisson’s equation reduces to Laplace's equation in a region in which there are no volume sources. That is, letting v = 0 in Poisson's equation yields ∇= 2 0 Φ Laplace's Equation It is important to recognize that the potential may be obtained by solving these equations with the correct sources and subject to the appropriate boundary conditions. EXAMPLE: Parallel-Plate Capacitor o V Φ = z 0 Φ = ρ s - ρ s ε x h V o Note a reference potential Φ= has been chosen on the bottom plate––this choice is arbitrary. But then the potential on the top conductor must be 0 o V Φ = to have the correct potential difference as determined by the impressed DC source. Since the region between the conductors can be considered source free (no free charge should exist between the plates, so 0 v = there), Poisson’s equation becomes Laplace's equation, 222 2 0, xyz + + = ∂∂∂
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This note was uploaded on 08/05/2011 for the course ECE 2317 taught by Professor Staff during the Spring '08 term at University of Houston.

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Notes6 - Poisson and Laplace's Equations Recall Gauss's law...

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