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329lect07 - 7 Poissons and Laplaces equations Summarizing...

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7 Poisson’s and Laplace’s equations Summarizing the properties of electrostatic fields we have learned so far, they satisfy the laws of electrostatics shown in the margin and, in addition, Laws of electrostatics: · E = ρ / o ∇ × E = 0 E = -∇ V as a consequence of ∇ × E = 0 . Using these relations, we can re-write Gauss’s law as · E = -∇ · ( V ) = ρ o , from which it follows that 2 V = - ρ o , (Poisson’s eqn) where 2 V 2 V x 2 + 2 V y 2 + 2 V z 2 is known as Laplacian of V . Poisson’s eqn: 2 V = - ρ o Laplace’s eqn: 2 V = 0 A special case of Poisson’s equation corresponding to having ρ ( x, y, z ) = 0 everywhere in the region of interest is 2 V = 0 . (Laplace’s eqn) 1
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Focusing our attention first on Laplace’s equation, we note that the equation can be used in charge free-regions to determine the electrostatic potential V ( x, y, z ) by matching it to specified potentials at boundaries as illustrated in the following examples: z x y z = d = 2 m V ( d ) = - 3 V V (0) = 0 z = 0 V ( z ) =? z V ( z ) V ( z ) = Az + B Example 1: Consider a pair of parallel conducting metallic plates of infinite extents in x and y directions but separated in z direction by a finite distance of d = 2 m (as shown in the margin). The conducting plates have non-zero surface charge densities (to be determined in Example 2), which are known to be responsible for an electrostatic field E = ˆ zE z measured in between the plates. Each plate has some unique and constant electrostatic potential V since neither E ( r ) nor V ( r ) can dependent the coordinates x or y given the geometry of the problem. Using Laplace’s equation determine V ( z ) and E ( z ) between the plates if the potential of the plate at z = 0 is 0 (the ground), while the potential of the plate at z = d is - 3 V. Solution: Since the potential function V = V ( z ) between the plates is only dependent on z , it follows that Laplace’s equation simplifies as 2 V = 2 V x 2 + 2 V y 2 + 2 V z 2 = 2 V z 2 = 0 .
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