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Unformatted text preview: Figure 7.25. A Non–Coaxial Cable. Example 7.39. A noncoaxial cable . The goal of this example is to determine the electrostatic potential inside a noncoaxial cylindrical cable, as illustrated in Figure 7.25, with prescribed constant potential values on the two bounding cylinders. Assume for definiteness that the larger cylinder has radius 1, and is centered at the origin, while the smaller cylinder has radius 2 5 , and is centered at z = 2 5 . The resulting electrostatic potential will be independent of the longitudinal coordinate, and so can be viewed as a planar potential in the annular domain contained between two circles representing the crosssections of our cylinders. The desired potential must satisfy the Dirichlet boundary value problem Δ u = 0 when  z  < 1 and vextendsingle vextendsingle z − 2 5 vextendsingle vextendsingle > 2 5 , u = a, when  z  = 1 , and u = b when vextendsingle vextendsingle z − 2 5 vextendsingle vextendsingle = 2 5 . According to Example 7.36, the linear fractional transformation ζ = 2 z − 1 z − 2 (7 . 92) maps this nonconcentric annular domain to the annulus A 1 / 2 , 1 = braceleftbig 1 2 <  ζ  < 1 bracerightbig , which is the crosssection of a coaxial cable. The corresponding transformed potential U ( ξ, η ) has the constant Dirichlet boundary conditions U = a, when  ζ  = 1 2 , and U = b when  ζ  = 1 . (7 . 93) Clearly the coaxial potential U must be a radially symmetric solution to the Laplace equation, and hence, according to (6.103), of the formequation, and hence, according to (6....
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 Fall '10
 Olver
 Fluid Dynamics, Differential Equations, Equations, Partial Differential Equations, Peter J. Olver

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