P24_025 - tributed uniformly along it. Thus, the total...

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25. (a) The diagram below shows a cross section (or, perhaps more appropriately, “end view”) of the charged cylinder (solid circle). Consider a Gaussian surface in the form of a cylinder with radius r and length ` , coaxial with the charged cylinder. An “end view” of the Gaussian surface is shown as a dotted circle. The charge enclosed by it is q = ρV = πr 2 ,where V = πr 2 ` is the volume of the cylinder. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ......................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . R r . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... ..... ..... .... . . . . . . . . . . . If ρ is positive, the electric Feld lines are radially outward, normal to the Gaussian surface and dis-
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Unformatted text preview: tributed uniformly along it. Thus, the total flux through the Gaussian cylinder is Φ = EA cylinder = E (2 πr` ). Now, Gauss’ law leads to 2 πε r`E = πr 2 `ρ = ⇒ E = ρr 2 ε . (b) Next, we consider a cylindrical Gaussian surface of radius r > R . If the external Feld E ext then the flux is Φ = 2 πr`E ext . The charge enclosed is the total charge in a section of the charged cylinder with length ` . That is, q = πR 2 `ρ . In this case, Gauss’ law yields 2 πε r`E ext = πR 2 `ρ = ⇒ E ext = R 2 ρ 2 ε r ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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