P24_019 - here), so the total electric ux through the...

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19. We assume the charge density of both the conducting cylinder and the shell are uniform, and we neglect fringing. Symmetry can be used to show that the electric Feld is radial, both between the cylinder and the shell and outside the shell. It is zero, of course, inside the cylinder and inside the shell. (a) We take the Gaussian surface to be a cylinder of length L , coaxial with the given cylinders and of larger radius r than either of them. The flux through this surface is Φ = 2 πrLE ,whe re E is the magnitude of the Feld at the Gaussian surface. We may ignore any flux through the ends. Now, the charge enclosed by the Gaussian surface is q 2 q = q . Consequently, Gauss’ law yields 2 πrε 0 LE = q ,so E = q 2 πε 0 Lr . The negative sign indicates that the Feld points inward. (b) Next, we consider a cylindrical Gaussian surface whose radius places it within the shell itself. The electric Feld is zero at all points on the surface since any Feld within a conducting material would
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Unformatted text preview: here), so the total electric ux through the Gaussian surface is zero and the net charge within it is zero (by Gauss law). Since the central rod is known to have charge q , then the inner surface of the shell must have charge q . And since the shell is known to have total charge 2 q , it must therefore have charge q on its outer surface. (c) inally, we consider a cylindrical Gaussian surface whose radius places it between the outside of conducting rod and inside of the shell. Similarly to part (a), the ux through the Gaussian surface is = 2 rLE , where E is the Feld at this Gaussian surface, in the region between the rod and the shell. The charge enclosed by the Gaussian surface is only the charge q on the rod. Therefore, Gauss law yields 2 rLE = q = E = q 2 Lr . The positive sign indicates that the Feld points outward....
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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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