# ch23_p27 - 27 We assume the charge density of both the...

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uniform, and we neglect fringing effect. Symmetry can be used to show that the electric field 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 2, rLE Φ= π where E is the magnitude of the field at the Gaussian surface. We may ignore any flux through the ends. Now, the charge enclosed by the Gaussian surface is q enc = Q 1 + Q 2 = – Q 1 = –3.40 × 10 12 C. Consequently, Gauss’ law yields 0e n c rL E q πε = or 12 enc 12 2 2 3 0 3.40 10 C 0.214 N/C, 2 2 (8.85 10 C / N m )(11.0 m)(20.0 1.30 10 m) q E Lr επ −− −× == = π× × × or | | 0.214 N/C. E = (b) The negative sign in E indicates that the field points inward. (c) Next, for r = 5.00 R 1 , the charge enclosed by the Gaussian surface is q enc Q 1 3.40 × 10 12 C. Consequently, Gauss’ law yields n c = or 12 enc 12 2 2 3 0 3.40 10 C 0.855 N/C. 2 2 (8.85 10 C / N m )(11.0 m)(5.00 1.30 10 m)
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