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 rthan either of them. The flux through this surface is 2,rLEΦ= πwhere Eis 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 qenc= Q1 + Q2= –Q1= –3.40×10−12 C. Consequently, Gauss’ law yields 0encrLE qπε=or 12enc1222303.40 10C0.214 N/C,22 (8.85 10C / N m )(11.0 m)(20.0 1.30 10 m)qELrεπ−−−−×===−π×⋅××or || 0.214 N/C.E=(b) The negative sign in Eindicates that the field points inward. (c) Next, for r= 5.00 R1, the charge enclosed by the Gaussian surface is qencQ13.40×10−12 C. Consequently, Gauss’ law yields nc=or 12enc1222303.40 10C0.855 N/C.22 (8.85 10C / N m )(11.0 m)(5.00 1.30 10 m)
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