ssm_ch23 - Chapter 23 1 ~ ~ The vector area A and the...

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Chapter 23 1 The vector area ~ A and the electric field ~ E are shown on the dia- gram to the right. The angle μ between them is 180 ± ¡ 35 ± = 145 ± , so the electric flux through the area is © = ~ E ¢ ~ A = EA cos μ = (1800 N = C)(3 : 2 £ 10 ¡ 3 m) 2 cos 145 ± = ¡ 1 : 5 £ 10 ¡ 2 N ¢ m 2 = C. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. μ A E 35 ± 9 Let A be the area of one face of the cube, E u be the magnitude of the electric field at the upper face, and E ` be the magnitude of the field at the lower face. Since the field is downward, the flux through the upper face is negative and the flux through the lower face is positive. The flux through the other faces is zero, so the total flux through the cube surface is © = A ( E ` ¡ E u ). The net charge inside the cube is given by Gauss’ law: q = ² 0 © = ² 0 A ( E ` ¡ E u )=(8 : 85 £ 10 ¡ 12 C 2 = N ¢ m 2 )(100 m) 2 (100 N = C ¡ 60 : 0N = C) =3 : 54 £ 10 ¡ 6 C=3 : 54 ¹ C : 19 (a) The charge on the surface of the sphere is the product of the surface charge density ¾ and the surface area of the sphere (4 ¼r 2 , where r is the radius). Thus q =4 ¼r 2 ¾ =4 ¼ μ 1 : 2m 2 2 (8 : 1 £ 10 ¡ 6 C = m 2 )=3 : 7 £ 10 ¡ 5 C : (b) Choose a Gaussian surface in the form a sphere, concentric with the conducting sphere and with a slightly larger radius. The flux through the surface is given by Gauss’ law: © = q ² 0 = 3 : 7 £ 10 ¡ 5 C 8 : 85 £ 10 ¡ 12 C 2 = N ¢ m 2 =4 : 1 £ 10 6 N ¢ m 2 = C : 23 The magnitude of the electric field produced by a uniformly charged infinite line is E = ¸= 2 ¼² 0 r , where ¸ is the linear charge density and r is the distance from the line to the point where the field is measured. See Eq. 23–12. Thus ¸ =2 ¼² 0 Er =2 ¼ (8 : 85 £ 10 ¡ 12 C 2 = N ¢ m 2 )(4 : 5 £ 10 4 N = C)(2 : 0m)=5 : 0 £ 10 ¡ 6 C = m : 142 Chapter 23
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27 Assume the charge density of both the conducting rod and the shell are uniform. Neglect fringing. Symmetry can be used to show that the electric field is radial, both between the rod and the shell and outside the shell. It is zero, of course, inside the rod and inside the shell since they are conductors. (a) and (b) Take the Gaussian surface to be a cylinder of length L and radius r ,con c en t r i c with the conducting rod and shell and with its curved surface outside the shell. The area of the curved surface is 2 ¼rL . The field is normal to the curved portion of the surface and has uniform magnitude over it, so the flux through this portion of the surface is © =2 ¼rLE ,whe re E is the magnitude of the field at the Gaussian surface. The flux through the ends is zero. The charge
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This note was uploaded on 01/19/2009 for the course PHYS 171.102 taught by Professor Heckman during the Fall '98 term at Johns Hopkins.

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ssm_ch23 - Chapter 23 1 ~ ~ The vector area A and the...

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