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1、此資專為教學用請勿傳-楊志信 Chapter 23, HRW04, NTOUcs960329 1 Chapter 23 Gauss law 01. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is =AE rr = EAcos = (1800 N/C)(3.2103 m)2cos145 = 1.5102 Nm2/C. 07. To ex
2、ploit the symmetry of the situation, we imagine a closed Gaussian surface in the shape of a cube, of edge length d, with a proton of charge q = 1.61019 C situated at the inside center of the cube. The cube has six faces, and we expect an equal amount of flux through each face. The total amount of fl
3、ux is net = q/0, and we conclude that the flux through the square is one-sixth of that. Thus, 1 =net/6 = q/60 = 3.01109 Nm2/C. 09. Let A be the area of one face of the cube, Eu be the magnitude of the electric field at the upper face, and El be the magnitude of the field at the lower face. Since the
4、 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(ElEu). The net charge inside the cube is given by Gauss law: q = 0 = 0A(ElEu) = (8.851012 C2/Nm
5、2) (100 m)2(100 N/C 60.0 N/C) = 3.54106 C. 16. Using Eq. 23-11, the surface charge density is = 0E = (2.3105 N/C)(8.851012 C2/Nm2) = 2.0106 C/m2. 32. According to Eq. 23-13 the electric field due to either sheet of charge with surface charge density = 1.771022 C/m2 is perpendicular to the plane of t
6、he sheet (pointing away from the sheet if the charge is positive) and has magnitude E = /20. Using the superposition principle, we conclude: (a) E = /0 = (1.771022)/(8.851012) = 2.00 1011 (N/C), pointing in the upward direction, or E = (2.001011 N/C)j. (b) E = 0; (c) and, E = /0, pointing down, or E
7、 = (2.001011 N/C) j. (cf.Pb.35) 39. The forces acting on the ball are shown in the diagram below. The gravitational force has magni- tude mg, where m is the mass of the ball; the elec- trical force has magnitude qE, where q is the charge on the ball and E is the magnitude of the electric field at th
8、e position of the ball; and, the tension in the thread is denoted by T. The electric field pro- duced by the plate is normal to the plate and points to the right. Since the ball is positively charged, the electric force on it also points to the right. The tension in the thread makes the angle (= 30)
9、 with the vertical. Since the ball is in equilibrium the net force on it vanishes. The sum of the horizontal components yields qE Tsin = 0 and the sum of the vertical components yields Tcos mg = 0. The expression T = qE/sin, from the first equation, is substituted into the second to obtain qE = mg t
10、an. The electric field produced by a large uniform plane of charge is given by E = /20, where is the surface charge density. Thus, 0 tan 2 q mg = and = q mgtan2 0 (= 5.0109 C/m2) = 8 612 100 . 2 30tan)80. 9)(100 . 1)(1085. 8(2 . 35. We use Eq. 23-13. (a) To the left of the plates: E r = R E r (from
11、the right plate) + L E r (from the left one) =) 2 ( 0 ) i ( + ) 2 ( 0 i = 0. (b) To the right of the plates: E r = R E r + L E r =) 2 ( 0 i + ) 2 ( 0 ) i (= 0. (c) Between the plates: E r = R E r + L E r =) 2 ( 0 ) i (+) 2 ( 0 ) i (=)( 0 ) i ( = 2212 222 /CmN1085. 8 C/m1000. 7 ) i (= 7.911011 N/C) i
12、 (. 43. Charge is distributed uniformly over the surface of the sphere and the electric field it produces at points outside the sphere is like the field of a point particle with charge equal to the net charge on the sphere. That is, the magnitude of the field is given by E = q/40r 2, where q is the
13、magnitude of the charge on the sphere and r is the distance from the center of the sphere to the point where the field is measured. Thus, q = 40r 2E = 9 32 1099. 8 )100 . 3()15. 0( = 7.5109 (C). The field points inward, toward the sphere center, so the charge is negative: 7.5109 C. 49. At all points
14、 where there is an electric field, it is radially outward. For each part of the problem, use a Gaussian surface in the form of a sphere that is concentric with the sphere of charge and passes through the point where the electric field is to be found. The field is uniform on the surface, so EdA = 4r
15、2E, where r is the radius of the Gaussian sur- face. For r c, the charge enclosed by the Gaussian surface is zero. Gauss law yields 4r 2E = 0 E = 0. Thus, E = 0 at r = 3.50a. (g) Consider a Gaussian surface that lies completely within the conducting shell. Since the electric field is everywhere zero on the surface, EdA = 0 and, according to Gauss law, the net charge enclosed by the surface is zero. If Qi is the charge on the inner surface of the shell, then q1+Qi = 0 and Qi = q1 = 5.00 fC. (h) Let Qo be the charge on the outer surface of the shell. Since the net charge on the shell i
