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Kapittel 27 Løsning på alle oppgavene finnes på CD. Lånes ut hvis behov. 27.1. Model: The electric field is that of the two charges placed on the y-axis. Visualize: Please refer to Figure EX27.1. We denote the upper charge by q1 and the lower charge by q2. Because both the charges are positive, their electric fields at P are directed away from the charges. Solve: The electric field from q1 is 9 2 2 9 1 q1 9.0 10 N m /C 3.0 10 C E1 , below x -axis cos iˆ sin ˆj 2 2 2 0.050 m 0.050 m 4 0 r1 Because tan 5 cm 5 cm 1, the angle 45. Hence, 1 ˆ 1 ˆ E1 5400 N/C i j 2 2 Similarly, the electric field from q2 is 1 q2 1 ˆ 1 E2 , above x -axis 5400 N/C i 2 4 r 2 2 0 2 ˆj 1 ˆ 3ˆ Enet at P E1 E2 2 5400 N/C i 7.6 10 i N/C 2 Thus, the strength of the electric field is 7.6103 N/C and its direction is horizontal. Assess: Because the charges are located symmetrically on either side of the x-axis and are of equal value, the ycomponents of their fields will cancel when added. 27.3. Model: The electric field is that due to superposition of the fields of the two 3.0 nC charges located on the yaxis. Visualize: Please refer to Figure EX27.3. We denote the top 3.0 nC charge by q1 and the bottom 3.0 nC charge by q2. The electric fields ( E1 and E2 ) of both the positive charges are directed away from their respective charges. With vector addition, they yield the net electric field E net at the point P indicated by the dot. Solve: The electric fields from q1 and q2 are 9 2 2 9 1 q1 9.0 10 N m /C 3.0 10 C ˆ E1 , along x -axis i 10,800iˆ N/C 2 2 0.05 m 4 0 r1 1 q2 E2 , above x -axis 2 4 r 0 2 Because tan 10 cm 5 cm, tan 1 2 63.43. So, 9.0 10 9 E2 N m 2 /C 2 3.0 109 C 0.10 m 2 0.050 m 2 cos63.43 iˆ sin 63.43 ˆj 966iˆ 1127 ˆj N/C The net electric field is thus Enet at P E1 E2 11,766iˆ 1127 ˆj N/C To find the angle this net vector makes with the x-axis, we calculate tan 1127 N/C 5.5 11,766 N/C Thus, the strength of the electric field at P is Enet 11,766 N/C 1127 N/C 2 and E net makes an angle of 5.5 above the x-axis. 2 11,820 N/C = 1.18103 N/C Assess: Because of the inverse square dependence on distance, E2 E1. Additionally, because the point P has no special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis. 27.7. Model: We will assume that the wire is thin and that the charge lies on the wire along a line. Solve: is From Equation 27.15, the electric field strength of an infinitely long line of charge having linear charge density Eline Eline r 5.0 cm 2 1 4 0 5.0 102 m 1 2 4 0 r Eline r 10.0 cm 2 1 4 0 10.0 102 m Dividing the above two equations gives 5.0 102 m 1 3 Eline r 10.0 cm Eline r 5.0 cm 2000 N/C 1.00 10 N/C 1000 N/C 2 2 10.0 10 m 27.10. Model: The rod is thin, so assume the charge lies along a line. Visualize: Solve: The force on charge q is F qErod . From Example 27.3, the electric field a distance r from the center of a charged rod is Erod 1 Q 4 0 r r 2 L / 2 2 iˆ 9.0 109 N m 2 /C 2 40 10 9 C iˆ 0.04 m 0.04 m 0.05 m 2 5 1.406 105 iˆ N/C Thus, the force is F 6.0 109 C 1.406 105 N/C iˆ 8.4 104 iˆ N More generally, F 8.4 104 N, away from the rod . 27.11. Model: Assume that the rings are thin and that the charge lies along circle of radius R. Visualize: The rings are centered on the z-axis. Solve: (a) According to Example 27.5, the field of the left (negative) ring at z 10 cm is E1 z 4 0 z R 2 9.0 10 9 zQ1 2 3/ 2 N m 2 /C2 0.10 m 20 109 C 2 3/ 2 0.10 m 0.050 m 2 1.288 104 N/C That is, the field is E1 1.288 104 N/C, left . Ring 2 has the same quantity of charge and is at the same distance, so it will produce a field of the same strength. Because Q2 is positive, E2 will also point to the left. The net field at the midpoint is E E1 E2 2.6 104 N/C, left (b) The force is F qE 1.0 109 C 2.6 104 N/C, left 2.6 10 5 N, right 27.22. Model: A uniform electric field causes a charge to undergo constant acceleration. Solve: Kinematics yields the acceleration of the electron. v12 v0 2 2ax a v12 v0 2 (4.0 107 m/s) 2 (2.0 107 m/s) 2 5.0 1016 m/s 2 2x 2 0.012 m The magnitude of the electric field required to obtain this acceleration is E Fnet me a (9.11 1031 kg)(5.0 1016 m/s 2 ) 2.8 105 N/C. e e 1.6 1019 C 27.23. Model: The infinite charged plane produces a uniform electric field. Solve: (a) The electric field of a plane of charge with surface charge density is E 2 0 F ma 2 0 E 2 0 2 0 q q where m is the mass, q is the charge, and a is the acceleration of the electron. To obtain we must first find a. From the kinematic of motion equation v12 v02 2ax, a v12 v02 (1.0 107 m/s)2 (0 m/s)2 2.5 1015 m/s2 2x 2(2.0 102 m) 2(8.85 1012 C2 /N m2 )(9.11 1031 kg)(2.5 1015 m/s2 ) 2.5 107 C/m2 1.60 1019 C (b) Using the kinematic equation of motion v1 v0 at , t v1 v0 1.0 107 m/s 0 m/s 4.0 109 s a 2.5 1015 m/s2 27.30. Model: The electric field is that of three point charges q1, q2, and q3. Visualize: Please refer to Figure P27.30. Assume the charges are in the x-y plane. The 10 nC charge is q1, the 10 nC charge is q3, and the –5.0 nC charge is q2. The net electric field at the dot is Enet E1 E2 E3 . The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components. Solve: (a) The electric field produced by q1 is E1 9 2 2 9 q1 9.0 10 N m /C 10 10 C 100,000 N/C 2 4 0 r12 0.030 m 1 E1 points away from q1, so in component form E1 100,000 ˆj N/C . The electric field produced by q2 is E2 18,000 N/C . E2 points towards q2, so E2 18,000iˆ N/C. Finally, the electric field produced by q3 is E3 9 2 2 9 q3 9.0 10 N m /C 10 10 C 26,471 N/C 2 2 4 0 r32 0.030 m 0.050 m 1 E3 points toward q3 and makes an angle tan 1 3/5 31 with the x-axis. So, E3 E3 cos iˆ E3 sin ˆj 22,700iˆ 13,634 ˆj N/C Adding these three vectors gives Enet E1 E2 E3 40,700iˆ 86,366 ˆj N/C 4.1104 iˆ 8.6 104 ˆj N/C This is in component form. (b) The magnitude of the field is Enet Ex2 Ey2 40,700 N/C 86,366 N/C 2 2 95,476 N/C 9.5 104 N/C and its angle from the x-axis is tan 1 Ex Ey 25 . We can also write Enet (9.5 104 N/C, 25 CW from the x-axis). 27.38. Model: The fields are those of two infinite lines of charge with linear charge density . Visualize: Solve: The field at distance r from an infinite line of charge is E 1 2 rˆ 4 0 r It points straight away from the line. With two perpendicular lines, the field due to the line along the x-axis points in the y-direction and depends inversely on distance y. Similarly, the field due to the line along the y-axis points in the xdirection and depends inversely on distance x. That is Ex 1 2 4 0 x Ey E Exiˆ E y ˆj 1 2 4 0 y 2 1 ˆ 1 ˆ i j 4 0 x y The field strength at this point in space is E Ex2 E y2 2 4 0 1/x 2 1/ y 2 27.42. Model: The electric field is that of a line of charge of length L. Visualize: The origin of the coordinate system is at the center of the rod. Divide the rod into many small segments of charge q and length x. Solve: (a) Segment i creates a small electric field Ei at point P that points to the right. The net field E will point to the right and have Ey Ez 0 N/C. The distance to segment i is x, so Ei Ei x q 4 0 x xi 2 Ex Ei x 1 q 4 0 x xi 2 q is not a coordinate, so before converting the sum to an integral we must relate the charge q to length x. This is done through the linear charge density Q/L, from which q x Q x L With this charge, the sum becomes Ex Q / L 4 0 x x x i 2 i Now we let x dx and replace the sum by an integral from x 12 L to x 12 L. Thus, Ex Q / L L / 2 4 0 dx x x 2 L / 2 Q / L 1 Q 1 Q / L L E iˆ 4 0 x 2 L2 4 4 0 x x L / 2 4 0 x 2 L2 4 L/2 The electric field strength at x is E 1 Q 4 0 x L2 4 2 (b) For x L , E 1 Q 4 0 x 2 That is, the line charge behaves like a point charge. (c) Substituting into the above formula 3.0 109 C E 9.0 109 N m 2 /C2 9.8 104 N/C 2 2 2 2 1 3.0 10 m 5.0 10 m 2 27.45. Model: Assume that the ring of charge is thin and that the charge lies along circle of radius R. Solve: (a) From Example 27.5, the on-axis field of a ring of charge Q and radius R is Ez 1 zQ 4 0 z R 2 3/ 2 2 For the field to be maximum at a particular value of z, dE dz 0. Taking the derivative, z 3/ 2 2 z dE Q 1 1 3z 2 0 2 3 / 2 5 / 2 3 2 5/ 2 dz 4 0 z R 2 z 2 R2 z 2 R2 z 2 R2 1 (b) The field strength at the point z R Ez max Q 4 0 R 2 R 2 R 2 2 3/ 2 3z 2 R z 2 2 z R 2 2 is 2 Q 2 3 3 4 0 R 27.49. Motion: The very wide charged electrode is a plane of charge. Solve: For a plane of charge with surface charge density , the electric field is E plane z 2 0 2 0 Eplane z Q Q A R2 Q 2 R 2 0 Eplane 2 0.0050 m 8.85 1012 C2 /N m 2 1000 N/C 1.39 1012 C 1.39 103 nC 2 z Assess: Note that the field strength of a plane of charge does not depend on z and is a constant. We did not have to use the distance z 5.0 cm given in the problem.