ch22 - Chapter 22 1 We note that the symbol q2 is used in...

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904 Chapter 22 1. We note that the symbol q 2 is used in the problem statement to mean the absolute value of the negative charge that resides on the larger shell. The following sketch is for 1 2 q q . The following two sketches are for the cases q 1 > q 2 (left figure) and q 1 < q 2 (right figure). 2. (a) We note that the electric field points leftward at both points. Using F q E 0 , and orienting our x axis rightward (so ˆ i points right in the figure), we find   19 18 N ? 1.6 10 C 60 i ( 9.6 10 N)i C F     which means the magnitude of the force on the proton is 9.6 10 –18 N and its direction ˆ ( i) is leftward. (b) As the discussion in Section 22-2 makes clear, the field strength is proportional to the crowdedness of the field lines. It is seen that the lines are twice as crowded at A than at B , so we conclude that E A = 2 E B . Thus, E B = 30 N/C. 3. Since the charge is uniformly distributed throughout a sphere, the electric field at the surface is exactly the same as it would be if the charge were all at the center. That is, the magnitude of the field is 2 0 4 q E R 

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905 where q is the magnitude of the total charge and R is the sphere radius. (a) The magnitude of the total charge is Ze , so       9 2 2 19 21 2 2 15 0 8.99 10 N m C 94 1.60 10 C 3.07 10 N C. 4 6.64 10 m Ze E R  (b) The field is normal to the surface and since the charge is positive, it points outward from the surface. 4. With x 1 = 6.00 cm and x 2 = 21.00 cm, the point midway between the two charges is located at x = 13.5 cm. The values of the charge are q 1 = – q 2 = – 2.00 10 –7 C, and the magnitudes and directions of the individual fields are given by:     9 2 2 7 5 1 1 2 2 0 1 9 2 2 7 5 2 2 2 2 0 2 | | (8.99 10 N m C )| 2.00 10 C| ? ? i i (3.196 10 N C)i 4 ( ) 0.135 m 0.060 m (8.99 10 N m C )(2.00 10 C) ? ? i i (3.196 10 N C)i 4 ( ) 0.135 m 0.210 m q E x x q E x x             Thus, the net electric field is 5 net 1 2 ˆ (6.39 10 N C)i E E E   5. Since the magnitude of the electric field produced by a point charge q is given by 2 0 | |/ 4 E q r , where r is the distance from the charge to the point where the field has magnitude E , the magnitude of the charge is     2 2 11 0 9 2 2 0.50m 2.0 N C 4 5.6 10 C. 8.99 10 N m C q r E 6. We find the charge magnitude | q | from E = | q |/4 0 r 2 :    2 2 10 0 9 2 2 1.00 N C 1.00m 4 1.11 10 C. 8.99 10 N m C q Er 7. The x component of the electric field at the center of the square is given by
CHAPTER 22 906   3 1 2 4 2 2 2 2 0 1 2 3 4 2 0 | | | | | | | | 1 cos45 4 ( / 2) ( / 2) ( / 2) ( / 2) 1 1 1 | | | | | | | | 4 / 2 2 0. x q q q q E a a a a q q q q a Similarly, the y component of the electric field is     3 1 2 4 2 2 2 2 0 1 2 3 4 2 0 9 2 2 8 5 2 | | | | | | | | 1 4 ( / 2) ( / 2) ( / 2) ( / 2) 1 1 1 | | | | | | | | 4 / 2 2 8.99 10 N m / C (2.0 10 C) 1 1.02 10 N/C.

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ch22 - Chapter 22 1 We note that the symbol q2 is used in...

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