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Unformatted text preview: 909 Chapter 22 1. (a) We note that the electric field points leftward at both points. Using r r F q E = , and orienting our x axis rightward (so ˆ i points right in the figure), we find ( 29 19 18 N ? 1.6 10 C 40 i ( 6.4 10 N) i C F = + × =  × r which means the magnitude of the force on the proton is 6.4 × 10 –18 N and its direction ˆ ( i) is leftward. (b) As the discussion in §222 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 = 20 N/C. 2. We note that the symbol q 2 is used in the problem statement to mean the absolute value of the negative charge which 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). 3. Since the magnitude of the electric field produced by a point charge q is given by 2   / 4 E q r πε = , where r is the distance from the charge to the point where the field has magnitude E , the magnitude of the charge is CHAPTER 22 910 ( 29 ( 29 2 2 11 9 2 2 0.50m 2.0 N C 4 5.6 10 C. 8.99 10 N m C q r E ε = π = = × × ⋅ 4. We find the charge magnitude  q  from E =  q /4 πε r 2 : ( 29( 29 2 2 10 9 2 2 1.00 N C 1.00m 4 1.11 10 C. 8.99 10 N m C q Er = π = = × × ⋅ ε 5. 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 E q R = 4 2 π ε 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 E Ze R = = × ⋅ × × = × 4 8 99 10 94 160 10 6 64 10 307 10 2 9 2 2 19 15 2 21 π ε . . . . . N m C C m N C c hb gc h c h (b) The field is normal to the surface and since the charge is positive, it points outward from the surface. 6. 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: ( 29 ( 29 9 2 2 7 5 1 1 2 2 1 9 2 2 7 5 2 2 2 2 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 πε πε × ⋅ × =  =  =  × × ⋅ × =  =  =  × r r Thus, the net electric field is 5 net 1 2 ˆ (6.39 10 N C)i E E E = + =  × r r r 7. At points between the charges, the individual electric fields are in the same direction and do not cancel. Since charge q 2 =  4.00 q 1 located at x 2 = 70 cm has a greater magnitude than q 1 = 2.1 × 10 8 C located at x 1 = 20 cm, a point of zero field must be closer to q 1 than to q 2 . It must be to the left of q 1 . 911 Let x be the coordinate of P , the point where the field vanishes. Then, the total electric , the point where the field vanishes....
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
 Summer '10
 Prof.Yang
 Physics

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