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

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1. 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 12 qq = . The following two sketches are for the cases q 1 > q 2 (left figure) and q 1 < q 2 (right figure).
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2. (a) We note that the electric field points leftward at both points. Using GG Fq E = 0 , and orienting our x axis rightward (so ˆ i points right in the figure), we find G F =+ × F H G I K J =− × −− 16 10 40 6 4 10 19 18 . # . # C N C iN i c h 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 §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 = 20 N/C.
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3. The following diagram is an edge view of the disk and shows the field lines above it. Near the disk, the lines are perpendicular to the surface and since the disk is uniformly charged, the lines are uniformly distributed over the surface. Far away from the disk, the lines are like those of a single point charge (the charge on the disk). Extended back to the disk (along the dotted lines of the diagram) they intersect at the center of the disk. If the disk is positively charged, the lines are directed outward from the disk. If the disk is negatively charged, they are directed inward toward the disk. A similar set of lines is associated with the region below the disk.
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4. We find the charge magnitude | q | from E = | q |/4 πε 0 r 2 : () ( ) 2 21 0 0 92 2 1.00 N C 1.00m 4 1.11 10 C. 8.99 10 N m C qE r = = × ×⋅ ε
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5. Since the magnitude of the electric field produced by a point charge q is given by 2 0 || / 4 Eq 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 () 2 21 1 0 92 2 0.50m 2.0 N C 4 5.6 10 C. 8.99 10 N m C qr E ε = = × ×⋅
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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: 5 1 1 2 01 5 2 2 2 || ˆˆ i (3.196 10 N C)i 4( ) i (3.196 10 N C)i ) q E xx q E πε =− × × G G Thus, the net electric field is 5 net 1 2 ˆ (6.39 10 N C)i EE E =+= − × GG G
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7. 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 0 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 10 664 10 307 10 0 2 92 2 1 9 15 2 21 π .. . Nm C C m NC 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.
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8. (a) The individual magnitudes G E 1 and G E 2 are figured from Eq. 22-3, where the absolute value signs for q 2 are unnecessary since this charge is positive. Whether we add
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ch22 - 1. We note that the symbol q2 is used in the problem...

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