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HW_3_solutions - HW 3 Solutions(due online Wednesday...

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HW 3 Solutions (due online Wednesday, January 26 at 1 AM) 1. Chapter 22 Problem 12 Figure 22-34 shows an uneven arrangement of electrons (e) and protons (p) on a circular arc of radius r = 2.10 cm, with angles θ 1 = 33.0 ˚ , θ 2 = 48.0 ˚ , θ 3 = 31.0 ˚ , and θ 4 = 26.0 ˚ . What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the net electric field produced at the center of the arc? Solutions: The field of each charge has magnitude 19 9 2 2 6 2 2 2 1.60 10 C (8.99 10 N m C ) 3.6 10 N C. (0.020 m) 0.020m kq e E k r The directions are indicated in standard format below. We use the magnitude-angle notation (convenient if one is using a vector-capable calculator in polar mode) and write (starting with the proton on the left and moving around clockwise) the contributions to E net as follows: E E E E E 20 130 100 150 0 b g b g b g b g b g . This yields 393 10 764 6 . . c h , with the N/C unit understood. (a) The result above shows that the magnitude of the net electric field is 6 net | | 3.93 10 N/C. E (b) Similarly, the direction of E net is 76.4
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