HWCh23Solutions

# HWCh23Solutions - PHYSICS 133 HOMEWORK SOLUTIONS CHAPTER 23...

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Unformatted text preview: PHYSICS 133 HOMEWORK SOLUTIONS CHAPTER 23 P23.7 F 1 = k e q 1 q 2 r 2 = 8.99 × 10 9 N ⋅ m 2 C 2 ( 29 7.00 × 10- 6 C ( 29 2.00 × 10- 6 C ( 29 0.500 m ( 29 2 = 0.503 N F 2 = k e q 1 q 2 r 2 = 8.99 × 10 9 N ⋅ m 2 C 2 ( 29 7.00 × 10- 6 C ( 29 4.00 × 10- 6 C ( 29 0.500 m ( 29 2 = 1.01 N F x = 0.503cos60.0 ° + 1.01cos 60.0 ° = 0.755 N F y = 0.503sin 60.0 ° - 1.01sin 60.0 ° = - 0.436 N F = 0.755 N ( 29 ˆ i - 0.436 N ( 29 ˆ j = 0.872 N at an angle of 330 ° FIG. P23.7 P23.13 For equilibrium, F e = - F g or q E = - mg- ˆ j ( 29 . Thus, E = mg q ˆ j . (a) E = mg q ˆ j = 9.11 × 10- 31 kg ( 29 9.80 m s 2 ( 29- 1.60 × 10- 19 C ( 29 ˆ j = - 5.58 × 10- 11 N C ( 29 ˆ j (b) E = mg q ˆ j = 1.67 × 10- 27 kg ( 29 9.80 m s 2 ( 29 1.60 × 10- 19 C ( 29 ˆ j = 1.02 × 10- 7 N C ( 29 ˆ j P23.15 The point is designated in the sketch. The magnitudes of the electric fields, E 1 , (due to the - 2.50 × 10- 6 C charge) and E 2 (due to the 6.00 × 10- 6 C charge) are E 1 = k e q r 2 = 8.99 × 10 9 N ⋅ m 2 C 2 ( 29 2.50 × 10- 6 C ( 29 d 2 (1) E 2 = k e q r 2 = 8.99 × 10 9 N ⋅ m 2 C 2 ( 29 6.00 × 10- 6 C ( 29 d + 1.00 m ( 29 2...
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HWCh23Solutions - PHYSICS 133 HOMEWORK SOLUTIONS CHAPTER 23...

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