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Chapter 16 – Problem 13

Given Q=6.00x10 -3 C r=1.00m r=1.00m 2 k=9.0x10 9 N 2 m 2 /C 2. Find Force on the top right charge. Chapter 16 – Problem 13. Given Q=6.00x10 -3 C r=1.00m r=1.00m 2 k=9.0x10 9 N 2 m 2 /C 2. Find Force on the top right charge. Chapter 16 – Problem 13. Given Q=6.00x10 -3 C

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Chapter 16 – Problem 13

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  1. Given Q=6.00x10-3C r=1.00m r=1.00m2 k=9.0x109 N2m2/C2 Find Force on the top right charge Chapter 16 – Problem 13

  2. Given Q=6.00x10-3C r=1.00m r=1.00m2 k=9.0x109 N2m2/C2 Find Force on the top right charge Chapter 16 – Problem 13

  3. Given Q=6.00x10-3C r=1.00m r=1.00m2 k=9.0x109 N2m2/C2 Find Force on the top right charge Chapter 16 – Problem 13

  4. Given Q=6.00x10-3C r=1.00m r=1.00m2 k=9.0x109 N2m2/C2 Find Force on the top right charge Chapter 16 – Problem 13

  5. Given Q=6.00x10-3C r=1.00m r=1.00m2 k=9.0x109 N2m2/C2 Find Force on the top right charge Chapter 16 – Problem 13

  6. Chapter 16 – Problem 13 • F = KQQ / r2= 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2 • Fx=3.24x105N for top left on top right • Fy=3.24x105N for bottom right on top right • F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2 • F=1.62x105N • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

  7. Chapter 16 – Problem 13 • F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2 • Fx=3.24x105N for top left on top right • Fy=3.24x105N for bottom right on top right • F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2 • F=1.62x105N • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

  8. Chapter 16 – Problem 13 • F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2 • Fx=3.24x105N for top left on top right • Fy=3.24x105N for bottom right on top right • F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2 • F=1.62x105N • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

  9. Chapter 16 – Problem 13 • F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2 • Fx=3.24x105N for top left on top right • Fy=3.24x105N for bottom right on top right • F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2 • F=1.62x105N • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

  10. Chapter 16 – Problem 13 • F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2 • Fx=3.24x105N for top left on top right • Fy=3.24x105N for bottom right on top right • F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2 • F=1.62x105N • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

  11. Chapter 16 – Problem 13 • F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2 • Fx=3.24x105N for top left on top right • Fy=3.24x105N for bottom right on top right • F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2 • F=1.62x105N • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

  12. Chapter 16 – Problem 13 • F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2 • Fx=3.24x105N for top left on top right • Fy=3.24x105N for bottom right on top right • F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2 • F=1.62x105N • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

  13. Chapter 16 – Problem 13 • F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2 • Fx=3.24x105N for top left on top right • Fy=3.24x105N for bottom right on top right • F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2 • F=1.62x105N • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

  14. Chapter 16 – Problem 13 • Fx=3.24x105N for top left on top right • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Total x = 4.39x105N • Fy=3.24x105N for bottom right on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right • Total y = 4.39x105N

  15. Chapter 16 – Problem 13 • Fx=3.24x105N for top left on top right • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Total x = 4.39x105N • Fy=3.24x105N for bottom right on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right • Total y = 4.39x105N

  16. Chapter 16 – Problem 13 • Fx=3.24x105N for top left on top right • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Total x = 4.39x105N • Fy=3.24x105N for bottom right on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right • Total y = 4.39x105N

  17. Chapter 16 – Problem 13 • Fx=3.24x105N for top left on top right • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Total x = 4.39x105N • Fy=3.24x105N for bottom right on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right • Total y = 4.39x105N

  18. Chapter 16 – Problem 13 • Fx=3.24x105N for top left on top right • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Total x = 4.39x105N • Fy=3.24x105N for bottom right on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right • Total y = 4.39x105N

  19. Chapter 16 – Problem 13 • Fx=3.24x105N for top left on top right • Fx=1.62x105N cos45= 1.15x105N for bottom left on top right • Total x = 4.39x105N • Fy=3.24x105N for bottom right on top right • Fy=1.62x105N sin45= 1.15x105N for bottom left on top right • Total y = 4.39x105N

  20. Chapter 16 – Problem 13 • Total x = 4.39x105N • Total y = 4.39x105N • total (4.39x105N)2 + (4.39x105N)2 • Total = 6.2x105N away from the center

  21. Chapter 16 – Problem 13 • Total x = 4.39x105N • Total y = 4.39x105N • total (4.39x105N)2 + (4.39x105N)2 • Total = 6.2x105N away from the center

  22. Chapter 16 – Problem 13 • Total x = 4.39x105N • Total y = 4.39x105N • total (4.39x105N)2 + (4.39x105N)2 • Total = 6.2x105N away from the center

  23. Chapter 16 – Problem 13 • Total x = 4.39x105N • Total y = 4.39x105N • total (4.39x105N)2 + (4.39x105N)2 • Total = 6.2x105N away from the center

  24. Given Q=1.6x10-19C m=9.11x10-31kg r=.53x10-10m k=9.0x109 N2m2/C2 G=6.67.0x10-11 N2m2/kg2 Find FE Fg Chapter 16 – Problem 15

  25. Chapter 16 Problem 15 • F = KQQ / r2 = 9.0x109 N2m2/C2(1.6x10-19)2 • (.53x10-10m)2 • F=8.2x10-8N • F = Gmm / r2 = 6.67x10-11 N2m2/C2(9.11x10-31kg)2 • (.53x10-10m)2 • F=3.6x10-47N • Ratio=8.2x10-8N/3.6x10-47N=2.3x1039 times greater

  26. Chapter 16 Problem 15 • F = KQQ / r2 = 9.0x109 N2m2/C2(1.6x10-19)2 • (.53x10-10m)2 • F=8.2x10-8N • F = Gmm / r2 = 6.67x10-11 N2m2/C2(9.11x10-31kg)2 • (.53x10-10m)2 • F=3.6x10-47N • Ratio=8.2x10-8N/3.6x10-47N=2.3x1039 times greater

  27. Chapter 16 Problem 15 • F = KQQ / r2 = 9.0x109 N2m2/C2(1.6x10-19)2 • (.53x10-10m)2 • F=8.2x10-8N • F = Gmm / r2 = 6.67x10-11 N2m2/C2(9.11x10-31kg)2 • (.53x10-10m)2 • F=3.6x10-47N • Ratio=8.2x10-8N/3.6x10-47N=2.3x1039 times greater

  28. Chapter 16 Problem 19 • Given • The charge must be placed beyond one of the charges in order for the net force to equal zero

  29. Given Q=5.7x10-6C r=.25m+x Q2=3.5x10-6C r2=x k=9.0x109 N2m2/C2 Find X position a positive or negative particle would experience no force Chapter 16 Problem 19

  30. Chapter 16 Problem 19 • F=KQQ / r2 • K(5.7x10–6 C)Q / (.25m + x)2 =K(3.5x10–6 C)Q / x2 • (5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C)/ x2 • X=.91m beyond the negative charge • The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

  31. Chapter 16 Problem 19 • F=KQQ / r2 • K(5.7x10–6 C)Q / (.25m + x)2 =K(3.5x10–6 C)Q /x2 • (5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C)/ x2 • X=.91m beyond the negative charge • The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

  32. Chapter 16 Problem 19 • F=KQQ / r2 • K(5.7x10–6 C)Q / (.25m + x)2 =K(3.5x10–6 C)Q / x2 • (5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C)/ x2 • X=.91m beyond the negative charge • The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

  33. Chapter 16 Problem 19 • F=KQQ / r2 • K(5.7x10–6 C)Q / (.25m + x)2 =K(3.5x10–6 C)Q / x2 • (5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C)/ x2 • X=.91m beyond the negative charge • The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

  34. Chapter 16 Problem 19 • F=KQQ / r2 • K(5.7x10–6 C)Q / (.25m + x)2 =K(3.5x10–6 C)Q / x2 • (5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C)/ x2 • X=.91m beyond the negative charge • The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

  35. Chapter 16 Problem 19 • F=KQQ / r2 • K(5.7x10–6 C)Q / (.25m + x)2 =K(3.5x10–6 C)Q / x2 • (5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C)/ x2 • X=.91m beyond the negative charge • The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

  36. Given q=1.6x10-19C m=9.11x10-31kg E=600 N/C Find a=?m/s2 Direction=? Chapter 16 Problem 21

  37. Chapter 16 Problem 21 • F=qE= ma • F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a • a=1.04x1014m/s2 • Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric field • The direction of the acceleration is independent of the velocity

  38. Chapter 16 Problem 21 • F=qE= ma • F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a • a=1.04x1014m/s2 • Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric field • The direction of the acceleration is independent of the velocity

  39. Chapter 16 Problem 21 • F=qE= ma • F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a • a=1.04x1014m/s2 • Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric field • The direction of the acceleration is independent of the velocity

  40. Chapter 16 Problem 21 • F=qE= ma • F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a • a=1.04x1014m/s2 • Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric field • The direction of the acceleration is independent of the velocity

  41. Chapter 16 Problem 21 • F=qE= ma • F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a • a=1.04x1014m/s2 • Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric field • The direction of the acceleration is independent of the velocity

  42. Chapter 16 Problem 21 • F=qE= ma • F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a • a=1.04x1014m/s2 • Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric field • The direction of the acceleration is independent of the velocity

  43. Given q=1.6x10-19C m=9.11x10-31kg a=125m/s2 Find E=? N/C Chapter 16 Problem 27

  44. Chapter 16 Problem 27 • F=qE= ma • F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) • E=7.12x10-10N/C south • Because the charge on the electron is negative, the direction of force,and thus the acceleration, is opposite to the direction of the electric field, so the electric field is south

  45. Chapter 16 Problem 27 • F=qE= ma • F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) • E=7.12x10-10N/C south • Because the charge on the electron is negative, the direction of force,and thus the acceleration, is opposite to the direction of the electric field, so the electric field is south

  46. Chapter 16 Problem 27 • F=qE= ma • F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) • E=7.12x10-10N/C south • Because the charge on the electron is negative, the direction of force,and thus the acceleration, is opposite to the direction of the electric field, so the electric field is south

  47. Chapter 16 Problem 27 • F=qE= ma • F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) • E=7.12x10-10N/C south • Because the charge on the electron is negative, the direction of force,and thus the acceleration, is opposite to the direction of the electric field, so the electric field is south

  48. Chapter 16 Problem 27 • F=qE= ma • F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) • E=7.12x10-10N/C south • Because the charge on the electron is negative, the direction of force,and thus the acceleration, is opposite to the direction of the electric field, so the electric field is south

  49. Chapter 16 Problem 27 • F=qE= ma • F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) • E=7.12x10-10N/C south • Because the charge on the electron is negative, the direction of force,and thus theacceleration, is opposite to the direction of the electric field, so the electric field is south

  50. Chapter 16 Problem 30a Given 9.0x10-6C 9.0x10-6C .05m .10m Find the Electric field strength at (0,.05m)

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