anonymous one year ago What are the magnitude and direction of the electric field produced at P by the three charges in the figure below?

1. anonymous

|dw:1443486683805:dw|

2. anonymous

$E=\frac{ kQ }{ r^2 }$ where E is electric field k is Coulomb constant Q is charge

3. anonymous

$E=\frac{ (8.9*10^9)Q }{ r^2 }$ how exactly do you apply the charges? I actually have no much idea

4. anonymous

Give me a minute. Be back.

5. anonymous

ok!

6. anonymous

|dw:1443487357841:dw|

7. anonymous

Had to switch computers earlier Electric fields goes away from positive charge in the opposite direction. (repel) Electric fields goes toward negative charge in the same direction. (attract) So your diagram of the electric field would be |dw:1443487695840:dw|

8. anonymous

Understood that?

9. anonymous

wait how is it known that P is positive?

10. anonymous

P is an electric field. The electric field relative to the charges.

11. anonymous

Alright, about the electric field |dw:1443487887846:dw| is it something like that?

12. anonymous

|dw:1443488158566:dw|

13. anonymous

Yes

14. anonymous

It moves in opposite direction from the charges. So if there is angle it move diagonally

15. anonymous

Dont worry about the movement of the charges. The question only concerns with the field not the charges. But the charges should be like this if you want to know |dw:1443488272987:dw|

16. anonymous

Opposite charge attract Same charges repel.

17. anonymous

Why not on that side? |dw:1443488417223:dw|

18. anonymous

Oh yeah, I forgot.

19. anonymous

But thats not important in this problem. Lets label the electric field vectors now. E1, E2, E3 |dw:1443488495113:dw|

20. anonymous

Alright.

21. anonymous

First convert the cm to meters. Then use $E _{1}=\frac{ kQ }{ r^2 }$

22. anonymous

Solve for electric field one first.

23. anonymous

|dw:1443488673433:dw|

24. anonymous

$E=\frac{ kQ }{ r^2 }$ $E=\frac{ (8.9*10^9)Q }{r^2 }$ I am unsure what to plug for Q and r

25. anonymous

Yes, right units of meters and charges

26. anonymous

Remember, that electric field move away from the positive charge?

27. anonymous

Which charge is E1 referring to?

28. anonymous

the 5nC right?

29. anonymous

|dw:1443488995264:dw|

30. anonymous

Yes.

31. anonymous

And what is the distance from P to 5.0nC?

32. anonymous

$E=\frac{ (8.9*10^9)(5.0*10^{-9}) }{ r^2 }$ what is r? distance between P and 5nC?

33. anonymous

or rather, the radius so 0.02m, correct?

34. anonymous

$E=\frac{ (8.9*10^9 )(5.0*10^{-9})}{(0.02)^2 }$$E=111250$ seems odd

35. anonymous

Where did you get 0.02m?

36. anonymous

|dw:1443489283742:dw|

37. anonymous

|dw:1443489266073:dw|

38. anonymous

isn't r = radius? 0.04m = diameter

39. anonymous

|dw:1443489310614:dw| r is just distance

40. anonymous

oh... physics..

41. anonymous

$E=\frac{ (8.9*10^9)(5.0*10^{-9}) }{ (0.04)^2 }$

42. anonymous

No circles involved :)

43. anonymous

Do the same for E2

44. anonymous

$E_1= 27812.5$

45. anonymous

Show me the formula.

46. anonymous

$E_2= 988.9$

47. anonymous

aww

48. anonymous

$E_2=\frac{ (8.9*10^9)(1.0*10^{-10}) }{ (0.03)^2 }$

49. anonymous

Correct. Sorry I was lagging.

50. anonymous

Now do E3, which is the hardest one.

51. anonymous

|dw:1443489911056:dw|

52. anonymous

|dw:1443489917026:dw|$(0.04m)^2 + (0.03m)^2 = c^2$$c=0.05m$$E_3=\frac{ (8.9*10^9)(-1.0*10^{-10}) }{ (0.05)^2 }$$E_3=-356$

53. anonymous

Right :D?

54. anonymous

Yes, nice job.

55. anonymous

Since it is diagonal and makes an angle of 45 degrees |dw:1443490189435:dw| E3cos45 to get x component. E3sin45 to get y component

56. anonymous

uh?

57. anonymous

x= -251.7 ; y=-251.7

58. anonymous

|dw:1443490394186:dw|

59. anonymous

oh ok, this is to find it's magnitude?

60. anonymous

or you can call it $E _{3x}$

61. anonymous

Give me a second to write this.

62. anonymous

|dw:1443490526909:dw| |dw:1443490697051:dw|

63. anonymous

oooooo I see

64. anonymous

Add all the x and y component together $x^2+y^2=r^2$\ R is the resultant field or magnitude.

65. anonymous

Direction is $\tan^{-1}( \frac{ y }{ x })$

66. anonymous

$E_1= 27812.5$$E_2=988.9$$E_3=-356$

67. anonymous

$x=-251.7 , y=-251.7$

68. anonymous

Wait so you said E_1 and E_2 are -x and -y?

69. anonymous

E1x=-27812.5 E1y=0 E2x=0 E2y=-988.9 E3x=251.7 E3Y=251.7

70. anonymous

ah ok I see

71. anonymous

If it goes left or down it is negative, just like the coordinate system. Up or right is positive. Refer tot eh diagram for the direction of the field. brb

72. anonymous

$r=\sqrt{(251.7-27812.5)^2+(251.7-988.9)^2}$

73. anonymous

$r=27570.65$

74. anonymous

Yes

75. anonymous

Now find direction.

76. anonymous

direction?

77. anonymous

oh

78. anonymous

Direction is the angle in which it moves. Read what i wrote several post above.

79. anonymous

well since it's magnitude its positive then the direction should be a positive type of slope right? |dw:1443491606328:dw|

80. anonymous

No, you find the direction to know whether its positive or negative. It looks more negative to me. Find the direction first It can have a magnitude of 100000 and moves down diagonally (negative slope)

81. anonymous

I got to go. Go find your direction.

82. anonymous

y is the resultant y and x is the resultant x Same as the one you would use to find the magnitude.

83. anonymous

|dw:1443491766486:dw|

84. anonymous

I'll be back in like 75 minutes. I can check your calculations when I get back on.

85. anonymous

ok :[

86. anonymous

What are you doing? Use the formula.

87. anonymous

Read every post I post. I am heading off for sure now.

88. anonymous

$\tan^{-1}(\frac{ y }{ x }) = \tan^{-1}(\frac{ -737.2 }{ -27560.3 }) = 0.027$

89. anonymous

90. anonymous

91. anonymous

Sorry I did not catch the mistake you made earlier. 10nC is not 1.0 x 10^-10 C It is 1.0 x 10^-8 C

92. anonymous

oh....

93. anonymous

For the arctan it should not be in radians, not degree. When I used it on degree mode I got your answer 0.027

94. anonymous

My calculator is in radians and I get 0.027, and in degrees 1.53

95. anonymous

Strange. Well I am using the online right now. There might be a bug. But just use the one that doesnt have a 0.0XX something

96. anonymous

Redo your calculator for E2 and E3 since you wrote the C incorrectly.

97. anonymous

calculations*

98. anonymous

Alright but once I get the answer, how can I tell the direction by just the number?