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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?

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  1. anonymous
    • one year ago
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    |dw:1443486683805:dw|

  2. anonymous
    • one year ago
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    \[E=\frac{ kQ }{ r^2 }\] where E is electric field k is Coulomb constant Q is charge

  3. anonymous
    • one year ago
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    \[E=\frac{ (8.9*10^9)Q }{ r^2 }\] how exactly do you apply the charges? I actually have no much idea

  4. anonymous
    • one year ago
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    Give me a minute. Be back.

  5. anonymous
    • one year ago
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    ok!

  6. anonymous
    • one year ago
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    |dw:1443487357841:dw|

  7. anonymous
    • one year ago
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    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
    • one year ago
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    Understood that?

  9. anonymous
    • one year ago
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    wait how is it known that P is positive?

  10. anonymous
    • one year ago
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    P is an electric field. The electric field relative to the charges.

  11. anonymous
    • one year ago
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    Alright, about the electric field |dw:1443487887846:dw| is it something like that?

  12. anonymous
    • one year ago
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    |dw:1443488158566:dw|

  13. anonymous
    • one year ago
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    Yes

  14. anonymous
    • one year ago
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    It moves in opposite direction from the charges. So if there is angle it move diagonally

  15. anonymous
    • one year ago
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    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
    • one year ago
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    Opposite charge attract Same charges repel.

  17. anonymous
    • one year ago
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    Why not on that side? |dw:1443488417223:dw|

  18. anonymous
    • one year ago
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    Oh yeah, I forgot.

  19. anonymous
    • one year ago
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    But thats not important in this problem. Lets label the electric field vectors now. E1, E2, E3 |dw:1443488495113:dw|

  20. anonymous
    • one year ago
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    Alright.

  21. anonymous
    • one year ago
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    First convert the cm to meters. Then use \[E _{1}=\frac{ kQ }{ r^2 }\]

  22. anonymous
    • one year ago
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    Solve for electric field one first.

  23. anonymous
    • one year ago
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    |dw:1443488673433:dw|

  24. anonymous
    • one year ago
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    \[E=\frac{ kQ }{ r^2 }\] \[E=\frac{ (8.9*10^9)Q }{r^2 }\] I am unsure what to plug for Q and r

  25. anonymous
    • one year ago
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    Yes, right units of meters and charges

  26. anonymous
    • one year ago
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    Remember, that electric field move away from the positive charge?

  27. anonymous
    • one year ago
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    Which charge is E1 referring to?

  28. anonymous
    • one year ago
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    the 5nC right?

  29. anonymous
    • one year ago
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    |dw:1443488995264:dw|

  30. anonymous
    • one year ago
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    Yes.

  31. anonymous
    • one year ago
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    And what is the distance from P to 5.0nC?

  32. anonymous
    • one year ago
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    \[E=\frac{ (8.9*10^9)(5.0*10^{-9}) }{ r^2 }\] what is r? distance between P and 5nC?

  33. anonymous
    • one year ago
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    or rather, the radius so 0.02m, correct?

  34. anonymous
    • one year ago
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    \[E=\frac{ (8.9*10^9 )(5.0*10^{-9})}{(0.02)^2 }\]\[E=111250\] seems odd

  35. anonymous
    • one year ago
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    Where did you get 0.02m?

  36. anonymous
    • one year ago
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    |dw:1443489283742:dw|

  37. anonymous
    • one year ago
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    |dw:1443489266073:dw|

  38. anonymous
    • one year ago
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    isn't r = radius? 0.04m = diameter

  39. anonymous
    • one year ago
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    |dw:1443489310614:dw| r is just distance

  40. anonymous
    • one year ago
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    oh... physics..

  41. anonymous
    • one year ago
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    \[E=\frac{ (8.9*10^9)(5.0*10^{-9}) }{ (0.04)^2 }\]

  42. anonymous
    • one year ago
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    No circles involved :)

  43. anonymous
    • one year ago
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    Do the same for E2

  44. anonymous
    • one year ago
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    \[E_1= 27812.5\]

  45. anonymous
    • one year ago
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    Show me the formula.

  46. anonymous
    • one year ago
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    \[E_2= 988.9\]

  47. anonymous
    • one year ago
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    aww

  48. anonymous
    • one year ago
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    \[E_2=\frac{ (8.9*10^9)(1.0*10^{-10}) }{ (0.03)^2 }\]

  49. anonymous
    • one year ago
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    Correct. Sorry I was lagging.

  50. anonymous
    • one year ago
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    Now do E3, which is the hardest one.

  51. anonymous
    • one year ago
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    |dw:1443489911056:dw|

  52. anonymous
    • one year ago
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    |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
    • one year ago
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    Right :D?

  54. anonymous
    • one year ago
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    Yes, nice job.

  55. anonymous
    • one year ago
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    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
    • one year ago
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    uh?

  57. anonymous
    • one year ago
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    x= -251.7 ; y=-251.7

  58. anonymous
    • one year ago
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    |dw:1443490394186:dw|

  59. anonymous
    • one year ago
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    oh ok, this is to find it's magnitude?

  60. anonymous
    • one year ago
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    or you can call it \[E _{3x}\]

  61. anonymous
    • one year ago
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    Give me a second to write this.

  62. anonymous
    • one year ago
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    |dw:1443490526909:dw| |dw:1443490697051:dw|

  63. anonymous
    • one year ago
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    oooooo I see

  64. anonymous
    • one year ago
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    Add all the x and y component together \[x^2+y^2=r^2\]\ R is the resultant field or magnitude.

  65. anonymous
    • one year ago
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    Direction is \[\tan^{-1}( \frac{ y }{ x })\]

  66. anonymous
    • one year ago
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    \[E_1= 27812.5\]\[E_2=988.9\]\[E_3=-356\]

  67. anonymous
    • one year ago
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    \[x=-251.7 , y=-251.7\]

  68. anonymous
    • one year ago
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    Wait so you said E_1 and E_2 are -x and -y?

  69. anonymous
    • one year ago
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    E1x=-27812.5 E1y=0 E2x=0 E2y=-988.9 E3x=251.7 E3Y=251.7

  70. anonymous
    • one year ago
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    ah ok I see

  71. anonymous
    • one year ago
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    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
    • one year ago
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    \[r=\sqrt{(251.7-27812.5)^2+(251.7-988.9)^2}\]

  73. anonymous
    • one year ago
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    \[r=27570.65\]

  74. anonymous
    • one year ago
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    Yes

  75. anonymous
    • one year ago
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    Now find direction.

  76. anonymous
    • one year ago
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    direction?

  77. anonymous
    • one year ago
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    oh

  78. anonymous
    • one year ago
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    Direction is the angle in which it moves. Read what i wrote several post above.

  79. anonymous
    • one year ago
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    well since it's magnitude its positive then the direction should be a positive type of slope right? |dw:1443491606328:dw|

  80. anonymous
    • one year ago
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    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
    • one year ago
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    I got to go. Go find your direction.

  82. anonymous
    • one year ago
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    y is the resultant y and x is the resultant x Same as the one you would use to find the magnitude.

  83. anonymous
    • one year ago
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    |dw:1443491766486:dw|

  84. anonymous
    • one year ago
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    I'll be back in like 75 minutes. I can check your calculations when I get back on.

  85. anonymous
    • one year ago
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    ok :[

  86. anonymous
    • one year ago
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    What are you doing? Use the formula.

  87. anonymous
    • one year ago
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    Read every post I post. I am heading off for sure now.

  88. anonymous
    • one year ago
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    \[\tan^{-1}(\frac{ y }{ x }) = \tan^{-1}(\frac{ -737.2 }{ -27560.3 }) = 0.027\]

  89. anonymous
    • one year ago
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    Your calculator should be radian mode and not degree.

  90. anonymous
    • one year ago
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    its in radian

  91. anonymous
    • one year ago
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    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
    • one year ago
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    oh....

  93. anonymous
    • one year ago
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    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
    • one year ago
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    My calculator is in radians and I get 0.027, and in degrees 1.53

  95. anonymous
    • one year ago
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    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
    • one year ago
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    Redo your calculator for E2 and E3 since you wrote the C incorrectly.

  97. anonymous
    • one year ago
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    calculations*

  98. anonymous
    • one year ago
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    Alright but once I get the answer, how can I tell the direction by just the number?