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E field

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the magnitude of the collective contribution to the E field at the origin from both the + and - charges located 5 cm above the horizontal axis?

    2. Relevant equations
    kQ/d^2



    3. The attempt at a solution
    Im just having a problem setting up this problem. Could someone help me on getting started
     

    Attached Files:

  2. jcsd
  3. Mar 8, 2012 #2

    tiny-tim

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    hi sasuke07! :smile:

    (try using the X2 button just above the Reply box :wink:)

    first, find the magnitude of each contribution,

    from that, find the x and y components …

    what do you get? :smile:
     
  4. Mar 8, 2012 #3
    I already figured out the magnitued of the the positive charge 5 cm above the axis and got 5.4X10^4. What do you mean by the x and y components. WOuldn't the negative charge have the same magnitude of the positive charge.
     
  5. Mar 8, 2012 #4

    tiny-tim

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    yes :smile:
    draw an arrow at the origin showing the direction of the field from the positive charge

    then find the x and y components of that arrow

    (and finally you'll add them to the x and y components of the arrow from the negative charge :wink:)
     
  6. Mar 8, 2012 #5
    so to figure out the x component wouldn't it be Kq/d^2 where k=9X10^9, q is 30X10^9= charge and d would be .05m^2. Or would i have to use cosine and sine to figure out the x and y components and if so could you show me how to set it up.
     
  7. Mar 8, 2012 #6

    tiny-tim

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    no, you must use cos and sin, of the angle the arrow makes
     
  8. Mar 8, 2012 #7
    so would it be Kq/r^2sintheta
    and Kq/r^2Costheta. Where theta would be 90 degrees?
     
  9. Mar 8, 2012 #8
    or would theta be 45
     
  10. Mar 8, 2012 #9

    tiny-tim

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    θ is the angle of the line from the charge to the origin
     
  11. Mar 8, 2012 #10
    awesome so 45 degrees.
     
  12. Mar 8, 2012 #11
    So were the equations correct though?
     
  13. Mar 8, 2012 #12

    tiny-tim

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    looks ok :smile:

    so what is the total field from both the positive and the negative charge? ​
     
  14. Mar 8, 2012 #13
    i got 7.6X10^4 by doing KQ/r^2sin45 and the answer is the same for cos45. BUt i checked the answer and its supposed to be 7.1X10^4. Any suggestions
     
  15. Mar 8, 2012 #14

    tiny-tim

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    your answer looks right to me :confused:

    (are you sure it's 5 cm that they're asking about?)
     
  16. Mar 8, 2012 #15
    i think its 5cm because the length of the x and y components is 5cm.
     
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