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Electric Force/ Electrical Fields

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Two electric charges, q1 = +21.3 nC and q2 = +11.0 nC, are located on the x-axis at x
    = 0 m and x = 1.00 m, respectively. What is the magnitude of the electric field at the
    point x = 0.506 m, y = 0.506 m?
     
  2. jcsd
  3. Feb 6, 2009 #2

    LowlyPion

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    How would you think to approach the problem?
     
  4. Feb 7, 2009 #3
    21.3+11.0 /1 = .506/.506 ???????????????????????
     
  5. Feb 7, 2009 #4

    LowlyPion

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    Based on what?
     
  6. Feb 7, 2009 #5

    LowlyPion

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    They want you to find the E-field vector at .506,.506.

    First calculate the |E1| and |E2| based on the geometric distance to both charges.

    Then separate those vectors - they are vectors - into their components.

    Then add the x-components and y components separately.

    Happily they only want the magnitude of the E-field - |E| - so just use Pythagoras to figure the |E| of the resulting vector.
     
  7. Feb 7, 2009 #6
    theres no angle though, so it would be .506 sin____ and .506cos___ or is it 90 or 180 degrees
     
  8. Feb 7, 2009 #7

    LowlyPion

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    You have the coordinates of the point.

    and you know - or should know that a2 + b2 = c2

    So ... figure it out.
     
  9. Feb 7, 2009 #8
    .506 sq + .506 sq = 574.69 then sq root = .715,
     
  10. Feb 7, 2009 #9

    LowlyPion

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    It is .715 but your intermediate result is nonsense.

    So that's the r for 1 charge.

    Now figure the r for the other.
     
  11. Feb 7, 2009 #10
    21.3 sq + 11 sq= 574.7 equals 23.97 after sq rt, then take the 2 r's and add them up and tan y/x?
     
  12. Feb 7, 2009 #11

    LowlyPion

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    We're looking at distances, not charges.
     
  13. Feb 7, 2009 #12
    would it be : E1= ke q1/ r sq? then E2= ke q2/r sq but how would the 2nd r be found would the x= 0, and 1 or 1, 0.506 since it for q2?
     
  14. Feb 7, 2009 #13

    LowlyPion

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    They want you to find the distance from 1,0 to .506,.506 for the second charge. That triangle then is .494,.506.
     
  15. Feb 7, 2009 #14
    once i get r, i solve for e, then once i get both e's do i add them up or leave it as 2 seperate answers?
     
  16. Feb 8, 2009 #15

    LowlyPion

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    Once you get the |E| (magnitude of E) then you must resolve them into their x,y components and add the components - being careful of the signs - and then determine the magnitude of the result.

    They want the magnitude of the Total E field at that point.
     
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