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Electric Field finding zero

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data
    In the figure below (q1 = -2.3 µC, q2 = 6.80 µC), determine the point (other than infinity) at which the electric field is zero. (Figure is attached)

    2. Relevant equations
    E=k*Q/r^2
    F-k*q_1*q_2/r^2
    F-Q*E

    3. The attempt at a solution
    I thought about finding where the force is zero and then the electric field would be zero because of F=Q*E. But I did not think that that was correct.
     

    Attached Files:

  2. jcsd
  3. Jan 18, 2010 #2

    tiny-tim

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    Hi Keith! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    No, that's fine … F = qE, so they're zero together. :smile:

    (btw, i think it's usually little q for charge, except big Q for the charge of a black hole :wink:)
     
  4. Jan 18, 2010 #3
    Alright I get that part.
    So then I set the two forces equal to each other and k*q_1*q_2/x^2=k*q_2*q_1/(d-x)^2
    Is that the correct set up?
     
  5. Jan 18, 2010 #4

    tiny-tim

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    erm :redface: … what about q (or Q) ? :wink:
     
  6. Jan 18, 2010 #5
    Um you mean the F=qE part?
    If that is the case then the q is just the overall charge right and I am given two qs so I am not sure what that translates as...
     
  7. Jan 18, 2010 #6

    tiny-tim

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    No, q isn't "the overall charge", it's a hypothetical "test" charge.

    Your …
    … cancels out to give 1/x2 = 1/(d-x)2.

    You need the forces on a hypothetical charge q to be equal and opposite.
     
  8. Jan 18, 2010 #7
    Yeah I did that earlier and I got x to be .5 using d=1 meter.
    However this answer is not correct. Was my algebra just bad or is there something more to the problem. Sorry I am just having a tough time.
     
  9. Jan 18, 2010 #8

    tiny-tim

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    That's because your k*q_1*q_2/x^2=k*q_2*q_1/(d-x)^2 is wrong.

    You need the force on q from q1 to be equal to the force on q from q2
     
  10. Jan 18, 2010 #9
    So that leaves me with q_1/x^2=q_2/(d-x)^2?
     
  11. Jan 18, 2010 #10

    tiny-tim

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  12. Jan 18, 2010 #11
    Alright good. I got x=.57. This does not seem to be the right answer and I think this leaves me to be in between the two charges instead of to the left of them. What is the next step or did I do something incorrectly again?
     
  13. Jan 18, 2010 #12

    tiny-tim

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    ah, they have opposite charge, so the equilibrium position will be outside, so you need to use d+x instead of d-x. :smile:
     
  14. Jan 18, 2010 #13
    Do I include the negative signs for the charges or is it the absolute value of those charges I cannot remember?
     
  15. Jan 19, 2010 #14

    tiny-tim

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    Good morning Keith! :smile:

    (just got up :zzz: …)

    To find the direction of a force, you must include the sign of the charge. :wink:
     
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