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Electric Field

  1. Jan 5, 2016 #1
    1. The problem statement, all variables and given/known data

    IF the resultant electric field at D due to the three point charges is zero, find the magnitudes of q1 q2 in terms of q


    https://scontent-sin1-1.xx.fbcdn.net/hphotos-xft1/v/l/t34.0-12/12498556_10205376500763571_410670163_n.jpg?oh=f586685c3d06d630fad4469546c534dc&oe=568DB0E8
    2. Relevant equations
    Electric Field

    3. The attempt at a solution
    Ex = E(q1) + E(q2)x = k q1 / AB^2 + k q2 cosθ / BD^2
    Ey = E(q2)y + E(q) = k q2 sinθ / BD^2 + k q / CD^2
    must be
    k q2 cosθ / BD^2 = - k q1 / AB^2
    k q2 sinθ / BD^2 = - k q / CD^2
    then
    q1 = q AB^2 / (tanθ CD^2)
    AB = CD
    tanθ = AB / BC
    then
    q1 = q BC / AB = 2 q
    and
    q2 = - q1 BD^2 / (cosθ AB^2)
    cosθ= BC / BD
    then
    q2 = - q1 BD^3 / (BC AB^2) = - q (BD / AB)^3 = -11,2 q

    Did i make some mistake here?
     
  2. jcsd
  3. Jan 5, 2016 #2

    mfb

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    Staff: Mentor

    I don't think this should be AB here.

    The rest looks fine, but you should define the coordinate axes and the angle in some way.
     
  4. Jan 5, 2016 #3
    Oopss!! :woot::woot::woot::woot::woot:

    Correction
    Ex = E(q1) + E(q2)x = k q1 / AD^2 + k q2 cosθ / BD^2
    Ey = E(q2)y + E(q) = k q2 sinθ / BD^2 + k q / CD^2
    must be
    k q2 cosθ / BD^2 = - k q1 / AD^2 ......(1)
    k q2 sinθ / BD^2 = - k q / CD^2 .....(2)
    (2)/(1) I got
    tanθ = (q/CD^2)(AD^2/q1)
    q1 = q. (AD^2/tanθ.CD^2)
    q1=8q
    Did i make some mistake here?
     
  5. Jan 5, 2016 #4

    mfb

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    Looks right.

    You also need q2.
     
  6. Jan 5, 2016 #5
    Yea i am stucked here T_T
    Can i just sub 8q into eq(1)?
     
  7. Jan 5, 2016 #6

    mfb

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    Sure.
     
  8. Jan 5, 2016 #7
    So i got q2 = (-8q/AD^2)(BD^2/cos)
    q2=-11.18q am i doin right?
     
  9. Jan 5, 2016 #8

    mfb

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    I get the same result.
     
  10. Jan 5, 2016 #9
    thanks!!
     
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