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Finding the electric field vector along the bisector

  1. Apr 18, 2012 #1
    Two equal positive charges are held fixed and separated by distance D. Find the electric field
    vector along their perpendicular bisector. Then find the position relative to their center where
    the field is a maximum.

    so i started with symmetry and principle of superposition

    and got to this answer
    E = (1/(4πε))*q*D i^/( y^2 + (D^2)/4)^(3/2)

    y stand for the distance from the field to the origin

    is this equation the right answer for the first question? Also for the second question what changes when E is a maximum?
     
  2. jcsd
  3. Apr 18, 2012 #2

    tiny-tim

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    welcome to pf!

    hi dk321! welcome to pf! :smile:

    (try using the X2 and B buttons just above the Reply box :wink:)
    nooo, draw a diagram … it isn't along i, i'ts along j, isn't it? :wink:
     
  4. Apr 18, 2012 #3
    why is it j ? the two charges line in horizontal direction so along the perpendicular bisector would be in vertical direction right?
     
  5. Apr 18, 2012 #4

    tiny-tim

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    yes, j :confused:
     
  6. Apr 18, 2012 #5
    The first part is right.

    Hint for second part: E is a function of 'd'. And you have to find the maxima of E(d). How about using some calculus!
     
  7. Apr 18, 2012 #6
    And the vertical direction is 'j'. Isnt it?
     
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