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Homework Help: Electric field problem

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data
    what is the potential at point P, located at the centre of the square of the point charges as shown in the Fig 10? Assume that d = 1.3m and that the charges are q1 = +12 nC,
    q2 = -24 nC, q3 = +31 nC and q4 = +17 nC

    http://i923.photobucket.com/albums/ad73/look416/tege.jpg

    2. Relevant equations
    F = Eq
    E = V/d
    F = [tex]\frac{q1q2}{4\pi\epsilon r^2}[/tex]


    3. The attempt at a solution
    at first i thought of getting resultant force at the center then using f=Eq to find the E, later using E=V/d to find the V
    however, i have to find the charges of P 1st if im doing so
    then think of electric field, E is applicable to everywhere, thinking of something else
    conclusion : cant solve it
     
  2. jcsd
  3. Jan 24, 2010 #2

    tiny-tim

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    Hi look416! :smile:
    uhh? :confused:

    just find the individual potentials, and add them. :wink:
     
  4. Jan 24, 2010 #3
    lolz, calculated ans wrong =.=
    i found out tha for q1 = 63.8, q2 = 127.7, q3 = 164.9, q4 = 90.5
    and totaled up to 446.9,
    using E = V/d
    V = Ed
    V = 446.9 x 1.3
    = 580.97
    omg thats not the answer XD
    the answer is 352V
    aw probably i need more detailed instruction =.=
     
  5. Jan 25, 2010 #4
    push push
     
  6. Jan 25, 2010 #5

    tiny-tim

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    Shouldn't the potential for q2 be negative?
     
  7. Jan 26, 2010 #6
    even if the q2 is negative
    Enett=63.8+(-127.7)+163.9+90.5
    =191.5
    Convert it into V using E=V/d
    V = 248.95
    it still not correct
     
  8. Jan 26, 2010 #7
    The relation V=Ed is true only for a region of uniform electric field and does not apply here. Assume a test charge Q at the centre of the arrangement, find the potential energy of this test charge , a multiple of Q. Divide by Q to get the potential at the centre. Use
    potential energy =
    (Qq1 + Qq2 + Qq3) + Qq4[tex]/[/tex]4[tex]\pi[/tex][tex]\epsilon[/tex](d[tex]/[/tex][tex]\sqrt{2}[/tex] )

    I'm getting 352.4 V
     
  9. Jan 26, 2010 #8
    impressive, coz in my memory what i have alrdy learnt is all about uniform parallel electric field
    too sad no idea you are talking about the equation given
    mind giving some website that can clearly express out what you have meant?

    Btw, since we dont know the charge of Q,how are we calculating it?
     
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