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Electric field finding third charge

  1. Jun 25, 2009 #1
    1. The problem statement, all variables and given/known data
    a point charge of +5c is located at x=-2cm, and a second point charge of -6c is located at x=1cm
    where should a thid charge of +4c be placed so that electric field at x=0cm is zero?


    2. Relevant equations

    E(total)=E1+E2+E3

    E=k*q/(r^2)

    3. The attempt at a solution

    E(total)=0
    E3=-E1-E2

    what I did first is divide the k out so now the equation will look like this
    q3/(r3^2)=-(q1/(r1^2))-(q2/r2^2))
    and then solve for r3 and put a negative sign because the force is going to the right to the electric field at x=0
    i got r3=-.917cm
    It is saying I am getting this wrong can anyone tell me what i am doing rong
     
  2. jcsd
  3. Jun 25, 2009 #2

    ideasrule

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    Homework Helper

    You forgot that electric field adds vectorially, not arithmetically. Think about the configuration: if you put a positive test charge at x=0, both the +5c charge and the -6c charge would push it to the right. The third charge's field has to counter the combined effect of these two charges. Try writing an equation with only the magnitudes of the electric fields; forget about +/- signs.
     
  4. Jun 25, 2009 #3
    Well the first charge is pushing to the right and the second charge is pushing to the left thus making first plus and second negative because of their displacement and the position of the eletric field at x=0
    but if we do forget about the +/- then this cannot be solve
    because the E3=-E1-E2
    since we have negavtives
    R^2 is in E3 we have to square root negatives number...if you see where i am getting at.
    thus we need +5 and -6
     
  5. Jun 26, 2009 #4

    ideasrule

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    No, both charges push to the right: the one on the left repels while the one of the right attracts, so their effects combine.

    Also, when I said we should ignore all +/- signs, I meant all of them. So don't use E3=-E1-E2; use E3=E1+E2. Don't use q=-3; use q=3.
     
  6. Jun 26, 2009 #5
    ok i got the answer correct but i am not understand when you mean by both charges are pushing to the right. This i what i see as.

    q1=5c----->E=0<------q2=-6c

    to me this diagram makes sense because q1 is positive and q2 is negative
    and the other one that doesnt make sense to me is that
    how does E3=E1+E2
    where the standard equation was E1+E2+E3=0
    and when you move E1 and E2 to the right they become negative.
    E3=-E1-E2
    thanks for your help though
     
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