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X direction and -X direction

  1. Jan 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A negative point charge q1 = -4.00 nC is on the x axis at x = 0.60 m. A second point charge q2 is on the x axis at x=-1.20 m.

    What must the sign and magnitude of q2 be for the net electric field at the origin to be 50.0 N/C in the +x direction be?

    What must the sign and magnitude of q2 be for the net electric field at the origin to be 50.0 N/C in the -x direction be?


    2. Relevant equations
    Coulomb's Law
    E = q1 + q2


    3. The attempt at a solution

    Ok, i got the part when it is in the + x direction but how would i do it for the - x direction? I was doing 50 = 100-q2, but i don't think it will work?? Any help would be appreciated. Thanks in advance.
     
  2. jcsd
  3. Jan 16, 2008 #2

    Kurdt

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  4. Jan 16, 2008 #3
    Thanks for the sites, which I understand. But I know how to get the electric force in the +x direction but not in the -x direction. Could I do that to make it go in the -x direction, the charge of q1 could be +4.00 nC since electric field lines go away from positive charges? Any help would be appreciated. Thanks.
     
  5. Jan 16, 2008 #4

    HallsofIvy

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    Since q1 have negative charge, it will attract a positive charge and repel a positive charge. Since q1 is to the right of q2 ( 0.6> -1.2) an attraction, toward q1, will be "in the positive direction" (from -1.2 toward +0.6) while a repulsion, away from q1, will be "in the negative direction" (from 0.6 toward -1.2).
     
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