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Two charges create an electric field-electric field strenght at a point above h.fiel

  1. Jan 26, 2010 #1
    Two charges create an electric field--electric field strenght at a point above h.fiel

    1. The problem statement, all variables and given/known data
    Two charges are located on a horizontal axis. The Coulomb constant is 8.98755x10^9 Nm^2/C^2.

    a) Determine the electric field at p on a vertical axis as shown in the attachment. Up is the positive direction. Answer in units of V/m.

    b) Calculate the vertical component of the electric force on a -3.1e-6C charge placed at point p. Answer in units of N


    2. Relevant equations

    a) E=kQ/r^2
    b) F=Eq


    3. The attempt at a solution

    a) I'm fairly certain I know how to find the field strength from the two charged particles, if point p was directly on the field and in the center.

    E=E1+E2
    E=k/r^2(Q+q)
    E=8.98755e9/3^2(2.2e-6+2.2e-6)
    E=4393 V/m

    But I think I also need to take into account that p is above the two particles, but now I'm stuck...

    b) I think this part would be easier, I just can't do it since I haven't gotten the answer to part a.

    F=Eq
    F=E(that would be found in part a)(-3.1e-6)
     

    Attached Files:

  2. jcsd
  3. Jan 26, 2010 #2

    tiny-tim

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    Hi dgl7! :smile:
    That's correct, if p was on the x-axis, you would use r = 3 for each charge, and add.

    Instead, use r = the actual distance between p and each charge.

    That will give you the force, so remember force is a vector, and add the two vectors.
     
  4. Jan 26, 2010 #3
    Re: Two charges create an electric field--electric field strenght at a point above h.

    AHHH that makes sense. Thanks very much!
     
  5. Jan 26, 2010 #4
    Re: Two charges create an electric field--electric field strenght at a point above h.

    Ok nope. Nevermind, I'm still confused.
    This is what I've been doing and attempting:
    E=Eleft+Eright
    E=kQleft/r^2+kQright/r^2
    (r=r and Qleft=Qright)
    E=2kQ/r^2
    E=2*8.98755e9*2.2e-6/(1.8^2+3^2)
    E=3230.818627 V/m

    Not sure what I'm doing wrong.
     
  6. Jan 26, 2010 #5

    tiny-tim

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    The electric field is a vector,

    so the electric field from each charge has the magnitude kQ/r2, but it also has a direction.

    The direction is different for each charge, so you can't just add the magnitudes, can you? :smile:
     
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