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Electric field due to two point charge

  1. Jan 23, 2008 #1
    Two point charges are placed on the x axis. The first charge, q1 = 8.00nC , is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.0nC , is placed a distance 9.00m from the origin along the negative x axis.

    Find the electric field at the origin, point O.
    What is Eox, Eoy=?

    Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.

    So here is what i did....
    Using Coulombs equation to frind electric field E=kQ/r(sqrd)
    I converted from nC to C and found E for both charges.
    Then since they're asking for vector component. I plugged the number(E) in to i^hat of both charges, and the j^hat is zero. Then i added it. But its saying that my answer is wrong. can somebody please help me.
    What am i missing here? I am probably skipping the most important part of this problem......
     

    Attached Files:

  2. jcsd
  3. Jan 23, 2008 #2

    Doc Al

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    Staff: Mentor

    They are looking for an answer like: (a, b)
    Show exactly what you did and maybe we can spot the error.
     
  4. Jan 23, 2008 #3
    I'm working on that EXACT same problem (we're probably working on the same masteringphysics assigment). This is what I got so far.

    I start by "placing" a test charge at the origin.

    Here's what I think I should do:

    E = [tex]\frac{kQ_1}{r^2}[/tex] + [tex]\frac{kQ_2}{r^2}[/tex]

    E_1 = [tex]\frac{(9x10^9)(8x10^-9)}{16^2}[/tex] = -0.281 N/C (it's negative because it's giving a force towards the negative x - axis)

    E_2 = [tex]\frac{(9x10^9)(6x10^-9)}{9^2}[/tex] = 0.667 N/C

    so 0.667 - 0.261 = 0.451.

    Probably wrong.
     
    Last edited: Jan 24, 2008
  5. Jan 24, 2008 #4
    Hey Doc Al did you get this problem figured out. As far as i can tell we're all using the same approach...but in the end no success. Any more ideas or approaches to this problem?
     
  6. Jan 24, 2008 #5
    did you get the same answer as I did for Part A?
     
  7. Jan 24, 2008 #6

    Doc Al

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    Staff: Mentor

    The approach spelled out by cse63146 is correct, but the answers are wrong. Two arithmetic mistakes were made: the calculation for E_1 has an error, as does the final subtraction.
     
  8. Jan 24, 2008 #7
    Found the mistake for E_1; it's 0.281 instead of 0.261

    so the final subtraction would be 0.667 - 0.281 = 0.386
     
  9. Jan 25, 2008 #8
    So in the second case, where q2 is negative (q2 = -6 nC), since it's negative, it would attract the test charge toward the negative x - axis, and thus the electrical field would be negative as well.

    So it would be: - 0.667 - 0.281 = - 0.948
     
    Last edited: Jan 25, 2008
  10. Jan 25, 2008 #9
    Thanks everyone!
     
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