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Electric Field Help

  1. Sep 3, 2008 #1
    1. The problem statement, all variables and given/known data
    At a particular moment, three small charged balls, one negative and two positive,are located as shown in Figure 13.46. Q1 = 3 nC, Q2 = 7 nC, and Q3 = -6 nC.

    What is the electric field at the location of Q1, due to Q2?

    2. Relevant equations
    [tex]\vec E = q1q2\hat{r} / 4\pi\varepsilon r^2[/tex]

    3. The attempt at a solution

    r = <0, .04, 0> m
    rmag = .04
    rhat = <0, 1, 0>

    E = [3e-9(7e-9)(9e9)] / .0016

    Which gives me:

    E = <0, 1.18125e-4, 0>

    What am I doing wrong here? At first I figured I missed a negative, so I tried making my answer negative, but apparently that's not the case.
    Last edited: Sep 3, 2008
  2. jcsd
  3. Sep 3, 2008 #2


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    Isn't the electric field at Q1 due to Q2 given by
    E = kQ/r2 = kQ2/(.04)2 ?
  4. Sep 3, 2008 #3
    The way we've been doing it in class is the way I posted. We haven't even used k...what does it represent?
  5. Sep 3, 2008 #4


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    Hi cowmoo32! :smile:

    No, that's Coulomb's law for the force

    the field is just [tex]q2\hat{r} / 4\pi\varepsilon r^2[/tex] :smile:

    (I think LowlyPion is saying the same thing, and his k is your [itex]1/ 4\pi\varepsilon[/itex])
  6. Sep 3, 2008 #5


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    k is Coulomb's constant

    [tex]k = \frac{1}{4 \pi \epsilon_0}[/tex]

    Edit: Tiny-Tim beat me to it.
  7. Sep 3, 2008 #6


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    I wrote it as k because Tex is more awkward to pound out.

    Sorry if I confused you.
  8. Sep 4, 2008 #7
    Thanks for the help, guys.
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