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Homework Help: Electric Field between 2 Point Charges

  1. Jan 27, 2008 #1
    [SOLVED] Electric Field between 2 Point Charges

    1. The problem statement, all variables and given/known data
    http://session.masteringphysics.com/problemAsset/1011305/15/1011305A.jpg

    Two point charges are placed on the x axis. The first charge, q1 = 8.00 nC, is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.00 nC, is placed a distance 9.00 m from the origin along the negative x axis.

    Find the electric field at the origin, point O.

    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.

    2. Relevant equations
    E = F/q
    F = kq1q2/r^2
    E = p / (2pi * e0 * x^3)
    p = qr

    q1 = 8*10^-9
    q2 = 6*10^-9
    r = 9+16 = 25 (The distance between the two point charges)
    e0 = 8.85*10^-12
    x = r (Assumably, not for certain)


    3. The attempt at a solution
    Since both charges are along the x axis, I conclude that they do not pose any influence on the y coordinate of the field, therefore the y coordinate is 0.

    The x coordinate can be computed via E = F/q.
    We are trying to calculate E.
    F can be calculated via F = kq1q2/r^2
    Issue is, what do we use for q?

    Alternatively, we could use p = qr to find p (But again, which q to use?)
    Then we can use E = p / (2pi * e0 * x^3)
    (But then which x do we use? 9, 16, or 25? Distance between a point and (0,0), or distance between both points?)
     
  2. jcsd
  3. Jan 27, 2008 #2
    I think you're making this too hard. All you need to do is find the electric field at the origin due to [tex]q_1[/tex] and add it to the electric field at the origin due to [tex]q_2[/tex]. In other words,

    [tex]\vec{E} = \vec{E}_1 + \vec{E}_2[/tex]
     
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