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What is the magnitude of the electric field?

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Three point charges, q1 = -4.19 nC, q2 = 5.41 nC and q3 = 2.51 nC are aligned along the x axis as shown in the figure below.
    [URL]http://imgur.com/RaRML[/URL]
    Assume that L1 = 0.496 m and L2 = 0.816 m. Calculate the electric field at the position (0, 2.02).


    2. Relevant equations

    E = kq/r^2

    3. The attempt at a solution

    Horizontal electric field = Ex = k(-4.19*10^-9)cos76/2.08^2 + k(2.51*10^-9)cos68/2.18^2 = -0.3276 N/C

    Vertical electric field = Ey = k(-4.19*10^-9)sin76/2.08^2 + k(5.41*10^-9)/2.02^2 + k(2.51*10^-9)sin68/2.18^2 = 7.88 N/C

    Net field = square root of Ex^2 + Ey^2 = 7.88 N/C

    Please help! :(
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Jul 17, 2011 #2
  4. Jul 17, 2011 #3

    Redbelly98

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    That equation gives the magnitude of the electric field, but not its direction. To tell what direction it points in, use the rule the E points away from positive charges or towards negative charges.

    You're close. Here are two questions for you:

    In what direction does the horizontal part point due to the -4.19 nC charge? (Hint, E points towards this negative charge, therefore it's in the (+x, -x?) direction.)

    In what direction does the the horizontal part point due to the +2.51 nC charge? (Hint, E points away from this positive charge, therefore it's in the (+x, -x?) direction.)

     
    Last edited by a moderator: Apr 26, 2017
  5. Jul 18, 2011 #4
    The components of the negative charge point in the opposite direction relative to the two positive charges, right? So the horizontal part of the negative charge points in the -x direction while that of the positive charge points in the +x direction. Isn't that accounted for with the negative sign in front of the 4.19 nC charge? So I ended up subtracting the magnitudes of the negative charge's components from those of the positive charges...is that not right?
     
  6. Jul 18, 2011 #5

    Redbelly98

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    Since the point in question is to the left of the positive charge q3, and E points away from positive charges, then the horizontal component of E due to q3 points to the left, i.e. it's negative also.

    If you still don't see that, label the point (0, 2.02) in the diagram. Then draw a vector pointing directly away from charge q3. In what direction is the horizontal component of this vector?
     
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