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Two charged beads on a plastic ring

  1. Sep 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Two charged beads are on the plastic ring in Figure (a). Bead 2, which is not shown, is fixed in place on the ring, which has radius R = 63.3 cm. Bead 1 is initially on the x axis at angle θ = 0o. It is then moved to the opposite side, at angle θ = 180o, through the first and second quadrants of the xy coordinate system. Figure (b) gives the x component of the net electric field produced at the origin by the two beads as a function of θ, and Figure (c) below gives the y component. The vertical axis scales are set by Exs = 5.80 × 104 N/C and Eys = -10.44 × 104N/C. (a) At what positive angle θ is bead 2 located? (Note: bead 2 is negative charged). What are the charges of (b) bead 1 and (c) bead 2?

    2. Relevant equations
    E = k (Q/R^2)

    3. The attempt at a solution

    I understand Part 1. I realize that bead 2 must be located at +270 degrees. Because of this fact, we know that @180 degrees the only field acting on the origin in the X direction is from Bead 1. The same applies for the y-direction in @180 and 0 degrees with Q2 being the only field acting on it. So, for the Q1 I did as follows:


    Ex = 5.8E4 = k (Q1/R^2)
    Q1 = (5.8E4)(1/k)(R^2)
    Q1 = (5.8E4)(1/8.99E9)(.633^2)
    Q1 = 2.58E-6 C

    For Q2

    Ey = 4 (-10.44E4 / 9) = -4.64E4 = k (Q2/R^2)
    Q2 = (-4.64E4)(1/8.99E9)(.633^2)
    Q2 = -2.07E-6 C

    WileyPlus is asking for it in uC so I just changed both from E-6 to E-12. I'm not getting the correct answer, where am I making a mistake? I'm starting to think this problem isn't as straightforward as I had assumed, although the solution I was given from the book is pretty much the same to as what I've done.
  2. jcsd
  3. Sep 19, 2015 #2
    Your conversion is wrong.

    You have for example ##Q_2 = -2.07 \times 10^{-6} C = -2.07 \mu C##.
  4. Sep 19, 2015 #3
    Wow I'm an idiot hah. For some reason I was doing the conversion wrong the 3 times I redid the problem. Thank you!
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