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Homework Help: Charged mass on a string

  1. Feb 6, 2006 #1
    A charged mass on the end of a light string is attached to a point on a uniformly charged vertical sheet of infinite extent. The acceleration of gravity is 9.8 m/s^2 and the permittivity of free space is 8.85 x 10^-12 C^2/Nm^2. Find the angle [tex] \theta [/tex] the thread makes with the vertically charge sheet. Answer in units of degrees.
    Given:
    mass of ball= 1 g
    Areal charge density of the sheet= 0.23 [tex] \mu C/m^2[/tex]
    length of the string = 78.9 cm
    Then force of charge= qE= q[tex] \sigma [/tex] / 2E_0
    We did some of this problem in class and went through the long process of drawing a free body diagram and summing up the components, we found that it was easier to use the pythagorean theorem to solve for T.
    I found that T= [tex] \sqrt (mg)^2 + (qE)^2 [/tex]
    So T= [tex] \sqrt 96.04 + 1.32 x 10^-5 [/tex]
    So T= 9.8.
    Then I plugged it into what we got for the forces in the y-direction, which was [tex] \theta= cos^-1 (-mg/T) [/tex]
    So theta= cos ^-1 (-9.8/9.8)
    = 180 degrees which is wrong... can someone help me please?
     
  2. jcsd
  3. Feb 6, 2006 #2

    Doc Al

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    Staff: Mentor

    Check your calculation of [itex]q E[/itex] and [itex]mg[/itex] (note that m = 0.001 kg). What's [itex]q[/itex]?
     
    Last edited: Feb 6, 2006
  4. Feb 6, 2006 #3
    Well I stupidly forgot to change to kg, but I'm still getting the wrong answer.
    [tex] T= \sqrt (mg)^2 + (qE)^2 [/tex]
    mg= .001 * 9.8 = .0098
    qE= [tex] q \sigma/2 E_o [/tex]
    qE= 2.8 x 10^-7 * 2.3 x 10^-7 / 2 * 8.85 x 10^-12
    qE= .00364
    T= [tex] \sqrt (.0098)^2 + (.00364)^2 [/tex]
    T= .0104
    [tex] \theta= cos^-1 (-mg/T) [/tex]
    [tex] \theta = cos^-1 (-.0098/.0104) [/tex]
    [tex] \theta= 159 degrees [/tex]
     
  5. Feb 7, 2006 #4

    Doc Al

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    Staff: Mentor

    What's with the minus sign?
    [tex] \theta= \cos^{-1} (mg/T) [/tex]

    Your calculation would be a bit easier if you used:
    [tex] \theta = \tan^{-1} (qE/mg) [/tex]
    (This way you don't have to calculate T.)
     
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