1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    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

    User Avatar

    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.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?