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Static system + a second long distance force

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data

    When an electrically charged object (a proton, an electron, a balloon after you’ve rubbed it
    on your hair) is placed in an electric field, E , an electric force proportional to this field acts
    on the object. Specifically,

    F = qE

    where q is the charge on the object and is in units called Coulombs (C).

    [​IMG]

    The figure shows a pendulum, the bob of which is charged. The ambient electric field is
    uniform, directed toward the right, and has a magnitude of 2 × 10^5 N/C. The pendulum
    hangs at an angle of 20 degrees as a result of the forces acting on it. What is the charge on the bob of the pendulum? The string can be considered of negligible mass and uncharged.

    So:

    (theta) = 20 degrees
    E = 2 x 10^5 N/C
    m = 5.0g (or 0.005kg)
    a(g) (acceleration due to gravity) = -9.8m/s
    F(e) (force of electric... thing) = ?
    T(xy) (x and y component of tension in rope) = ?
    F(net) (net force) = 0

    2. Relevant equations

    F = ma
    F = qE
    probably like four I can't remember

    3. The attempt at a solution

    I know that F(net) is zero, 'cause it's a static system. So:

    F(net) = T(xy) - F(e) - F(g) = 0

    The force of gravity is 0.049N (0.005kg*9.8m/s^2).

    But I have no idea how to find the tension in the rope. If I could get it, the only thing left would be to solve for F(e) using the net force equation, and then rearrange the F(e) = qE equation to solve for q, but that's the easy part.

    The second long-distance force is throwing me off. Also, I forget all of high school trig (it's been like, four years) and even then I'm not sure how I would apply it. cos(theta)*T(xy) = T(y), sin(theta)*T(xy) = T(x) (which, I'm kind of grabbing at straws here, so I give myself a 50/50 chance on being right). So what - I can't see a way for me to get or apply either of those. I'm stuck. :<

    Thanks for reading.

    EDIT: Is the magnitude of T(x) just the same as F(g)? If it is I'm gonna be mad, 'cause this took way too long. :'(
     
    Last edited: Oct 7, 2008
  2. jcsd
  3. Oct 7, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi imzkris! Welcome to PF! :smile:

    You could find T by taking vertical components …

    but you don't need to find T …

    when there's an unknown force in a known direction, like T, just take components perpendicular to that unknown force. :wink:
     
  4. Oct 7, 2008 #3
    I'm... not following. :shy:
     
  5. Oct 7, 2008 #4

    tiny-tim

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    ok … the tension force, T, is along the string, at 20º to the vertical.

    So take components perpendicular to that (at 20º to the horizontal).

    Then the component of T in that direction will be Tcos90º = 0,

    and the components of the other forces are … ? :smile:
     
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