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Homework Help: Uniform electric field - tension & electric field strength

  1. Feb 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A ball with a mass of 3.0 x 10-4 kg hangs from a 15 cm long thread between two parallel plates. The thread is deflected to the side, as shown in the following diagram. The charge on the ball is +5.0 x 10-5 C.

    a) Find the tension in the thread
    b) Find the magnitude and direction of the electric field between the plates
    image1 (1).jpg

    2. Relevant equations
    a) FTy =Fg
    FTx = FE

    b) ε = FE/q

    3. The attempt at a solution

    a) I solved for the horizontal and vertical components of tension (which I believe are equivalent, respectively, to the electric force and force of gravity, since the object is at equilibrium). I then found the resultant force of tension and the corresponding angle:

    3.3 x 10-3 N [N 27° W]

    b) 29 N/C [East]
  2. jcsd
  3. Feb 13, 2016 #2


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    I think you worked it correctly. It appears you used an angle of 27o for the thread even though the picture says 35o.

    For (b) I get an answer closer to 30 N/C rather than 29 N/C for an angle of 27o for the thread.
  4. Feb 14, 2016 #3
    Thanks so much for your help. I've corrected some probable errors, but I'm still not sure if I'm using the correct steps so I've typed them out in full.


    FTy = Fg
    (3.0 x 10-4 kg)(9.8m/s2)
    = 2.94 x 10-3 N [North]

    FTx = FE
    = tan35°(2.94 x 10-3)
    = 2.05861 x 10-3 N [West]

    FT2 = (2.94 x 10-3)2 + (2.05861 x 10-3)2
    FT = 3.6 x 10-3 N [N 35° W]

    b) ε = FE/q
    = 2.05861 x 10-3 N/5.0 x 10-5
    = 41 N/C [East]

    I assume the field goes from East to West since the positive charge is being pushed East.
  5. Feb 14, 2016 #4


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    Looks good.
  6. Feb 15, 2016 #5
    Thank you
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