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Cork ball electrometer

  1. Jun 27, 2009 #1

    Bas

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    1. The problem statement, all variables and given/known data
    Two cork balls hang with a massless insulating thread to the same support point, both balls get charged equally and consequently spread apart to a new equilibrium position.
    Given variables:
    • weight cork ball: 0.2 g
    • length thread: 20 cm
    • charge ball: 3.0*10-8C
    T = tension in thread
    L = radius between the two balls
    Horizontal force: F = force from electrical charge (horizontal direction)
    Vertical force: mg

    Question (the part wich I can't solve) : find the tension in the threads after adding the charge and the angle θ


    2. Relevant equations
    horizontal force: T sin θ = F = kqq/r2 [1]
    vertical force: T cos θ = mg [2]
    r = 2L sin θ [3]
    Horizontal force/vertical force = tan θ = F/mg = kq2/r2mg = kq2/(2Lsinθ)2mg [4]


    3. The attempt at a solution
    See the equations above (they were not given and are part of the attempt at a solution).
    The problem is that I have to find out the value of sin θ, once I've got that I must calculate T which can be easiest found with equation 2: T = mg/(cos θ)
    How can I calculate sinθ when I don't know the values of F and r?


    Thanks to anyone who is so kind to help me with this.
     
    Last edited: Jun 27, 2009
  2. jcsd
  3. Jun 27, 2009 #2

    ideasrule

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    Re: electrometer

    There's a much easier way to get to the answer. Examine the equations:

    T sin θ = F = kqq/r2
    T cos θ = mg

    What happens if you square both sides of much equations?
     
  4. Jun 27, 2009 #3

    tiny-tim

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    Welcome to PF!

    Hi ideasrule! Welcome to PF! :smile:
    Your equation [4] is the correct one.

    It's of the form cosθ = Asin3θ, which boils down to a cubic equation in sin2θ:

    1 - sin2θ = A2(sin2θ)3

    I don't think you can get it any simpler than that. :redface:
    that won't help in this case, because r is a function of θ.
     
  5. Jun 27, 2009 #4

    Bas

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    Re: electrometer

    This is what the official solution manual of the book (Pearson) says (of the same edition as the book, I checked it), I only use it to check my answer after I tried it myself:

    tan θ = F/mg = kqq/r2mg = kqq/(2L sin θ)2mg
    = (9 × 109 N · m2/C2)(1 × 10–7 C)2/[2(0.20 m) sin θ]2(0.2 × 10–3 kg)(9.8 m/s2) = 0.0065/(sin2 θ).
    This equation has only one unknown, θ, but the presence of trigonometric functions makes the algebra a little messy. When we calculate both sides for a range of angles, we get
    sin θ = 0.19, θ = 11°.

    As you see they accidentally use another value for the charge, don't mind that. I realy don't understand how they came to this conclusion.
     
    Last edited: Jun 27, 2009
  6. Jun 27, 2009 #5

    ideasrule

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    Re: electrometer

    Calculating both sides for a range of angles means choosing random numbers for θ, plugging them into both sides of the equation, and seeing how the numerical values of the two sides compare. If the two sides are very nearly equal, the θ used is close to the solution. This trial-and-error approach has to be used because it's very difficult to solve an equation in the form tan θ=A/sin^2 (θ) via algebraic means.

    tiny tim: Oops, I didn't realize θ wasn't known. Guess I should read more carefully next time!
     
    Last edited: Jun 27, 2009
  7. Jun 28, 2009 #6

    Bas

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    Re: electrometer

    Both thank you for your answers.
    However, it seems a bit odd to me that you just have to try it randomly.
    I also don't get that 0,0065 when I calculate this so where does that number come from?
    (9 × 109 N · m2/C2)(1 × 10–7 C)2 = 900
    [2(0.20 m) sin θ]2(0.2 × 10–3 kg)(9.8 m/s2) = 0.0065/(sin2 θ) = 0,42 * 0.0002 * 9.8 * sin2 θ = 0.000032 sin2 θ
    So I get 900/0.00031 sin2 θ
     
    Last edited: Jun 28, 2009
  8. Jun 28, 2009 #7

    Redbelly98

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    Re: electrometer

    Well, it's not totally random. There is a procedure you can follow:

    1. You know the solution should be between 0° and 90°.
    2. Break the interval [0,90] into thirds, and test the equation at 30° and 60°.
    3. Based on the results of (2), you can figure out if the answer is in the interval [0,30], [30,60], or [60,90].
    4. Continue narrowing down the interval until you have an answer to sufficient accuracy, i.e. to the nearest 1° or 0.1° or ?

    (Using a spreadsheet would help with making the calculation go faster.)

    You forgot to square the charge: (1 × 10–7 C)2 Note
     
  9. Jun 28, 2009 #8

    Bas

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    Re: electrometer

    I understood that but thanks for the explanation.

    Yes, I looked over it :blushing: (I still have some difficulty with reading these postmessages while I'm typing them because of all that code), but it doesn't matter for the answer.
    I get (0,00009/(0,00031 * sin2 θ) , that is nowhere near to 0.0065/sin2 θ
    So where does that 0.0065 come from?
    Maybe I can upload the PDF-file on this forum so you can see the original?

    Edit: the file should be uploaded (I don't know if you can see it as long as a mod hasn't appoved the attachment). It's question 10 on page 5.
     

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    Last edited: Jun 28, 2009
  10. Jun 28, 2009 #9

    ideasrule

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    Re: electrometer

    I also get around 0.00009/0.00031, so maybe the answer sheet is wrong?

    Anyhow, if you use the simplification tiny tim suggested, you'll get a depressed cubic of the form Ax^3 + Bx = B. This type of equation is actually not very hard to solve analytically; see http://www.sosmath.com/algebra/factor/fac11/fac11.html ("Solving the depressed cubic" section) for a very simple way of doing it.
     
  11. Jun 29, 2009 #10

    Bas

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    Re: electrometer

    Well, I'll then assume that the answer sheet is indeed wrong since it also used the wrong number for the charge. I find that a bit dissapointing of Pearson, knowing that it's one of best known publishers for science books. :confused:
    Thanks for the help and thank you for the link.
     
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