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Magnitude of charge with common Transparent tape

  1. Mar 7, 2017 #1
    1. The problem statement, all variables and given/known data

    a) Common transparent tape becomes charged when pulled from a
    dispenser. If one piece is placed above another, the repulsive force can
    be great enough to support the top piece’s weight. Assuming equal point
    charges (only an approximation), calculate the magnitude of the charge if
    electrostatic force is great enough to support the weight of a 10.0 mg
    piece of tape held 1.00 cm above another.

    2. Relevant equations

    F = k |q1 q2| / r2

    3. The attempt at a solution
    I'm getting confused: I know r=1cm and the two pieces of tape have the same charge.
    I am certain I need to add the weight in, but how?
    I even tried using Newton's Law of gravitation with 10mg being the mass, but that brought no insight.
     
  2. jcsd
  3. Mar 7, 2017 #2

    gneill

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    Staff: Mentor

    Draw a free body diagram for the floating piece of tape. What forces are acting?
     
  4. Mar 7, 2017 #3
    For the piece of tape to be suspended in air, the sum of the forces has to equal zero - the same situation as if that piece of tape was lying on a table. Did you draw a free body diagram for the tape and sum the forces acting on the tape?

    Welcome to Physics Forums.

    Edit: @gneill, you are too fast for me. :)
     
  5. Mar 7, 2017 #4
    I did. You have gravity pulling down the top piece of tape and because of the pieces being equally charged the bottom tape is pushing the top one away.
     
  6. Mar 7, 2017 #5

    gneill

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    Okay, so you should be in a position to write an equation. What are the expressions for the two forces?
     
  7. Mar 8, 2017 #6
    Sorry, I lost internet yesterday.
    So would I use the equation of weight (w=mg) for the tape on the top and then use Coulomb's Law for the static force between the two?
     
  8. Mar 8, 2017 #7

    gneill

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    Yup.
     
  9. Mar 8, 2017 #8
    So m=0.00001kg and g=9.8m/s2
    which makes 0.000098N

    Should I then place that answer as F in F = k|q1 q2|/r2 ?
     
  10. Mar 8, 2017 #9

    gneill

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    What do you think? Try it. What do you get?
     
  11. Mar 8, 2017 #10
    So if I continue with r=0.01m and k=8.988x109

    then my equation becomes |q1 q2|=215.4895
    if this path is right, I would then square root the answer since both pieces are to have equal point charges.
    Right?
     
  12. Mar 8, 2017 #11

    gneill

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    That doesn't look right. It's far, far, far too large. Show your calculation in detail.
     
  13. Mar 8, 2017 #12
    First we change it equation to find what I don't have:
    F
    = k|q1 q2|/r2
    F/k = |q1 q2|/r2
    (F/k)r2 = |q1 q2|
    So we take F=0.000098N from the weight equation
    k is the constant =8.988x109 Nm2/C2
    r=0.01m
    Oh! I messed up it's times r2
    Which would make the solution=2.667x10-8
     
  14. Mar 8, 2017 #13

    gneill

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    Okay so far.
    Solution to what? What are the units? Also, it doesn't look like it would be a result coming from what you've shown so far.

    Maybe try doing the calculation one step at a time. What do you get for F/k?
     
  15. Mar 8, 2017 #14
    0.000098N/8.988x109 Nm2/C2 =1.09x10-14 m2/C2
    Times that with r2=0.0001m2 = 1.09x10-18 C2
    Then because the two pieces are equal we square root the answer to get 1.044x10-9 C

    So I don't know how the numbers changed.
     
  16. Mar 8, 2017 #15

    gneill

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    That's much better :smile: Your result looks good.
     
  17. Mar 8, 2017 #16
    But now I'm lost how do I figure out if this magnitude from the electrostatic force is great enough to support the top piece of tape?
     
  18. Mar 8, 2017 #17

    gneill

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    :confused: There's nothing to figure out; You found the charge using an equation that balanced the two forces. So the forces are exactly equal in magnitude.
     
  19. Mar 8, 2017 #18
    Oh, I see. Thank You very much now I understand what I did.
    Thank You for your help.
     
  20. Mar 8, 2017 #19

    gneill

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    Staff: Mentor

    You're welcome!
     
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