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Homework Help: How many electrons would it take for a balloon to stick to the ceiling?

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data
    How many electrons would it take for a typical balloon to stick onto a ceiling? State assumption for ceiling size or balloon size if necessary.

    2. Relevant equations
    e = 1.60 x 10 ^ -19 C
    q = Ne

    3. The attempt at a solution
    I thought that it took only one electron for a balloon to induce a slight charge of opposite polarity in the surface of the ceiling. But I do not believe I am right. Please, any help would be appreciated. Thanks
  2. jcsd
  3. Jan 26, 2010 #2
    Hi there,

    For the balloon to stick on the wall, the friction must compensate the weight of the balloon. Knowing that friction is dependent on the attraction between the wall and the balloon, you should be able to find this out.

  4. Jan 26, 2010 #3
    i still dont get it.. anyone else care to explain?
  5. Jan 27, 2010 #4
    Hi there,

    Let's try with a bit more details. The balloon has a weight, quite little but still:

    [tex]F_g = mg[/tex]

    For the balloon to hold on the wall, you must counter weight this force:

    [tex]\sum \vec{F} = 0[/tex]

    on the vertical axe (y):
    [tex]-F_g + F_{\text{counter}} = 0[/tex]

    This counteracting force is the friction on the wall created by the electrostatic interaction.

    [tex]F_{\text{counter}} = \mu \cdot N[/tex]

    I believe I gave more than enough information on your problem. The rest of the solution should be quite easy to find out.

  6. Jan 27, 2010 #5
    But the balloon should stick to the ceiling, so friction doesn't come into it, and the electrostatic
    force has to counter gravity directly.
  7. Jan 27, 2010 #6
    Hi there,

    The problem is a bit different when you stick the balloon to the ceiling. In this case, the electrostatic force should only counter the gravity directly.

    But the original question was on a wall, where only the friction of the balloon can counteract the weight.

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