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Homework Help: Pith ball and parallel plate capacitor prolem

  1. May 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A 10 gram pith ball having a charge of 1μC is electrostatically attracted to one of the inside surfaces
    of a large parallel plate capacitor by an electric field between the plates. The plates are vertically
    oriented so that the electric field between the plates is horizontal. The distance between the plates of
    the capacitor is 10 cm, and the coefficient of static friction between the pith ball and the plate is 0.5.
    What is the minimum voltage needed to keep the ball from sliding down the plate from gravity? Draw
    a picture to show all forces on the pith ball.

    2. Relevant equations

    I used F = μ(coef. of static)(mg) to find the amount of force it would take to overcome static friction then applied it to the equation V = (F/q charge)(d).

    3. The attempt at a solution

    I got 50 V as my answer but I don't know if this is right. It would be appreciated if someone could check behind me or tell me what I'm doing wrong.

  2. jcsd
  3. May 28, 2012 #2
    I think you may be neglecting some forces here. Think about reducing your electric field to zero, what would happen then and why? Remember the plates are vertical. Then restore your field. What voltage do you need to create a force strong enough to oppose the forces you just considered.

    I hope that helps.
  4. May 28, 2012 #3

    I don't really understand. I would think that if you made the electric field equal to zero the pith ball would just be falling straight down because there wouldn't be any forces acting on it in the x direction.
  5. May 28, 2012 #4
    Yes exactly. Think about why to understand the problem . The voltage doesn't need to create a force to overcome static friction it needs to overcome gravity. Think about the direction of the forces due to friction and gravity.

    I think you need to consider the net force in the y-direction. In this case however it does not alter your the result from your value. This is a special case for coefficient of friction 0.5.

    Also think about your units, remember your charge is 1 MICRO coulomb and distance ten CENTI-metres.
  6. May 28, 2012 #5
    But doesn't friction help in overcoming gravity?

    Let's say I find the force of gravity on the pith ball F = mg which is F = (.01 kg)(9.80 m/s^2). I calculated this out to be .098 N. Shouldn't the friction be added into this equation to lessen the amount of force needed to overcome gravity?

    After I added the friction the equation the amount of force on the pith ball went down to half it's orignal measure at .049 N. Now wouldn't we just calculate the magnetic field, E = F/charge, using the force on the ball? So, E = .049/ 1x10^-4 which is 490 N/C?

    Then once you get E can't you just plug it into change in V = Ed? We have the electric field and distance so wouldn't that be able to give us the voltage we need?
    Last edited: May 28, 2012
  7. May 28, 2012 #6
    yes so you have 0.49 in positive y direction from friction and 0.98 in negative y direction from gravity. F(total) = 0.98 - 0.49 = 0.49 in negative y direction.

    That is the total force acting on the sphere. Use the formula you used initially and don't forget to use S.I units.
  8. May 28, 2012 #7
    Which formula?
  9. May 28, 2012 #8
    For the voltage as you did initially if that is the way you have been shown to do in class. Or you could apply Coulomb's law and take it from there.
  10. May 28, 2012 #9
    I did and I got 49 V, but what is the way you're talking about? Is that where F = K ((q of a)(q of b)/ r2)?
  11. May 28, 2012 #10
    F = (0.49 / 1.0*10^-6 C) (0.10 kg) .
    Do it how you have been shown but yes that is coulomb's law and it will give electrostatic charge needed to balance the force, then you can find a voltage.
  12. May 28, 2012 #11
    But don't you need the charge of the plate as well?
  13. May 29, 2012 #12
    Yes you then need the charge of the plate, you can find this be rearranging Coulomb's law.
  14. May 29, 2012 #13


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    Homework Helper

    The charge of the plate is not needed. Coulomb's law apllies to point charges. Here you have a pair of charged plates with homogeneous electric field between them.

  15. May 29, 2012 #14


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    Homework Helper

    Have you drawn that picture?

    Look at the picture. The normal force is not mg!! It would be in case if the ball was supported by a horizontal table. But it is pressed against the plate of the capacitor by electrostatic force! The normal force from the vertical plate balances the electric force. The static friction balances gravity.

    If you find the electric force Fe your formula for the voltage V=(Fe/q)d is correct.


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