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Homework Help: Pith Ball's charge

  1. Apr 25, 2009 #1
    1. The problem statement, all variables and given/known data

    A charged pith ball (mass = 2 g) is brought close to an identical, equally charged pith ball that's suspended from a thread. When the pith balls are 3.8 cm apart, the thread sits in equilibrium at 14 degrees from vertical.

    What's the charge on each of the pith balls? and Calculate the deficit/excess electrons

    2. Relevant equations
    Coulomb's Law

    3. The attempt at a solution
    Ok, I know this is already wrong coz I'm getting an invalid answer but I don't rly know how to approach the question...

    So, I assumed that the balls have equal charge and the force = 0 so...

    q=square root of (Fr^2/k)

    Please HELP ME!
  2. jcsd
  3. Apr 25, 2009 #2


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    Welcome to PF.

    Consider the forces on the pith balls.

    The Forces are m*g down and Electrostatic Force horizontally.

    Since the string is at 14° and the forces act at right angles, that tells you the ratio of the forces doesn't it?

    Tan 14° = F_electro / F_gravity


    E_electro = kq2/r2 = m*g*tan14°
  4. Apr 25, 2009 #3


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    Split the tension into vertical and horizontal components.

    Now balance the vertical and horizontal forces noting that the force of repulsion between the two is the horizontal force.
  5. Apr 25, 2009 #4
    Thanks for the quick reply! ^^

    Ok...so I just wanna ask if it's right to assume that both pithballs have the same charge?

    Is what I did here correct?

    Fe = mg*tan 14
    = (2) (9.8) (tan 14)
    = 4.89 N

    q = square root of (F r^2 / k)
    = square root of (4.89*(0.038)^2/ 9*10^9)
    = 8.86 x 10^-7

    Since it's a positive charge > it has a deficit of electrons...

    n = 8.86 x 10^-7/1.602 x 10-19
    = 5.53 x 10^12

    Thanks a lotttt for this!!
  6. Apr 26, 2009 #5


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    Yes, you can assume both charges are equal, since they said the pith balls are "identical, equally charged".

    But watch your units. The mass is in grams, not kg.

    As to whether there is a deficit or excess of electrons, all you know is both balls are the same if they are tied at a common point on the vertical and are repelling each other. They would be opposite charges if they were separated and hanging in || before establishing the attraction between the two. You supplied no diagram, so ...
    Last edited: Apr 26, 2009
  7. Apr 26, 2009 #6
    ...oopz, clumsy me! thanx ^^

    ...here's a diagram provided...

    so how do we know if it has deficit or excess electrons?

    I thought if u get a positive charge then it has deficit and excess if negative...?
    Last edited: May 28, 2009
  8. Apr 26, 2009 #7


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    I don't think you need to. I think all that is relevant is that they are just saying it is one or the other, not asking which.
    After all it is a square root of q.
  9. Apr 26, 2009 #8
    Thanks again for the fast reply...the thing is that the second part of the question is asking me to calculate the deficit or excess electrons...?
  10. Apr 26, 2009 #9


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    You can determine the number from the value for the charge that you calculated.

    As to whether it is a deficit or excess, that depends on the method by which the pith balls were charged.
  11. Apr 26, 2009 #10
    so just to double check...do i just divide q by the charge of an electron?


    +/- 2.80 x 10^-8 divided by 1.602 x 10^-19 = +/- 1.75 x 10^11 ??

    is that how u do it?
  12. Apr 26, 2009 #11


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    You should have gotten an answer in Coulombs for charge. How many charges to a Coulomb?
    So basically yes, that's the right answer.
  13. Apr 26, 2009 #12
    I cannot thank you enough! I'm starting to love this forum! ^.^
  14. Apr 26, 2009 #13


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    Happy to help. Good luck.
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