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Coulombs Law of two metal balls

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Two metal balls A and B of negligible radius are floating at rest on Space Station Freedom between two metal bulkheads, connected by a taut nonconducting thread of length 1.30 m. Ball A carries charge q, and ball B carries charge 2q. Each ball is 1.66 m away from a bulkhead.
    (a) If the tension in the string is 4.00 N, what is the magnitude of q?
    (b)What happens to this system as time passes ?


    2. Relevant equations
    F=Ke q1xq2 / x2



    3. The attempt at a solution

    I have gotten as far as finding the magnitude of q, which i got 1.94 x 105 C

    Is that right ?

    I am confused as to what this system might do over time ?
     
  2. jcsd
  3. Sep 17, 2009 #2

    kuruman

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    I haven't done the calculation but 1.94x105 C sounds like a bit too much. Can you show the details of your calculation?
     
  4. Sep 17, 2009 #3
    I had somebody help me (hope they didnt mess me up ! )

    I was told that Frepulsion is the same as tension, so therefore

    4=8.9875x109 = q1xq2 / x2

    From here , we can use algebra to get the following equation.

    q2= 4N(1.3M)2 / 2 (8.987x109)

    ??
     
  5. Sep 17, 2009 #4

    kuruman

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    No, they didn't mess you up. The numbers as you have them are correct. If you redo the calculation you should get a different number from the one you quoted initially with negative powers of ten for the charge. So you need to redo it.
     
  6. Sep 17, 2009 #5
    I keep coming up with 194333.1241

    by multiplying out the numbers, i get to 6.76 / 1.797 x 10 10

    Then the square root of that
     
  7. Sep 17, 2009 #6

    kuruman

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    What you are calculating is

    6.76/1.791*1010 which your calculating program interprets as 6.76*1010/1.791

    What you should be calculating is

    6.76/(1.791*1010)

    The first expression puts 1010 in the numerator. the second expression puts it in the denominator where it belongs.
     
  8. Sep 17, 2009 #7
    I sort of figured that out, but thanks for confirming that. So do I need to store the 1.1791 x 10 to the tenth ? then divide it out ?
     
  9. Sep 17, 2009 #8

    kuruman

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    Yes, or take the inverse of the number first and then multiply what you get with 6.76, then take the square root.
     
  10. Sep 17, 2009 #9
    Ok, I now come up with 3.77 x 10 -10

    You know whats sad ? Is that I used a free trial of one of those tutor services and he came up with the same 1.94 x 10 10
     
  11. Sep 17, 2009 #10
    So what will this system do as time passes ? I cant even fathom a response to that part.
     
  12. Sep 17, 2009 #11
    So is the 1.66 M away from each bulkead basically a non used number thrown in to screw with us ?
     
  13. Sep 17, 2009 #12

    kuruman

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    You forgot to take the square root of that.

    What do you think happens when time goes on? These balls are near a conducting surface. What happens to the conducting surface when when you bring a whole of charge near it?
     
  14. Sep 17, 2009 #13
    1.9 x 10 -5 ?

    Will the charge of the balls be transferred to the bulkhead ?

    When a conducter is charged in a small region, the charge will be distributed across the entire surface.
     
  15. Sep 17, 2009 #14
    Re: Coulombs Law - SOLVED !

    Solved !
     
  16. Sep 17, 2009 #15

    kuruman

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    The charge on a ball induces charges of opposite sign on the bulkhead nearest it. Because one bulkhead pulls one way and the other the opposite way, the non-conducting string that connects the balls is under tension. This is my question: Is there a net force on the two-charged-balls system? If "yes" the balls will accelerate and hit a bulkhead; if no, the balls will just sit there.
     
  17. Sep 17, 2009 #16
    Well, because the string is under tension and is non conducting, wouldnt there have to be a net force acting upon it ? Its what is pushing the balls apart to begin with right ?
     
  18. Sep 17, 2009 #17

    kuruman

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    The tension is internal to the system. If ball A pulls on ball B with force F, then ball B pulls on ball A with force -F by Newton's 3rd Law. Therefore the tension does not affect the two-ball system. As I said, the tension is generated by the bulkheads pulling on the balls nearest them. How do the forces exerted by the bulkheads compare?
     
  19. Sep 17, 2009 #18
    I think I see where you are going, because opposites attract, the forces on each bulkhead are the same, which is pulling each ball towards the bulkheads. Meaning the forces on the ball are like forces ? I dont know if im confusing myself more.
     
  20. Sep 17, 2009 #19

    kuruman

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    The fact that there is tension in the string does not mean that there is a net force on the balls. If you don't believe me, grab your shirt collar and pull up with your hand. There is tension in your shirt as you pull, but there is no net force acting on you. If there were a net force, you will be flying. What external forces do you think generate the tension in the string?
     
  21. Sep 17, 2009 #20
    I really dont know.

    I just sat down with a tutor at school for 45 minutes and he couldnt explain any of my homework problems, so quite honestly, I give up.

    You can close this thread if you would like.
     
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