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Spring constant versus electric repulsion

Discussion in 'Electronics Homework Help' started by chopnhack, Jul 9, 2017.

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  1. chopnhack

    chopnhack

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    Is this question this simple to answer? I feel like some of these questions are too easy....Makes me uneasy like I am missing something!

    Two spherical point charges each carrying a charge of 40 mC are attached to the two ends of a spring of length 20 cm. If its spring constant is 120 Nm-1, what is the length of the spring when the charges are in equilibrium?


    k = constant of proportionality

    F = k ×(q1q2/r^2)

    9x10^9 N×m2×C-2 × [(40μC^2)/0.2m^2]

    F = 360N


    k = spring constant = 120 N×m-1


    F = kx
    360N = 120 N×m-1 × x
    x = 3m


    When in equilibrium, the springs are stretched to 3 meters.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    What is the force between the two charges when they are 3 metres apart?
     
    Arouse1973 likes this.
  3. chopnhack

    chopnhack

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    F = k ×(q1q2/r^2)
    9x10^9 N×m2×C-2 × [(40μC^2)/3m^2]
    F=1.6x10^-6N this is the force of the two particles acting against each other
    the force of the spring at 3m stretch would be 360N

    I need to rethink this. Thanks for pointing that out! Knew it was too easy.
     
    Arouse1973 likes this.
  4. Arouse1973

    Arouse1973 Adam

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    Hi John
    What is the C-2? Also is it micro or milli-Coulombs. You say mC in your post and then your calculations show uC.
    Cheers
    Adam
     
  5. Arouse1973

    Arouse1973 Adam

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    Ok just realised C-2 is part of the units. :)
    Adam
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    You have 2 issues. The force between the two charges falls as the distance between them increases, and the spring tension (which is a force acting in the other direction) increases as the distance between it's ends increases

    You need to find a distance where these forces balance.
     
    chopnhack likes this.
  7. chopnhack

    chopnhack

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    I think I have a lot more than two issues right now :mad::confused::p:D

    This seems like it needs to be solved by a system of equations.

    I was thinking of graphing the two against each other to see where they cross.

    first equation boils down to y = 14.4/x^2 the second equation is y = 120x.

    x = 0.493242m
     
    Last edited: Jul 10, 2017
  8. Laplace

    Laplace

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    Apr 4, 2010
    How do you account for the fact that when the charges are 20 cm apart the force in the spring is zero?
     
  9. chopnhack

    chopnhack

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    I didn't assume that when I read the question. I took it as the spring is 20cm long when unstretched, then the charges are added.
     
  10. Laplace

    Laplace

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    So was the question, "How far does the spring stretch?" or "How far apart are the charges in equilibrium?"
     
  11. chopnhack

    chopnhack

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    Two spherical point charges each carrying a charge of 40 mC are attached to the two ends of a spring of length 20 cm. If its spring constant is 120 Nm-1, what is the length of the spring when the charges are in equilibrium?
     
  12. Laplace

    Laplace

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    So the spring does not exert any force until the length of the spring exceeds 20 cm. Is that what your equation for the spring force indicates? (the second equation is y = 120x.)
     
  13. chopnhack

    chopnhack

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    I believe so, since the original length is 0.2m. It doesn't start exerting force until it is stretched beyond 0.2m. So it should be amended to y=120*(x-0.2m)?

    14.4/x^2 = 120(x-0.2)

    x=0.569715m
     
  14. Laplace

    Laplace

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    That's the answer I calculated.
     
    chopnhack likes this.
  15. chopnhack

    chopnhack

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    Thank you for helping me through that! I don't think I would have spotted that on my own.... physics is tricky!!
     
  16. Ratch

    Ratch

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    He should write 1/C^2 to avoid confusion and bemusement.

    Ratch
     
    Arouse1973 likes this.
  17. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    C^-2 or C-2 would both be clearer AND consistent with what he is required to use.
     
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