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Electrical Energy and Coulombs Law

  1. Dec 8, 2014 #1
    1. The problem statement, all variables and given/known data

    Charge q1 is fixed to a spring with constant k_s. It is at equilibrium initially and located r+x away from q2 (which is oppositely charged). The charge q1 is released and allowed to move closer to q2, pulling the spring and stopping. Now the charges are only r distance away. I would like to assume I can solve this using both force and energy approaches. Here is my attempt...



    2. Relevant equations


    3. The attempt at a solution
    Using forces...
    [tex]

    |F_{ e }|=|F_{ s }|\\ \frac { k_{ e }q_{ 1 }q_{ 2 } }{ r^{ 2 } } =k_{ s }x\quad (1)
    [/tex]

    Using Energy
    [tex]
    \\ \\ r^{ ' }=r+\Delta x\\ E_{ T }=E_{ e }+E_{ s }\\ \frac { k_{ e }q_{ 1 }q_{ 2 } }{ r^{ ' } } =\frac { k_{ e }q_{ 1 }q_{ 2 } }{ r } +\frac { 1 }{ 2 } k_{ s }\Delta x^{ 2 }\\ \frac { k_{ e }q_{ 1 }q_{ 2 } }{ \Delta x } (\frac { 1 }{ r^{ ' } } -\frac { 1 }{ r } )=\frac { 1 }{ 2 } k_{ s }\Delta x\\ \frac { k_{ e }q_{ 1 }q_{ 2 } }{ \Delta x } (\frac { r-r^{ ' } }{ r^{ ' }r } )=\frac { 1 }{ 2 } k_{ s }\Delta x\\ k_{ e }q_{ 1 }q_{ 2 }(\frac { 2 }{ (r+\Delta x)r } )=k_{ s }\Delta x\quad (2)\\

    [/tex]

    To me, equation (1) and equation (2) should be the same. They are not. The only way they can be is if r=x. Any help would be greatly appreciated...

    Ryan
     
  2. jcsd
  3. Dec 8, 2014 #2

    Bystander

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    What was the original question in the problem statement? You've stated your question, and answered it correctly, r = x.
     
  4. Dec 9, 2014 #3
    Thank you for taking the time to look at this. My question is whether or not I can solve a question like this using both force and energy. I am really surprised at the answer, that r will be x. Doesn't that imply that the charged sphere will always move halfway towards the other sphere despite charge or spring constant?
     
  5. Dec 9, 2014 #4

    gneill

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    The problem with the energy approach is that it leaves out kinetic energy. You're making the assumption that the charge will move from one place to the other and just stop there -- well, what's to prevent oscillation? There's some "unseen hand" stealing away kinetic energy in order to bring the charge to rest at its destination.

    There is a similar exercise using a spring, a mass, and gravitation: A mass m is attached to an unstreched spring with spring constant k which is fixed at one end to the ceiling. You want to find how much the spring stretches due to weight of the mass.

    Clearly if you just let the mass go the system will oscillate. The mass will pick up KE on the way down and pass the new equilibrium point and oscillate around the new equilibrium. In order to "kill" the oscillation you need to steal away the kinetic energy from the system. You'll often see words like "...the mass is lowered slowly to its new equilibrium position..." so that this unspecified mechanism steals the KE.
     
  6. Dec 9, 2014 #5
    That makes sense. Thank you all for your time on this. I appreciate it. I guess if I applied equations for damping this would also "steal" kinetic energy and make the energy interpretation more realistic.
     
  7. Dec 9, 2014 #6

    gneill

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    Staff: Mentor

    You could do that and solve the differential equation for the position with respect to time, then determine the new equilibrium position from that.
     
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