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Homework Help: Spring with charged objects

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

    Two equal and opposite charged objects of magnitude 2C and mass 2kg are attached to a spring of spring constant 4 N/m in the natural length of the spring. The system is suddenly placed in a uniform electric field of field strength 5 N/C in the direction of electric field such that the -ve charge is towards the origin of electric field. assuming the natural length of the spring to be large, find the maximum compression and elongation in the spring length.

    2. Relevant equations

    3. The attempt at a solution

    First I tried to find the maximum extension.

    Let charge -2 C be at a distance x1 and charge +2 be at a distance x2 from the origin .

    Let the length of the spring be l .

    The length of the spring at any instant is x2-x1 .

    The extension in the spring would be x2-x1-l .

    The force of attraction between the charges would be kq2/(x2-x1)2 i.e 16/(x2-x1)2

    Force due to the external electric field will be of magnitude 10N .

    EOM for mass 1 (i.e -2C ) will be k(x2-x1-l)-10+16/(x2-x1)2 = 2d2x1/dt2

    EOM for mass 2 (i.e +2C ) will be 10-k(x2-x1-l)-16/(x2-x1)2 = 2d2x2/dt2

    From the above we get ,


    or , d2x2/dt2-d2x1/dt2=10-k(x2-x1-l)-16/(x2-x1)2


    Now putting x2-x1 = z ,

    We have d2z/dt2=10-k(z-l)-16/z2 .Now I need to maximize z .

    Is it the correct way to approach the problem ?

    I would be grateful if someone could help me with the problem .

    Attached Files:

    Last edited: Jan 8, 2014
  2. jcsd
  3. Jan 8, 2014 #2
    Hello Tanya!

    I am not sure but I think you are required to use the assumption that the natural length of spring is large to cancel one of the factors.

    Also, how do you get kq^2=20?
  4. Jan 8, 2014 #3
    For this kind of problems (two interacting bodies) it is usually simplest to convert to the CoM frame and use conservation of total momentum in the CoM frame. This separates coordinates into the coordinates of the CoM and the distance between the bodies.
  5. Jan 8, 2014 #4
    Does that mean the force between the charges doesn't come in picture ? Do I have to neglect this force ?

    I have fixed the error .kq2=16

    Have I approached it correctly ?
  6. Jan 8, 2014 #5
    This is what I think but as I said, I am not sure. :)

    Still incorrect. It looks to me as if you are confusing the Coulomb's constant with the spring constant.
  7. Jan 8, 2014 #6
    :redface: . Forgive me for that .What next ?
  8. Jan 8, 2014 #7
    You can use the substitution ##x_2-x_1-l=z## but your substitution is fine too.

    What do you get if you solve the differential equation? I suggest to denote everything by symbols and substitute the values later.

    Btw, voko's method is a nice one, you can use energy conservation in the CM frame, no need of differential equations. :)
  9. Jan 9, 2014 #8
    Work Energy theorem is the best approach according to me.
    Since the magnitude of charge and the masses are same, these objects will move symmetrically in opposite directions. Therefore, when maximum compression and elongation occur, both the objects will be at rest. So you just have to consider
    (i)Spring potential energy
    (ii)Interaction potential energy
    (iii)Work done by external force
    while making the equations.
  10. Jan 9, 2014 #9
    I understand work energy approach is better .But since I started off with DE's ,want to finish it first.

    The DE is


    Now putting x2-x1 = z ,

    We have d2z/dt2=10-k(z-l)

    or , d2z/dt2+kz-(kl+10) = 0

    The solution is z = C1cos(√kt) + C2sin(√kt) + (l + 10/k)

    Does that make sense ?
  11. Jan 9, 2014 #10
    "assuming the natural length of the spring to be large" means that you are supposed to approach the problem assuming that ## s = z - l << l ##. It does not mean that the term for the electrical interaction between the charges is completely neglected. It means that it must be approximated with a term linear in ##s##.
  12. Jan 9, 2014 #11
    Hi voko...

    Suppose we neglect the force between the charges , does post #9 makes sense ?
  13. Jan 9, 2014 #12
    In the sense whether the solution of the equation in #9 is correct, yes.
  14. Jan 9, 2014 #13

    So ,z = C1cos(√kt) + C2sin(√kt) + (l + 10/k)

    Now ,using z(0) = l we get C1 = -10/k .

    How do I find C2 ?
  15. Jan 9, 2014 #14
    What about ##\dot z(0)##?
  16. Jan 9, 2014 #15
    That gives C2 = 0 .

    So,z =(-10/k)cos(√kt) + (l+10/k) . Is it correct ?
  17. Jan 9, 2014 #16
    The equation from #9, which you have solved, is not a correct equation for the original problem. Yet you used the initial conditions from the original problem to determine the coefficients in its solution. That does not look correct to me.
  18. Jan 9, 2014 #17
    Hi voko! :smile:

    This is the DE we have:

    If we go with Tanya's substitution ##x_2-x_1=z##, I don't see how we get z-l in the denominator of electric force. :confused:
  19. Jan 9, 2014 #18
    I suggest that you first rewrite that equation in terms of ##s##. Then think how you can simplify the equation if ##s \ll l##.
  20. Jan 9, 2014 #19
    OK, so I switch to the substitution ##x_2-x_1-l=s##, then I have ##\ddot{x_2}-\ddot{x_1}=\ddot{s}##, hence
    $$\Rightarrow -2ks-\frac{2Kq^2}{l^2}\left(1+\frac{s}{l}\right)^{-2}+2qE=m\frac{d^2s}{dt^2}$$
    $$\Rightarrow m\frac{d^2s}{dt^2}=-2ks+\frac{4Kq^2s}{l^3}+2qE-\frac{2Kq^2}{l^2}$$
    Do I have to solve the above DE?
  21. Jan 9, 2014 #20
    The problem does not want you to solve differential equations. It wants you to find the minimum and maximum elongations.

    Note that the equation you obtained can be re-interpreted as an equation for a massive particle in a force field, where the force depends linearly on the position. That force is potential, so you could find the corresponding potential energy. Then you can write down the equation for conservation of energy, and determine the min/max elongation from it.
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