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Homework Help: 1-D coulomb repelling of 3 charges

  1. Jun 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Three positive point charges (with charge and mass given) A (+3q, 2m),B (+q, m) ,C(+4q, 7m/31) can move in a line under the coulomb repelling of one another. The movement begins when all are at rest. Initial distance are (left to right lie A, B, C) AB = r, BC = 2r.
    Then after they are faraway from one another, what will be their velocities, respectively?
    (q,m,r are all given.)

    2. Relevant equations
    The coulomb potential energy of q1, q2 displaced r apart, is given by E = kq1 q2/r.

    3. The attempt at a solution
    I notice the energy momentum conservation but this only gives 2 equations in this 1D case. This makes it impossible to solve for 3 velocities. So is it the only way to simulate it in a computer? Or is there any symmetry in it I have missed?
    I have thought about it for weeks so pls help. Thanks.
     
  2. jcsd
  3. Jun 9, 2010 #2
    think the solution in this way...
    calculate potential energy of all charges at A,B,C
    (for example p.e for charge at A we 3q*q*k/r + 3q*4q*k/3r)... similarly for other charges at B,C

    this energy converts into kinetic energy of the each particle i.e 1/2(m*v2 )......you know mass find velocity at infinity

    explanation is work energy theorem
    that is work done = (change in kinetic energy)
    but for conservative field like electric field due to charges
    work done=-(change in potential energy)
    we take one point as infinity and the other points as A and B and C
     
    Last edited by a moderator: Jun 9, 2010
  4. Jun 9, 2010 #3
    thanks. I tried this but I have 3 quantities va, vb, vc to determine. The final total kinetic energy KE = PE(a,b)+PE(b,c)+PE(c,a),
    in addition to the momentum conservation equation
    ma va + mb vb + mc vc = 0
    only give 2 restrictions on them. So I come to a case in which conservation laws don' suffice to determine the final state uniquely...
     
  5. Jun 9, 2010 #4
    Actually if u find potential energy of A due to B and C.. that is what converts into kinetic energy
    so 1/2*mass*(velocity of A)*(velocity of A)=Potential energy of A we just found (this hold only in this case as we took potential at infinity as zero and the initial velocity is zero... conservation law)
    we know mass of every particle,we will find Potential energy of A,B,C invidually so we get Va Vb Vc invidually...3 equations 3 unknowns
    no need of conservation of momentum...even its difficult to apply conservation because even on the same line we do not know which direction they move so we don't have directions of velocities
     
  6. Jun 9, 2010 #5

    collinsmark

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    Hello bharath423,

    I don't think that's going to work for this problem. If you were to find the potential energy of A (electrical potential [due to the other two charges] times its own charge), that would give you the final kinetic energy of A, but only if A was allowed to move, and B and C stay fixed in place. But that's not the case with this problem.

    Allow me demonstrate my point in a different way. Just for the moment, allow me to modify the problem statement to make it even simpler (we can go back to the original problem in a second). Imagine 3 particles. All equal mass and all equal positive charge, all restricted to the x-axis. The distance between the left and middle particle is identical to the distance between the right and middle particle. Then the particles are allowed to move freely. Due to symmetry, it's obvious that the final velocity of the middle partial will be zero. It will never move. Yet before the particles were allowed to move, the middle particle had the largest electrical potential of the three (and thus the largest potential energy).

    My above modified problem could be solved could be solved using the conservation of energy and momentum as cracking proposed, because we know the final velocity of 1 of the three particles (leaving 2 equations and 2 unknowns). But things are not so simple in the original problem (in the original post).

    Cracking,

    One way this problem could be solved (theoretically) is set up a system of 3 simultaneous differential equations. That would give you 3 equations and 3 unknowns. Unfortunately, solving this system of equations, although perhaps possible, wouldn't be a walk in the park since If I'm not mistaken, that's more precisely a system of three, second order, nonlinear ordinary differential equations. Maybe there's an easier way that I'm missing, but I'm at a loss. :frown:

    By the way, what sort of class is this for? Is this a class that involves Hamiltonian and/or Lagrangian mechanics? Are differential equations appropriate for this class?
     
  7. Jun 9, 2010 #6
    yes collinsmark you are right thank u for correcting me
    if A was allowed to move, B and C stay fixed in place...i missed that point
    so now we can find the resultant forces on each charge,as we know direction of movement we can also apply conservation of moment,and the equation Cracking has mentioned
    total kinetic energy KE = PE(a,b)+PE(b,c)+PE(c,a)
    but totally as u said collinsmark.. its solving of differential equations....
     
  8. Jun 9, 2010 #7
    Yes I agree with collinsmark that for example PE(A,B) + PE(A,C) belongs to KE of A,B,C --- not A alone. Actually an easy calculation tells me that the center of mass, which remains fixed in this closed sytem, is at 12r/25 to the left of particle b (that is between a&b). Then obviouly the final speed is leftward for A and rightward for B&C; the sign of final velocities are hence clear.
    I'm a high school teacher so only elementary conservation laws and easy ODEs (linear or variable-seperable) will be enough background for it; analytical mechanical methods are definitely irrelevant.
    Actually I also suspects that the values like 7m/31 are implying some symmetry as mentioned by collinsmark. though I simply can't find it.
     
  9. Jun 10, 2010 #8

    tiny-tim

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    Hi cracking! :smile:

    Have you tried putting the acceleration equal to zero in the force equation?
     
  10. Jun 12, 2010 #9
    Thank you for your suggestion tiny-tim. I don't quite agree with your method. But I still tried it as follows.
    A, C have no chance to give a force balance (2 forces are in the same direction). I used the balance of B and got
    3/(xA - xB)^2 = 4/(xB - xC)^2, where x is the coordinate. Taking radical, noting A goes oppositely as B,C, and then differentiating to get velocities, I obtained
    sqrt(3) vC = 2 vA + (sqrt(3) + 2) vB, where each v is speed (>0).
    Combining this with momentum conservation -2vA + vB + 7vC/31 = 0, this gives finally
    vB = 0.318vC and vA = 0.272vC.
    To verify this result, I did a simple simulation (time step 0.01s) till they move as far as ~100r away. Then I find they quickly approach uniform speed with
    vA = 0.21vC and vB = 0.19vC. (See attach)
    Actually the point in which I disagree with you is that if one knows all 3 terms in
    Fba - Fca = Ftotal
    go to 0, one can say Fba = Fca in the limit but one can't say Fba/Fca = 1 because Fca is in the limit = 0 and not dividable! My compution contains such a division.
    By the way according to the simulation A,B,C approach uniform motion rather quickly (actually at once) and
    the final vA/vC = the ratio of their initial acceleration
    In other words, vA/vC (t = inf) = aA/aC (t = 0) and the same for B,C. I am still checking my simulation but still at lost...
     

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