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Electrostatics Problem -Two moving charged particles

  1. Jun 23, 2014 #1
    Two small particles have electric charges of equal magnitude and opposite signs. The masses of the particles are m and 2m. Initially, the distance between the particles is d, and the velocities of the particles have equal magnitude v. However, the velocity of particle 2m is directed away from particle m, whereas the velocity of particle m is directed perpendicular to the line connecting the particles. In the subsequent motion of the particles, they are found to be at a distance 3d from each other—twice. Find the possible values of the charge of each particle.


    2. Relevant equations



    3. The attempt at a solution

    My approach is to find the maximum distance ‘x ‘ between the charges and impose the condition that x>d ,as then the charges will be at a distance d- twice .

    I will work in CM frame of reference.
    ## V_{cm} = \frac{1}{3m}(2mv\hat{i}+mv\hat{j}) ##

    ## V_{cm} = \frac{2v}{3}\hat{i}+\frac{v}{3}\hat{j} ##

    Now , Let ##V_{2i}## be initial velocity of 2m and ##V_{1i}## be initial velocity of m in CM frame.

    ##V_{2i} = \frac{v}{3}\hat{i}-\frac{v}{3}\hat{j} ##

    ##V_{i} = \frac{-2v}{3}\hat{i}+\frac{2v}{3}\hat{j} ##

    Let ##V_{2f}## be initial velocity of 2m and ##V_{1f}## be velocity of m in CM frame ,when the particles are at maximum distance.

    Since no external forces are acting,linear momentum is conserved .

    Initial momentum in CM frame = Final momentum in CM frame = 0

    ##2mV_{2f} = mV_{1f}##

    Let ##V_{1f} = v'## ,so ##V_{1f} = \frac{v'}{2}##

    Since no external torques are present ,angular momentum is conserved in CM frame .

    ## 2m\frac{v}{3}\frac{d}{3} + m\frac{2v}{3}\frac{2d}{3} = (2m)\frac{v'}{2}\frac{x}{3} + (m)(v')\frac{2x}{3} ##

    This gives ##v' = \frac{2vd}{3x}##

    Now applying conservation of energy ,

    ##\frac{1}{2}m(\frac{2v}{3})^2+\frac{1}{2}2m(\frac{v}{3})^2 - \frac{kq^2}{d} = \frac{1}{2}mv'^2+\frac{1}{2}2m(\frac{v'}{2})^2 - \frac{kq^2}{x}##

    ##\frac{1}{2}m(\frac{2v}{3})^2+\frac{1}{2}2m(\frac{v}{3})^2 - \frac{kq^2}{d} = \frac{1}{2}m(\frac{2vd}{3x})^2+\frac{1}{2}2m(\frac{1}{2}\frac{2vd}{3x})^2 - \frac{kq^2}{x}##

    ##\frac{1}{3}mv^2- \frac{kq^2}{d} = \frac{1}{3}\frac{mv^2d^2}{x^2}- \frac{kq^2}{x}##

    ##\frac{mv^2d-3kq^2}{3d} = \frac{mv^2d^2-3kq^2x}{3x^2}##

    I am not sure whether I am approaching the problem correctly .I would be grateful if somebody could help me with the problem.
     
  2. jcsd
  3. Jun 24, 2014 #2

    haruspex

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    Good approach, but you mean 3d - twice.
    There is another constraint on x, though.
    Don't you mean ##2mV_{2f} = -mV_{1f}##
    This is confusing. v is a scalar, but v' is a vector?
    Not sure how you are deriving these expressions for angular momentum. In the initial condition, 2m has no moment about CM.
     
  4. Jun 24, 2014 #3
    Yes ,I meant 3d twice .

    I do not know .

    I was equating the speeds of the two particles at maximum separation.


    ##V_{2i} = \frac{v}{3}\hat{i}-\frac{v}{3}\hat{j}##

    Here ##V_{2i} ## is the initial velocity of 2m in CM frame .

    ##\vec{r} = \frac{d}{3}\hat{i}##

    By using the definition of angular momentum ##m\vec{r} \times {\vec{V}_{2i}} ## - Why would 2m not have angular momentum about the CM ?
     
  5. Jun 24, 2014 #4

    TSny

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    Typo: V2f = v'
    [EDIT: I should have written V2f = v'/2 as you noted in post #5. (Typo of a typo :redface:)]

    Did you use the correct values for the initial speeds?

    Otherwise, I think your work is OK. What do you get for the charge q if x is set equal to the critical value 3d?
     
    Last edited: Jun 24, 2014
  6. Jun 24, 2014 #5
    I think it should be ##V_{1f} = v'## and ##V_{2f} = \frac{v'}{2}##


    It should be

    $$ \frac{1}{2}m(\frac{2\sqrt{2}v}{3})^2+\frac{1}{2}2m(\frac{\sqrt{2}v}{3})^2 - \frac{kq^2}{d} = \frac{1}{2}mv'^2+\frac{1}{2}2m(\frac{v'}{2})^2 - \frac{kq^2}{x} $$
     
    Last edited: Jun 24, 2014
  7. Jun 24, 2014 #6
    $$ q = \sqrt{\frac{17mv^2d}{18k}} $$

    Edit:Removed error
     
    Last edited: Jun 24, 2014
  8. Jun 24, 2014 #7

    TSny

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    I think that's right except you have too many d's in your answer.
     
  9. Jun 24, 2014 #8
    OK.

    But this gives me the upper bound of the value of charge .How should I get the lower bound ?

    Another thing , how do I know whether the result just obtained is lower or upper bound ?
     
  10. Jun 24, 2014 #9

    TSny

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    You should be able to answer these questions if you think about the different types of motion that can occur for an inverse-square force of attraction.
     
  11. Jun 25, 2014 #10

    haruspex

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    As I wrote in post #2, there is another constraint on x. How large can x be? How large can any number be? Think big, really big!
     
  12. Jun 25, 2014 #11
    I haven't been able to think about it.

    Are you indicating to the symbol in your "Thanks" badge :smile:.
     
  13. Jun 26, 2014 #12

    ehild

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    Adding to TSny's hint, this two-body problem can be reduced to a one-body problem with the reduced mass approach, you are familiar with. You get the same equation for the difference of the position vectors as in case of the Kepler problem or the Hydrogen atom. A particle in an attracting, central, inverse-square force field. There are two conserving quantities, one is energy, what is the other one?
    What kind of orbits are possible? What is the condition that the orbit is bounded? When can the particle move infinitely far from the centre?

    It is not quite clear form the text of the problem, that the particles are really only twice at distance 3d, or they are twice at that distance along their repeating orbits.

    ehild
     
  14. Jun 26, 2014 #13

    haruspex

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    That's the one. What inequality relates x to that?
     
  15. Jun 28, 2014 #14
    x<∞ ,but I don't know the value of x ?

    Circular and Elliptical ,just as in planetary motion .

    Angular momentum .

    Circular and elliptical orbits are possible .The energy should be negative.When the particle has non negative energy .

    If the two particles collide with each other then it should be former .If they are orbiting around the common CM then it should be latter.
     
  16. Jun 28, 2014 #15
    If I put the initial energy as negative I get the lower bound i.e E1 <0

    $$\frac{1}{2}m(\frac{2\sqrt{2}v}{3})^2+\frac{1}{2}2m(\frac{\sqrt{2}v}{3}) ^2 - \frac{kq^2}{d} < 0 $$

    This gives $$q > \sqrt{\frac{2mv^2d}{3k}}$$

    Does this make sense ?
     
    Last edited: Jun 28, 2014
  17. Jun 28, 2014 #16

    ehild

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    It is less than infinity :devil:


    Yes, the planets are on circular and elliptical orbits, but what other orbits are possible?


    yes, the angular momentum is also conserved and it is not zero.

    And what happens if the energy is zero or positive?


    Can the particles collide if the angular momentum of the system with respect to the CM differs from zero? What would it be in the instant of collision?

    ehild
     
  18. Jun 28, 2014 #17
    Please have a look at post#15 :smile:
     
    Last edited: Jun 28, 2014
  19. Jun 28, 2014 #18

    ehild

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    Explain why the energy must be negative?


    ehild
     
  20. Jun 28, 2014 #19
    So that the two particles are under the influence of each others electrostatic force i.e are bound with each other just like earth and sun .
     
  21. Jun 29, 2014 #20

    ehild

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    What about a spaceship that leaves the solar system? Or a not-returning comet? When the attracting force is small and the speeds are great?

    ehild
     
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