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Charge particle in motion

  1. May 28, 2006 #1
    Another problem I'm struggling with this one I don't have a clue where to start. Any hints to a starting point would be greatly appreciated.

    A charge of -4.00uC is fixed in place. From a horizontal distance of 55.0cm a particle of mass 2.50 x 10^-3 kg and charge -3.00uC is fires with an intial speed of 15.0m/s directly toward the fixed charge. How far does the particle travel before it stops and begins to return back?
     
  2. jcsd
  3. May 28, 2006 #2
    The net force on the moving charge should be the force of repulsion it experiences. With this, you can set up an equation using Newton's 2nd Law.

    Take note though, that the net force (and hence the acceleration) of the moving particle is not constant, as it varies with the distance between the two particles. Hence, you will need to express acceleration as [tex]\frac{dv}{dt}[/tex], or some other equivalent expression.

    Once you have carried out integration, do not forget to add a constant of integration. You can solve for it by using the initial conditions provided in the quesion.
     
  4. May 29, 2006 #3
    You don't need to integrate. The basic idea is similar to the scattering experiment. The moving charge stops when all its kinetic energy has become zero. A simple energy conservation for the moving charge is to be done here. You need to figure out the distance of closest approach from the static charge.

    Pizzasky's approach will land you with the same answer but the basic idea is energy conservation of the moving charge in classical terms.
     
  5. May 29, 2006 #4
    So here's what I've come up with:

    Q = fixed charge
    q = moving charge

    kQq/(x-0.55m) = kQq/r +(1/2mv^2)

    Solving for x.

    I'm not sure if x-0.55 is correct? Or even if I've set up the equation correctly?
     
  6. May 29, 2006 #5

    Doc Al

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    It's almost right. What's "r"?
     
  7. May 29, 2006 #6
    r is the initial distance between the charges = 0.55m
     
  8. May 29, 2006 #7

    Hootenanny

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    You're almost there. Using conservation of energy;

    [tex]U_{final} = E_{k} + U_{intial}[/tex]

    [tex]\frac{kQq}{0.55 - x} = E_{k} + \frac{kQq}{0.55}[/tex]

    ~H

    Edit: Sorry Doc, guess I was typing for a long time
     
    Last edited: May 29, 2006
  9. May 29, 2006 #8

    Doc Al

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    Right. So, if the moving particle moves a distance x, what's the new separation distance between them?
     
  10. May 29, 2006 #9

    Doc Al

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    :eek: You might want to reconsider that, Hoot.
     
  11. May 29, 2006 #10
    I'm not sure, so if

    [tex]\frac{kQq}{0.55 - x} = E_{k}[/tex]

    then Ek is the PE + KE of q?

    So my equation would be:

    [tex]\frac{kQq}{0.55 - x} = \frac{kQq}{0.55} + 1/2mv^2[/tex]?
     
  12. May 29, 2006 #11

    Doc Al

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    That's correct.
     
  13. May 29, 2006 #12

    Hootenanny

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    Wait a minute Doc, I think that I am correct. Why would it be different? The moving particle is doing work against the electrostatic force, therefore kinetic energy will be converted into potential.

    ~H
     
    Last edited: May 29, 2006
  14. May 29, 2006 #13
    So I plugged in the numbers based on my final equation and got 1.05m. So this means that the fired particle goes past the fixed charge before it would turn back, does this make sense?
     
    Last edited: May 29, 2006
  15. May 29, 2006 #14

    Hootenanny

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    Just for reference, when using my formula I obtained 16.6cm for the distance of closest approach. I'm not sure whether it is correct though, I am waiting for Doc Al to respond.

    ~H
     
  16. May 29, 2006 #15

    Astronuc

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    Of course this does not make sense. That is one reason why it is important to 'develop an intuitive feeling' for the physics and understand the numbers.

    Also 1/r2 - 1/r1 does not equal 1/[itex]\Delta{r}[/itex] where [itex]\Delta{r}[/itex] = r2 - r1
     
  17. May 29, 2006 #16
    Hoot, I don't think Doc was referring to the conservation of energy, but rather how in your earlier post you used the change in distance to evaluate the change in pot. energy and thereby the change in KE .
    Compare the first part of that post ( where you wrote the proper expression ) and the second part, where you use change in distance .
    They don't match, do they ?:smile:

    Arun

    Edit : Astronuc's faster than me
     
  18. May 29, 2006 #17

    Hootenanny

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    Ahh, yes apologies everyone. I forgot that the particle started at a point in the electric field. I have edited by eroneous post.

    ~H
     
  19. May 29, 2006 #18
    So I am really confused now.

    The equation should read:

    [tex]\frac{kQq}{0.55} - \frac{kQq}{x} = E_{k}[/tex]?

    And Ek is [tex]\frac{kQq}{0.55} + 1/2mv^2[/tex]?

    Therefore [tex]\frac{kQq}{0.55} - \frac{kQq}{x} = \frac{kQq}{0.55} + 1/2mv^2[/tex] ?
     
  20. May 29, 2006 #19
    No, this is wrong.
    KE is always 1/2mv^2
    The expression should equate so :

    Initial PE + Initial KE = Final PE (no KE)
     
  21. May 29, 2006 #20

    Doc Al

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    Your equation that I quoted in post #11 is correct.

    Hoot: As arunbg and Astronuc have pointed out, it was your mathematical step that I was objecting to. Nothing that an extra cup of coffee wouldn't cure. :wink:
     
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