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Time of approach of two oppositely charged particle

  1. Jul 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Two point charges q1=1[itex]\mu[/itex]C and q2=-1[itex]\mu[/itex]C with mass 1g each are at distance 1m from each other. The charges are released from rest at . Find the time t in seconds at which they collide.

    2. Relevant equations
    F=kQ1*Q2/r^2.


    3. The attempt at a solution
    Let the dist. of separation be r and the mass of charged be m.
    I think they will meet at the middle of separation.Force on q1=kq^2/r^2 so the acceleration =kq^2/r^2*m.But in this case acceleration is not constant.So I have to use calculus here.
    Let the charge q1 be at origin and at a distance x from origin acceleration=kq^2/(r-2x)^2*m( 'r-2x' because if q1 will move x meter towards q2 then q2 will also move x metre towards q1 so the length of separation will decrease by 2x) .
    For a small displacement dx this acceleration is constant.I tried to use S=ut+at^2/2.But I am unable to find u.Please help.
     
  2. jcsd
  3. Jul 11, 2014 #2

    ehild

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    Have you studied differential equations?

    When setting up the solution, draw a picture first and show how you define the variables in that picture.

    The particles move symmetrically with respect to the centre. If one is at x m distance from the centre, how far is the other? And what is the distance between them?

    ehild
     
  4. Jul 11, 2014 #3
    The other should also be x m distance from the center.So the distance between them should be 2x.
     
  5. Jul 11, 2014 #4
    I should have choose origin at the center.Now at a distance x from the origin acceleration of q1 will be kq^2/(4x^2*m).Here m is for mass of particle.
    I have made this figure Physics problem.png .
     
  6. Jul 11, 2014 #5

    ehild

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    Good. How the acceleration is related to the second derivative of x?

    ehild
     
  7. Jul 11, 2014 #6
    [itex]\frac{d^{2}x}{dt^{2}}[/itex]=acceleration
     
  8. Jul 11, 2014 #7

    ehild

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    Write the equation of motion ma=F in terms of x. You wrote that the acceleration is kq^2/(4mx2). It is positive. That means x increasing with time. Is it true? Write the equation of motion in terms of x.

    ehild
     
    Last edited: Jul 11, 2014
  9. Jul 11, 2014 #8
    kq[itex]^{2}[/itex]/4x[itex]^{2}[/itex]=m*d^2x/dt^2
    I think x decreases with time because as the time passes both the particle move toward origin so their distance of separation decreases.
     
  10. Jul 11, 2014 #9

    ehild

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    Look at m1. The force acts to the right. The acceleration is to the right. Does x increase?

    ehild
     
  11. Jul 11, 2014 #10
    According to figure which I have made in post#4 it seems that x decreases and acceleration increases because x is measured from origin and in this figure origin is at the middle of the line of separation.
     
  12. Jul 11, 2014 #11

    ehild

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    Yes. So -kq2/4x2=m*d^2x/dt^2
    That is a differential equation, it is not easy to solve. You can go one step ahead by applying Conservation of Energy. Then you get the speed as function of x. Try.



    ehild
     
  13. Jul 12, 2014 #12
    I have not applied conservation of energy concept to system of particles earlier so this is my first time.
    Initially E(potential)=-[itex]\frac{kq^{2}}{r}[/itex] (I got -ve sign as the charge at the left is negative)
    Now at x kinetic energy of system = kinetic energy of the CM+ kinetic energy of all particles relative to the CM.
    E[itex]_{k}[/itex] at x=mv[itex]^{2}[/itex]
    E[itex]_{p}[/itex] at x = -[itex]\frac{kq^{2}}{2x}[/itex]
    so
    -[itex]\frac{kq^{2}}{r}[/itex]=mv[itex]^{2}[/itex]+ -[itex]\frac{kq^{2}}{2x}[/itex].

    v=sq.root of [itex]\frac{kq^{2}(r-2x)}{2mxr}[/itex].
    the figure is here View attachment 71218 .V is the velocities of particles at x.
     
    Last edited: Jul 12, 2014
  14. Jul 12, 2014 #13
    Now write v =dx/dt and integrate to find x as function of time .
     
  15. Jul 12, 2014 #14

    ehild

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    Be careful with the signs. x is the distance between the charges. It decreases with time, dx/dt<0. v is the speed of both particles, a positive quantity.

    ehild
     
  16. Jul 12, 2014 #15
    Well I got
    -[itex]\frac{√2x}{√(1-2x)}[/itex]dx=3*10^6 dt.
    I have some problem in solving LHS.I substituted 2x=a^2.
    And I got dx=ada.On putting values I got
    [itex]\frac{a^2}{√(1-a^2}[/itex]da.
    Don't know what to do next.
     
    Last edited: Jul 12, 2014
  17. Jul 12, 2014 #16
    This will not work . Try some trigonometric substitution and see which one fits in.
     
  18. Jul 12, 2014 #17
    If denominator is in the √a^2-x^2 and numerator is dx then we substitute x as asinθ.I don't have very much knowledge of calculus.I know only some basics.Will the substitution work.
     
  19. Jul 12, 2014 #18
    While performing integration using substitution method you need to substitute values of both x as well as dx in terms of θ.

    For example if you are using the substitution x= asinθ , then dx = (acosθ)dθ .
     
    Last edited: Jul 12, 2014
  20. Jul 12, 2014 #19

    ehild

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    You can cheat a bit... It is not an easy integral. Use wolframalpha.com

    Check the right-hand side, it does not look correct.

    http://www.wolframalpha.com/input/?i=int_0.5^0%28sqrt%282x%2F%281-2x%29%29dx%29

    ehild
     
    Last edited: Jul 12, 2014
  21. Jul 12, 2014 #20
    Yes . The integral is not easy .
     
    Last edited: Jul 13, 2014
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