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Charged particle near a grounded plane

  1. Feb 25, 2006 #1
    This one is from Griffiths. A toughie!!

    A point charge q of mass m is released at a distance d from an infinite grounded conducting plane. How long will it take for the charge to hit the plane?

    I chose a distance 'x' as a height above the plane. So the force on 'q' is given by:
    [tex]F = {-1\over 4\pi \epsilon_0}{q^2\over 4x^2}[/tex]

    And equal that to force equation:
    [tex]F = m\frac{d^2 x}{d t^2}[/tex]

    So,
    [tex]\frac{d^2 x}{d t^2} = \frac{-q^2}{16\pi \epsilon_0 x^2 m}[/tex]

    I need to solve this equation for 't'. Any clues?
     
  2. jcsd
  3. Feb 25, 2006 #2
    Oh, by the way, for once Griffiths has been kind enough to provide the solution for this problem :biggrin:.

    [tex]t = \left({\pi d \over q}\right) \sqrt{2\pi \epsilon_0 md}[/tex]

    This should be helpful.
     
  4. Feb 25, 2006 #3
    You will find the substitution y=ln x useful.

    -Dan
     
  5. Feb 25, 2006 #4

    StatusX

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    You can multiply both sides by 2 dx/dt. This will leave you with d/dt((dx/dt)2) on the left side and d/dt(something else) on the right side, which you can easily integrate to get a first order ODE.
     
  6. Feb 27, 2006 #5
    Did not work :frown: .

    OK, I multiplied it by v = dx/dt on both sides.

    [tex]v {dv\over dt} = \left({q^2\over 16\pi \epsilon_0 m}\right)\left({-1\over x^2}\right){dx\over dt}[/tex]

    Taking [itex]A = {q^2\over 16\pi \epsilon_0 m}[/itex]

    [tex]{d\over dt}\left({1\over 2} v^2\right) = {d\over dt}\left({A\over x}\right)[/tex]

    So,
    [tex]{1\over 2} v^2 = {A\over x} + \mbox{constant}[/tex]

    I need to bring this 'v' in terms of 'x' to integrate. How do I proceed?
     
    Last edited: Feb 27, 2006
  7. Feb 28, 2006 #6
    Oh dear, no replies....:rolleyes: .

    I proceeded with evalutions and I think I figured out the answer.

    When v = 0; x = d. So, constant = (A/d)
    Hence,
    [tex]v^2 = 2A\left({1\over x} - {1\over d}\right)[/tex]

    [tex]{dx\over dt} = \sqrt{{2A\over d}}\sqrt{{d-x\over x}}[/tex]

    [tex]\int_0^d {\sqrt{x}\over \sqrt{d-x}}dx = \sqrt{{2A\over d}}t[/tex]

    I just need to solve this for 't' but I'm not able to find a suitable substitution for the integrand terms to evaluate them. Any CLUES?
     
    Last edited: Feb 28, 2006
  8. Feb 28, 2006 #7
    put d-x = t square . u will get integeral of the form integeral(dt/(d-t^2)^1/2) which is a standard integeral . i dont know there might be a shorter way .... but u ll get the answer through this
     
  9. Feb 28, 2006 #8

    Physics Monkey

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    Hi Reshma,

    You are on the right track. Along the lines gandharva suggested, you might try a substitution of the form [tex] x = d \sin^2{\theta} [/tex].
     
  10. Mar 3, 2006 #9
    Thanks, Physics Monkey and gandharva. I got the answer.
     
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