1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]

    [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.

  5. Feb 25, 2006 #4


    User Avatar
    Homework Helper

    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]

    [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)
    [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

    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook