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Point charge in an electric field

  1. Sep 1, 2004 #1
    an electron with a speed of 5.00 x 10^8 cm/s enters an electric field of magnitude 1.00 x 10^3 N/C, traveling along field lines in the direction that retards its motion.

    a) How far will the electron travel in the field before stopping momentarily
    b) how much time will have elapsed
    c) in the region with the E field is only 8.00 mm long, what fraction of the electron's initial KE will be lost in that region?

    I really need to get jumpstarted on this problem. Newton's second law comes to mind, but without the mass of the electron its pretty useless. anyone?
     
  2. jcsd
  3. Sep 1, 2004 #2

    robphy

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    Check the inside cover of your textbook for the electron mass.
     
  4. Sep 1, 2004 #3
    k this is what i have so far

    [tex]\vec {F} = m \vec {a}[/tex]
    [tex]q\vec {E} = m \vec {a}[/tex]

    as far as [tex] \vec {a} [/tex] goes, what is a relationship I can use that includes the speed of the electron?
     
  5. Sep 1, 2004 #4
    here can [tex] \vec {a} [/tex] be represented by [tex] \frac{v^2 - v_o^2}{2(x - x_o)} [/tex], with [tex] v [/tex] being the velocity given?
     
  6. Sep 1, 2004 #5
    I think you only need the following equations to solve it:
    [tex]F=ma, v^2 = u^2 + 2as, v=u+at, F=qE, E=V/d, V=W/Q[/tex]
     
  7. Sep 1, 2004 #6
    could you define u, s, V, W, and Q
     
  8. Sep 1, 2004 #7
    Oh sorry. u is the initial velocity of the electron, s is its displacement, V is the electric potential, W is the work done (more like KE lost due to the field), and Q is charge of the electron.
     
  9. Sep 1, 2004 #8
    the thing is, is that we haven't studied potential...i was trying to work this problem using strictly kinematics...
     
  10. Sep 1, 2004 #9
    In that case then for part c, calculate the final velocity v of the electron, and the loss in KE is given by [tex]1/2m(v^2-u^2)[/tex] You can determine v by using the kinematics equations given, once you know the deceleration the electron is subjected to.
     
  11. Sep 2, 2004 #10
    Try This

    I'm not sure that is the best way to solve the problem. Let's start by listing what we know:

    Initial Velocity: [itex] \overrightarrow{V_0} [/itex] (remember to convert that to meters per second)

    The Electric field [itex] \overrightarrow{E} [/itex]

    Okay. Now lets do the problem. We know that:

    [tex] F = ma [/tex]

    and

    [tex] F = qE [/tex].

    The first part of the problem asks how far the electron will travel before momentarily coming to rest - in otherwords, how long will it take the electric field to decelerate this particle to a final velocity of zero:

    [tex] {V_f}^2 = {V_0}^2 +2a(x - x_0) [/tex]

    Substitute for a and solve for x with final velocity being zero. That should give you part (a). For part (b), we need to find the elapsed time. This equation comes to mind:

    [tex] \sqrt{\dfrac{2\Delta x}{a}} = t [/tex]

    Okay, now lets see. How much of the particle's KE will be lost after travelling 8mm in the electric field. Remember to convert this to meters. Find the initial KE - this should be easy enough for you. Then find the KE after the particle travels this distance 8mm. I'll not go in to this deeply, but we know accelleration, we know [itex] x - x_0 [/itex], and all we need to find is the velocity at that point to be able to calculate KE.

    Hope this helps.
     
    Last edited: Sep 2, 2004
  12. Sep 2, 2004 #11
    i figured it out earlier today

    [tex]F = ma [/tex]

    [tex]qE = ma [/tex]

    [tex]\frac{eE}{m}= a [/tex]

    find a, use it to find the displacement (x) in the simple mechanics equation

    [tex] \frac{-v_{0}^2}{2a}= x [/tex]

    then use x to find t :

    [tex] \sqrt{\frac{2x}{a}}= t [/tex]

    now simply use:
    [tex] \Delta KE = \frac{1}{2}mv_{0}^2 - \frac{1}{2}mv^2 [/tex]

    thanks for the suggestions!
     
  13. Sep 2, 2004 #12
    Cool, glad you figured it out.
     
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