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Electric Field and electron speed

  1. Feb 5, 2005 #1
    An electron (m = 9.11 x 10^-31 kg) is accelerating in the uniform field E (E = 1.85 x 10^4 N/C) between 2 parallel charged plates. The separation of the plates is 1.2 cm (.012 m). The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate.

    a.What speed does it leave the hole?

    b.Show that gravitational force can be ignored.

    a. I know that E = F/q, where F = ma.

    Given: E = 1.85 x 10^4 N/C, r (or x) = .012 m, m = 9.11 x 10^-31 kg, but what is q?

    Using mathematical definitions, I can solve for v.

    E = ma/q, where a = Eq/m

    v = x/t a = v/t
    a*t = x/t
    t = sqrt(x/a)

    Then I can plug in x and t values for v = x/t. Correct???

    b. I have no idea how to carry out this proof. Need help!

  2. jcsd
  3. Feb 5, 2005 #2
    Is q=1.60 x 10^-19? Is there a diagram dictating whether the electron is moving in a downward motion???
  4. Feb 5, 2005 #3
    Start by finding the voltage between both plates. Can you think of a formula for this?
  5. Feb 5, 2005 #4
    The velocity vector is pointing straight to the right (parrallel to the x-axis). It is not pointing upward or downward.

    ----------------------> V


    moves from neg. to pos. plate

    The voltage relates to part a, right ?
  6. Feb 5, 2005 #5


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    Why would he need the voltage???IIRC,the electric force is proportional to the field...

  7. Feb 5, 2005 #6
    Come again? Could someone explain what I am doing incorrectly?

    Thanks for your patience.
  8. Feb 5, 2005 #7


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    What's the equation u need to find the final velocity...?

  9. Feb 5, 2005 #8
    For part (a), it would serve you best to find the potential difference across the two plates of this capacitor. Do you see how this question involves potential energy? Remember those annoyingly easy Conservation of Energy questions, where kinetic was converted to potential and vice-versa? This question can be made analogous to something like that if you find the voltage across the plates.
    As for part (b), I think it would suffice to show that the electric force (and subsequent acceleration of the electron) far overpowers the gravitational force (and the normal acceleration of [tex]g[/tex]).
  10. Feb 5, 2005 #9
    That's right. Once you find the potential difference across the 2 plates, you know that the voltage from this ( [tex]V = \frac{E}{q}[/tex] ) will cause the acceleration. The "E" in the equation represents energy. Before the electron starts accelerating, it will only have potential energy. At the end, all of this potential energy will be converted to kinetic energy. So the equation becomes: [tex]V = \frac{\frac{1}{2}mv^2}{q}[/tex]

    All you gotta do now is solve for v...

    BTW, q is the charge of the electron
  11. Feb 5, 2005 #10
    As for the second question, it is very simple. Remember the gravitational law:
    [tex] F_g = \frac{Gm_1m_2}{r^2}[/tex]
    Here, G has a value of [tex] 6.67 x 10^{-11}[/tex]

    Now, compare this to Coulomb's law:

    [tex] F_e = \frac{kq_1q_2}{r^2}[/tex]
    Here, k has a value of [tex] 8.99 x 10^9[/tex]

    Can you see how gravitational force is so much smaller than electrical force? It is virtually insignificant in this particular problem.
  12. Feb 5, 2005 #11


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    HINT:That "x" is expressed as "\times" or [itex] \times [/itex].



    EDIT:Alternatively "\cdot" or [itex] \cdot [/itex].
    Last edited: Feb 5, 2005
  13. Feb 5, 2005 #12
    oh yeah... :blushing: I won't forget next time.
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