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Elecric charge and electric field

  1. Aug 24, 2005 #1

    My problem is:

    An electron( mass m=9.11 x 10^-31 kg) is accelerated in the uniform field E( E=1.45 x 10^4 N/C) between two parallel charged plates. The seperation of the plates is 1.10 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. With what speed does it leave the hole?


    m=9.11 x 10^-31 kg
    E=1.45 x 10^4 N/C
    distance= 1.10 x 10^.2 m
    q= -1.6 x 10^-19 C

    qE(s) = ½ mv^2 - ½ mv^2

    -1.6 x 10^-19 C= ½ (9.0 x 10^-31 kg)(v)^2

    solve for v

    v=7.14 x10^-7 m/s

    Is this the correct way of approaching the problem, or answer?
    Please help

    Thank You
    Last edited: Aug 25, 2005
  2. jcsd
  3. Aug 24, 2005 #2
    Energy is the correct method, but you made an error in your calculation.

    The electron goes through the E field acquiring kinetic energy equal to the loss in potential energy. Its total potential energy is given by [itex] PE = qE\Delta x [/itex], with delta x being the separation of the two plates. Since it starts at rest and experiences no other forces, the final kinetic energy will be equal the initial potential energy.

    [tex] \frac{1}{2}mv^2 = qE\Delta x[/tex]

    You seem to have equated the charge with the final kinetic energy on the step before "solve for v".
  4. Aug 25, 2005 #3
    A funnier way of writing the potential energy of an electric field:
  5. Aug 25, 2005 #4

    Whozum said that I should use this equation:
    [tex] \frac{1}{2}mv^2 = qE\Delta x[/tex]

    So all I would have to do is plug everything in right? :smile:

    My work

    (1/2)(9.11 x 10 ^31)V^2=(1.6 x 10 ^-19)(1.45 X 10^4)(.11)

    V^2=((1.6 x 10 ^-19)(1.45 X 10^4)(.011)
    )/((1/2)(9.11 x 10 ^31))

    V=7.49 X 10^-25 m/s

    But I looked at the back of the book and gives the answer of 7.49 X10^6 m/s :yuck:

    Did I forget something???

    Please help and Thank You
    Last edited: Aug 25, 2005
  6. Aug 25, 2005 #5
    Quantum electrodynamics?

    And uhh... check your electron mass again, I'm pretty sure an electron is lighter than the sun. :rofl:
  7. Aug 26, 2005 #6

    I see what I did wrong. :blushing:


    (1/2)(9.11 x 10 ^31)V^2=(1.6 x 10 ^-19)(1.45 X 10^4)(.11)

    V^2=((1.6 x 10 ^-19)(1.45 X 10^4)(.011)
    )/((1/2)(9.11 x 10 ^-31))

    V=7.49 X10^6 m/s

    Thanks :smile:
  8. Aug 26, 2005 #7

    The second portion of this question asks:

    Show that the gravitational force can be ignored.

    I found that I can ignore gravitational force as long as the size of qE / m relative to g. As long as qE / m is much larger than g, gravity can be ignored. Gravity is very easy to account for, of course : simply add mg to the free-body diagram and go from there.

    So if that's the case couldn't I just find the acceleration of the problem, since I already found the velocity and see if this is true.

    Thank You
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