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Homework Help: Velocity of an electron in an electric field

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data

    In a picture tube (eg. a TV set) electrons are accelerated of an electric field. If the electric field strength is 100 kV / m and the electrons travel a distance 0.2 m from rest, what speed will they achieve?

    2. Relevant equations
    E=U/d where E is electric field intensity, U is the voltage and d is the distance.



    (mv^2)/2 = kinetic energy

    3. The attempt at a solution

    I think that the potential electrical energy of the electron will be converted into kinetic energy. Hence:

    W=qU=(mv^2)/2 or v=sqrt((2qU)/m). The charge and mass of an electron is known and so is the electrical field intensity. If I insert these values in the equation above I will get:

    sqrt((2*1.602*10^(-19)*100*10^3)/9.11*10^(-31))=1.87..*10^8 m/s. The correct answer is 8*10^7 m/s.
  2. jcsd
  3. May 5, 2012 #2


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    hi zeralda21! :smile:
    what about the distance? :wink:
  4. May 6, 2012 #3
    You are right, I didn´t take that into consideration. Here is another try:

    If W=qU and E=U/d, then of course W=qEd. If I set this equal to the kinetic energy and solve for v i end up with:

    W=qEd=(mv^2)/2 ----> v=sqrt((2qEd)/m). When I insert values for q,E,d and m(which are all known) the answer is 6*10^7 m/s, much more accurate but I think the error is too large for an "approximation error". No? What do you guys say?
  5. May 6, 2012 #4


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    you got 18.7 (times 107) before, and you're multiplying by √(0.2) …

    how do you get 6 ? :confused:
  6. May 6, 2012 #5
    I am ashamed :redface:. The answer is 8.36*10^7 m/s. Thanks tiny-tim :smile:
  7. May 6, 2012 #6


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    you are humble yet contrite

    o:) go in peace! o:)
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