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Potential difference of an electron

  1. Feb 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Through what potential difference would an electron need to be accelerated for it to achieve a speed of 46.0% of the speed of light, starting from rest? The speed of light is
    c= 3.00 x 10^8 m/s
    m_e = 9.11 x 10^-31 kg
    |e| = 1.60 x 10^-19 C
    Round your answer to three significant figures in units kV.

    2. Relevant equations
    V= Ed

    3. The attempt at a solution

    1. I tried to break down E into E= F/q.
    2. F=ma
    3. V = (ma/q)d
    4. V = mad/q
    5. V = m(d/t^2)d/q
    6. V = m(d^2/t^2)/q
    7. V = m(d/t)^2/q
    8. V = mv^2/q

    I plugged in the numbers using this and didn't get the right answer. Am I just going at this all wrong?
    Last edited: Feb 12, 2008
  2. jcsd
  3. Feb 11, 2008 #2


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    Homework Helper
    Gold Member

    You are going about this problem in a convoluted way, and may be making some mistakes along the way. Try this:

    Remember that the kinetic energy gained by an electron going through a potential difference, V, is equal to the potential energy lost:

    [tex]K_{gained}= U_{lost}=qV[/tex]

    Now, how much energy would an electron need to gain to go the given speed?
  4. Feb 12, 2008 #3
    I just don't feel like I'm wrapping my head about this correctly.

    Using your K=qV formula, I worked the other way around.



    1/2mv^2 = qV
    V = (mv^2)/(2q)

    Using this, I still didn't get the correct answer.
  5. Feb 12, 2008 #4
    Try using M = m/[tex]\sqrt{1 - (v/c)^2}[/tex]
  6. Feb 12, 2008 #5


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    Homework Helper
    Gold Member

    This is probably your source of error. You need to use the relativistic mass of the electron, since you are moving at a significant fraction of the speed of light.

    Nice catch michalll.
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