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Electricity problem

  1. Feb 6, 2007 #1
    electricity problem :(

    so heres the problem:

    through what potential difference would an electron need to be accelerated for it to achieve a speed of 42.0% of the speed of light, starting from rest? The speed of light is c = 3.00e8 m/s

    so i thought the equation i would use would be Vf-Vi=deltaPE / q

    so i thougth the change in potential energy would be equal to the opposite of change in kinetic energy, so delta PE would = .5mv^2, and v is 3e8 X .42, so:

    .5 X 9.11e-31 X (1.26e8)^2 / 1.6e-19 = 4520V, but this is wrong, so yea, no clue. any help would be much appreciated, thanks
     
  2. jcsd
  3. Feb 6, 2007 #2

    Tom Mattson

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    You've got the right idea. All of the potential energy should show up as kinetic energy of the electron. Where you went wrong is right here:

    The expression [itex]K=\frac{1}{2}mv^2[/itex] is the nonrelativistic kinetic energy. In relativity there is a different expression:

    [tex]K=\gamma\left(mc^2-1\right)[/tex]
    [tex]\gamma=\frac{1}{\sqrt{1-(v/c)^2}}[/tex]

    If you use the correct expression, you should get the correct answer.
     
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