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Electric Potential : First concepts

  1. Sep 20, 2007 #1
    I submitted this last night, but for some reason it didn't go through... so I don't have time to make it as awesome as I had before.

    Through what potential difference would an electron need to be accelerated for it to achieve a speed of 39.0% of the speed of light, starting from rest?

    C = 3.00 x 10^8
    M(e-) = 9.11x10^-31
    abs(e) = 1.60x10^-19 [I have been using this value as q)

    Ka + Ua = Kb + Ub where K is kinetic energy and U is the potential. Ka = 0 and the equation can be simplified to something like -Kb = Ub - Ua.

    I have all the variables for Kb as well as the charge. What I worked things down to last night was along the lines of Kb/q = (Ub/q - Ua/q) = [tex]\Delta[/tex]V. The units worked out (I'm pretty certain) and I got a reasonable answer in kV, something like 39.0kV.

    Apologies for the lack of depth here, I actually have to run off to my physics class! Any help is appreciated, thanks!
     
  2. jcsd
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