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Potential difference to bring a proton to rest

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A proton has an initial speed of 3.5×10^5 . What potential difference is required to reduce the initial speed of the proton by a factor of 2?


    2. Relevant equations
    KE=.5*m*v^2
    EPE=q*V


    3. The attempt at a solution
    I did KE=.5(1.67x10^-27)(3.5x10^5/2)^2 and then divided that by the charge of a proton to get V, and I got 160 V which is wrong. Then I tried squaring the velocity first, then dividing it by 2 and I got 320 V, which is also not right. I'm stuck.
     
  2. jcsd
  3. Sep 18, 2012 #2
    this is just conservation of energy

    electric potential starts at zero plus kinetic energy starts at some KEi

    this must be equal to some electric potential energy plus half of KEi
     
  4. Sep 18, 2012 #3
    Thank you so much! I didn't even think of that. I got the answer.
     
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