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Final velocity of a proton through parallel plates

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data

    A moving proton has 6.4x10^-16 J of kinetic energy. The proton is accelerated by a potential difference of 5000 V between parallel plates.

    http://members.shaw.ca/barry-barclay/Self-Tests/test09/elecst17.gif


    The proton emerges from the parallel plates with what speed?

    A. 8.8x10^5 m/s B. 9.8 x10^5 m/s C. 1.3x10^6 m/s D. 1.8x10^6 m/s

    I know that the answer is C, I just don't know how to get there.

    2. Relevant equations
    q=1.60x10^-19 C
    m=1.67x10^-27 kg
    Ek=6.4x10^-16 J
    5000V electric potential difference (I don't really fully grasp what this means. I have something that says electric potential difference = work done in moving test charge/magnitude of test charge [V=ΔE/q])


    I know that Kinetic Energy=q|E|d
    V=ΔE/q
    W=Vq
    Fe=kq/r2
    |E|=V/d
    Then I know the basic kinematics formulas

    3. The attempt at a solution
    I found work by using W=Vq. I get (5000V)(1.60x10-19C)= 8.00x10-16J

    Then I'm lost as to what to do. But:

    Using Ek=0.5mv2 I get the answer A. Is that actually the correct answer and the answer key from the thing he copied/pasted is wrong?

    I saw this thread when googling https://www.physicsforums.com/showthread.php?t=168634 , which made me think that.
     
  2. jcsd
  3. Feb 3, 2012 #2

    wukunlin

    User Avatar
    Gold Member

    5000V of potential difference means for every coulomb of your charged particle, they will gain 5000J of energy so....

    how much energy is gained/loss for a single particle?
    how much energy does the particle end up with?
    what speed does that translate to?
     
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