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Velocity from electric potential

  1. Feb 27, 2008 #1

    tony873004

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    1. The problem statement, all variables and given/known data
    Can someone double check that this is right? I only multiplied charge of a proton by given potential difference to make the units work, without actually knowing if this is the right way to do this problem.

    A proton is accelerated from rest through a potential difference of 0.100 MV. What is the final speed of the proton?


    3. The attempt at a solution

    [tex]
    \begin{array}{l}
    E_k = \frac{1}{2}mv^2 \,\,\,\, \Rightarrow \,\,\,\,v = \sqrt {\frac{{2E_k }}{m}} = \sqrt {\frac{{2 \cdot 100\,{\rm{MV}} \cdot 10^6 {\rm{V/MV}} \cdot 1.602 \cdot 10^{ - 19} {\rm{C}}}}{{1.67 \times 10^{ - 27} {\rm{ kg}}}}} \\
    \\
    = \sqrt {\frac{{2 \cdot 100\, \cdot 10^6 {\rm{V}} \cdot 1.602 \cdot 10^{ - 19} {\rm{C}}}}{{1.67 \times 10^{ - 27} {\rm{ kg}}}}} = \sqrt {\frac{{2 \cdot 100\, \cdot 10^6 \frac{{{\rm{kg}} \cdot m^2 }}{{\frac{C}{s}s^3 }} \cdot 1.602 \cdot 10^{ - 19} {\rm{C}}}}{{1.67 \times 10^{ - 27} {\rm{ kg}}}}} \\
    = \sqrt {\frac{{2 \cdot 100\, \cdot 10^6 \frac{{{\rm{ \times m}}^2 }}{{\frac{{}}{{}} \cdot {\rm{s}}^2 }} \cdot 1.602 \cdot 10^{ - 19} }}{{1.67 \times 10^{ - 27} {\rm{ }}}}} =1.39x10^{8}m/s \\
    \end{array}
    \[/tex]

    The final speed is rather high. This is one problem with E&M physics as opposed to mechanics. If a ball was tossed in the air and landed with this speed, I could use my intuition to tell me I did something wrong.

    This velocity is half the speed of light. Wouldn't relativity come into play in these types of problems?
     
  2. jcsd
  3. Feb 27, 2008 #2

    olgranpappy

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    to get the exactly correct answer you would use the relativistic expressions, but as long as [tex]mc^2[/tex] is a lot bigger than the potential energy difference (V) you can get away with using the nonrelativistic formula. That is, you should use the relativistic
    [tex]
    v=\frac{pc^2}{V+mc^2}
    [/tex]
    and
    [tex]
    p^2 c^2 = V^2 + 2Vmc^2
    [/tex]

    ... but if [tex]mc^2 >> V[/tex] then the first equation reduces to
    [tex]
    v\approx p/m
    [/tex]
    and the second reduces to
    [tex]
    p^2\approx 2mV
    [/tex]

    so...
     
  4. Feb 27, 2008 #3
    Ever hear of the unit of energy called an electron volt?(eV?)it's the energy gained by an electron accelerated through 1 volt, so it's 1.602x10^-19 joules. So something with the charge of an electron(a proton!)through .1 MV should give .1 MeV, right?

    http://www.google.com/search?hl=en&rls=com.microsoft:en-US&q=100000+ev+to+joules

    Neato

    Anyways, the point of that is yes you were right to do so. However you have an error while calculating it, it's not 1 MV, it's .1 MV don't forget

    You found it for, well, one HUNDRED megavolts, accelerating a little proton through that potential difference would be pretty hefty indeed
     
  5. Feb 27, 2008 #4

    Dick

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    Your v is in the region of the speed of light. If that were correct, you should be using the relativistic form, as olgranpappy said. On the other hand, it looks like your calculation is using 100MV for the potential, but the problem you quoted says 0.100MV.
     
  6. Feb 27, 2008 #5

    tony873004

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    Gold Member

    Thanks for catching that. I hate losing points for dumb mistakes like that. Thanks for the info on the eV. I'll look into that.
     
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