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Electric Potential - first concepts

  1. Sep 20, 2007 #1
    We just started applying the concept of potential energy and conservative forces in my electromagnetism class, so I might have a few of these up soon.

    Question : 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? The speed of light is 3.00e8m/s; Mass of electron = 9.11e-31; abs.value(e) = 1.60e-19 C

    Ka + Ua = Kb + Ub where K is kinetic energy and U is potential energy of the charge.

    Object is at rest, so equation may be simplified to -Kb = Ub - Ua or Kb = -(Ub - Ua).

    U = V*q where q is the charge in coulombs and V is the potential. The change in V (from rest to .39 the speed of light) is what I'm looking for...

    -[tex]\Delta[/tex]U/q = -(Ub/q - Ua/q) = -[tex]\Delta[/tex]V

    I can start substituting now to solve for [tex]P\Delta[/tex]V. I take the above simplified equation based on conservation of energy and divide both sides by the charge q to get

    Kb/q = -Ub/q + Ua/q = -[tex]\Delta[/tex]V

    Kb = .5(m)v^2 where v = .39(3.00e8).

    So my final attempt was .5(9.11e-31(.39(3.00e8)^2))/1.60e-19 = -[tex]\Delta[/tex]V

    My answer was around 39000... I'm asked for kV so dividing by 1000 I get 39.0, which seems appropriate, but my answer isn't accepted. The units check out... I have put a negative on the answer, which also seems appropriate according to my above logic, but that often fails me around this time of night!

    That abs(e) = 1.6 etc. coulombs seemed a little weird... any idea where I'm going wrong?

    Any help is appreciated. :)
     
  2. jcsd
  3. Sep 20, 2007 #2
    scratch that large, obtrusive P there... not sure where that sucker came from.
     
  4. Sep 20, 2007 #3

    dynamicsolo

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    I come up with the final kinetic energy of 38,922 electron-volts; since this *is* an electron, the magnitude of the potential difference is 38,922 V. Electrons accelerate in the opposite direction that the electric field points, so they accelerate from lower to higher electric potentials. The potential difference should thus be +38,922 V or about
    +38.9 kV.

    You're quite sure the computer asks for kilovolts (I'm presuming from your remarks that you are on a problem file system, like WebAssign)? What is the accepted level of precision? [A really horrible case we ran into a few years back was that the instructor forgot to set a precision tolerance on one problem and the default tolerance was 0%. Good luck guessing the computer's answer to eight decimal places...]
     
  5. Sep 20, 2007 #4
    It says three significant figures... I'm fairly sure I tried 38.9 too, that was my exact answer. We are using ThomsonNow, which is considerably less effective than masteringPhysics.com. Hopefully it's not screwed up like you're saying... I have to try it again anyway because it reset to 34% the speed of light. Bah. *cross fingers*
     
  6. Sep 20, 2007 #5

    dynamicsolo

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    I just had a thought: are you supposed to use "Newtonian" mechanics to find the electron's kinetic energy or relativistic mechanics? At speeds higher than 0.1c, the difference between the two results starts to become rather significant...
     
  7. Sep 21, 2007 #6
    I ended up guessing results and discovered the answer. To achieve 33% the speed of light, the acceleration of potential is 30.4 kV. I worked the problem backwards, and found the value I was using for q (|e| = 1.60e-19) should have been something like 1.458e-16. I'm not quite sure how to achieve that value, though.

    Thanks for the advice D.
     
  8. Sep 21, 2007 #7

    dynamicsolo

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    Aha, it is relativistic mechanics! You don't want to use the "classical" formula for kinetic energy, K = (1/2)m(v^2). Instead you have to find the total energy of the electron, E = (gamma)(m)(c^2), where

    gamma = 1/sqrt[ 1 - {(v/c)^2} ]. The kinetic energy of the electron is then

    K = E - (m)(c^2) = (gamma - 1)(m)(c^2).

    At 0.33c, gamma = 1/sqrt[ 1 - (0.33)^2 ] = 1.05934 . To find the kinetic energy of the electron in suitable units, you want to give its rest mass as 511,000 electron-volts. Then

    K = (1.05934 - 1)(511,000 eV) = 30,324 eV . As before, the potential difference accelerating the electron from rest will then be +30.32 kV.

    I guess I should have asked what course you are in. Most of the people asking me about this are taking first-year physics, which is all classical; they don't talk about relativistic physics, though the problems are generally chosen so as not to deal with speeds where that model must be used.
     
  9. Sep 22, 2007 #8
    Awesome! I'm in a -fairly- basic physics course, electromagnetism and waves, so in class we haven't gotten much into relativistic equations... though he has spent a good deal of time breaking all of the mathematics into conceptual work, and emphasizing the alterations from the classical model that separate the physics we knew in high school with... well, reality!!!

    Thanks again for the help - I am thoroughly impressed.
     
  10. Sep 23, 2007 #9

    dynamicsolo

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    You may well do so someways down the road: a paradox in classical electromagnetic theory was one of the wellsprings of special relativity.

    The problem we've been looking at here is also relevant. Somewhere along in the 1880s to '90s, people accelerating electrons in Crookes' tubes (the ancestor of the no-longer-fashionable CRT) noticed that the speed of the electron didn't jibe with the expected kinetic energy change for selected potential differences in the tube. You could increase KE all you liked, but the velocity just seemed to asymptotically approach c. Hmmm...
     
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