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Homework Help: Distance between electrons in electron beam?

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    A beam of 9.5 MeV electrons (gamma = 20) amounting as a current to 0.05 microamperes, is traveling through a vacuum. The transverse dimensions of the beam are less than 1 mm, and there are no positive charges in or near it.

    (a) In the lab frame, what is approximately the electric field strength 1cm away from the beam, and what is the average distance between an electron and the next one ahead of it, measured parallel to the beam?

    (b) answer the same question for the electron rest frame.

    2. Relevant equations

    3. The attempt at a solution
    I used kinetic energy to find the electric field, so i did something like this:

    K = .5mv^2 = eV = eEd, where m was the mass of the electron, v was its velocity, e is the electron charge, E is the electric field strength, and d is the distance away from it. V is voltage. I ended up with something like this:

    E = (mv^2)/(2ed) = 2.55 x 10^7 V/m (velocity was found using gamma).

    I'm having a little bit more trouble with the interelectron difference, though. I'm just stuck. Can anyone just give me a hint to get me started?
  2. jcsd
  3. Nov 6, 2011 #2


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    Homework Helper

    Welcome to PF, mooneyp.
    I don't remember much relativity, but maybe I can help with this charge distance. I'm thinking of one second of charge flow, a cylinder of charges moving at nearly the speed of light, so the cylinder will be 3 x 10^8 m long. The current is 5 x 10^-8 Coulombs/s or 3.1 x 10^11 electrons/s. So in that one second cylinder region, you have 3.1 x 10^11 electrons over a distance of 3 x 10^8 meters. Looks like 1000 electrons per meter of length.

    The electric field due to a long line of charge is E = 2kλ/z, where λ is the linear charge density, z the distance away from it. See
    That way, the E field works out to about 90 000.

    But I don't know how relativity impacts this. There would be one heck of a Lorentz contraction for starters.
    Last edited: Nov 6, 2011
  4. Nov 6, 2011 #3
    Thanks for the help on the distance between electrons, though I think you're confusing the current with linear charge density. I don't really need much help for the relativistic effects.
  5. Nov 6, 2011 #4


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    I would be most interested in understanding why you think I have confused current with charge density . . . please elaborate!

    I'm worried about your use of electric potential. You start out with
    .5mv^2 = eV
    which gives the energy or speed of an electron accelerated through a potential difference of V. That V could be used to find the E field inside the accelerator if we knew how long it was, but has nothing to do with the electric field caused by the electrons after they have left their accelerator and are moving at constant speed.
  6. Nov 6, 2011 #5
    I just worked backwards from what you got as your final result of 90000 N/C to get λ, and got 5x10^-8 C/m when, if it is really 1000 e-1/m, it should be 1000 * 1.6x10^-19 C/m. You're right about the potential difference.
  7. Nov 6, 2011 #6


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    Yah, looks like I mixed up my electron charges and Coulombs in that estimate.
    I = 5 x 10^-8 C/s so 5 x 10^-8 C spread over the line 3 x 10^8 meters long.
    λ = 1.67 x 10^-16 C/m
    It is this charge that cause the electric field. From basics, you would use
    E = kq/r² for each charge. Integrated over a long line that works out to
    E = 2kλ/z = 2*9x10^9*1.67 x 10^-16/.01 = 3 x 10^-4 V/m
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