1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Motion of charged particles in E and B fields

  1. Mar 29, 2007 #1
    1. The problem statement, all variables and given/known data

    Photoelectrons are emitted with negligible initial velocity from one plate of a parallel-plate capacitor which is at zero potential. The other plate is held at a potential +V and the capacitor is in a uniform magnetic field of flux density B parallel to the plane of the plates. Show that no electron can reach the positive plate if V is less than e(DB)^2/2m where D is the separation of the plates and e/m is the specific charge of the electron.

    2. Relevant equations


    3. The attempt at a solution

    Ok well I've found the force due to the electric field created by the capacitor by saying:

    F=qE and E=V/D
    so F=qV/D
    The charge of the electron is e/m so:

    Then you've got the force due to the magnetic field which is trying to pull the electron in a circle. I'm not sure where you go from here. All I know is that:

    F=evB but what do I use for v as obviously it's accelerating?
  2. jcsd
  3. Mar 29, 2007 #2
    Pointing out that specific charge is NOT charge of the electron.
    So F/m = eV/mD
    So accleleration towads the plate by the electric force is = eV/mD

    Try relating it with centripetal acceleration caused by the magnetic field.
  4. Mar 29, 2007 #3
    It might be best to use potential energy/kinetic energy relation for a particle that just grazes the plate, eg 1/2mv^2=eV. What else must v be? (use quantums idea about centripetal force from the magnetic field)
  5. Apr 5, 2007 #4
    Ok so at the plate with zero potential the photoelectron has negligible kinetic energy and potential energy of eV. At the other plate the photoelectron has no potential energy and all kinetic energy?

    So: 1/2mv^2=eV Is this correct?

    The radius of the circle if the photoelectron just grazes the plate will be D so I can say that the centripetal force is mv^2/D and this is equal to the force due to the magnetic field (?)

    So mv^2/D=evB
    and so v=eBD/m

    Substituting into: 1/2mv^2=eV gives:

    V=e(DB)^2/2m so have I done it?!

    If this is right then thanks everyone for your help :smile:
    Last edited: Apr 5, 2007
  6. Apr 5, 2007 #5
    methinks so. But remember this is for particle that never makes it. so for particle that reaches other side, v> that quantity as asked for in statement in problem.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Motion of charged particles in E and B fields