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Homework Help: Electron traveling between two plates

  1. Jul 24, 2009 #1
    1. The problem statement, all variables and given/known data
    An electron travels with speed 4.91×107 m/s between the two parallel charged plates. The plates are separated by s=1.47 cm and are charged by a 203 V battery. What magnetic field strength will allow the electron to pass between the plates without being deflected?
    ----------+------- < postive plate
    >>>>O < electron
    ----------(-)------- < negative plate

    2. Relevant equations

    I thought that

    Fe= electric force
    Fm=magnetic force

    3. The attempt at a solution

    Set it equal to zero o=Fe+Fm

    solve for B....that didn't work what am I doing wrong?
  2. jcsd
  3. Jul 24, 2009 #2
    Fe is not (Kq1)/(r^2)

    (Kq1)/(r^2) is E ( electric field )

    Fe = E x q
  4. Jul 25, 2009 #3
    So then my equation should be (Kq1q2/R^2)/qVB
  5. Jul 25, 2009 #4
    You mean (Kq1q2/R^2) = qVB :wink:

    Yes that's right, but it's better to use Eq = qVB
    Find E, then solve for B
  6. Jul 25, 2009 #5
    What charges do I use, I can tell from that equation that the charge of the plates are going to cancel out. So then....

    do i assume is half of the distance between the plates?
  7. Jul 25, 2009 #6
    For two parallel charged plates, E = V/d, where V = voltage across the parallel plate and d = distance between the plate
  8. Jul 26, 2009 #7
    oh so instead of using E=KQ1Q1/R^2 I should use E=V/D?
  9. Jul 26, 2009 #8
  10. Jul 27, 2009 #9
    For some reason it didn't worl
  11. Jul 27, 2009 #10
    Why? Have you converted 1.47 cm to m?
  12. Jul 27, 2009 #11
    Here is my work....
    Voltage/s(distance between plates)=qV(velocity)B

  13. Jul 27, 2009 #12
    The charge (q) has cancelled out and even if they haven't, you can subs it with the charge of an electron which is 1.6 x 10^(-19) C
  14. Jul 27, 2009 #13
    right thats is what q I used.....My professor had his c++ language wrong online so it wouldn't take my answer but he fixed it and my answer is right now.
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