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Finding the electric field magnitude in a parallel plate compacitor.

  1. Sep 24, 2008 #1
    Problem:
    The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is 9.42 x 10^6 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

    What i did:

    x=.5(Vo + V)t
    .02m=.5(9.42 x 10^6 m/s + 9.42 x 10^6 m/s)t
    solve for t=2 x 10^-9
    then:
    x=Vot + .5at^2
    .0015m=0 + .5(a)(2 x 10^-9)^2
    solve for a=8.87 x 10^15
    then:
    F=ma
    F=(9.11 x 10^-31)(8.87 x 10^15)
    F=8.08 x 10^-15
    then:
    E=F/q
    E=8.08 x 10^-15/(1.6 x 10^-19)
    E=50524.3 N/C

    But that is the wrong answer, i was just wondering what i did wrong and how to correct myself.


    In Addition i was wondering how to do this one?:
    A long, thin rod (length = 3.0 m) lies along the x axis, with its midpoint at the origin. In a vacuum, a +7.0 C point charge is fixed to one end of the rod, and a -7.0 C point charge is fixed to the other end. Everywhere in the x, y plane there is a constant external electric field (magnitude 2.0 x 103 N/C) that is perpendicular to the rod. With respect to the z axis, find the magnitude of the net torque applied to the rod.

    Thank you in advanced!!
     

    Attached Files:

  2. jcsd
  3. Sep 24, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Redo this arithmetic.

    Also: Don't round off intermediate values (such as your value for t); wait until the last step.
     
  4. Sep 24, 2008 #3

    Doc Al

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    Staff: Mentor

    Start by finding the force on each end of the rod. Once you have the forces, you should be able to compute the torque about the z-axis. (What direction does the field point?)
     
  5. Sep 24, 2008 #4
    How do you find the force on each end of the rod?
    Should i just us the equation: E=kq/r^2 and then use E=F/q ?
     
  6. Sep 24, 2008 #5

    Doc Al

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    Staff: Mentor

    No, that formula gives you the field from a point charge. Not relevant here, since you are given the field.
    That's the one you want. (F = Eq.)
     
  7. Sep 24, 2008 #6
    ahh ok, i got it right, thank you so much for your help :)
     
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