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Determining angular frequency in an electric field using velocity.

  1. Nov 16, 2014 #1
    • Member warned about not using the formatting template for homwework questions
    1. There are 2 negatively charged plates opposite each other. In between them, there is a vacuum tube (50 cm long), containing only 1 electron. Assume it is completely isolated.
    The charge value for the plates is equivalent to 10000 electrons.
    Initially the single electron is directly in the center, after the initial displacement by an external electric field the electron is displaced 10 cm.
    r1=0.15 r2=0.25 r3=0.35
    k=9.0*10^9
    q1 as well as q2 = - (1.6*10^-19)
    n = 10001 (electrons in 1 plate + electron in the vacuum)

    2. Q1: How much kinetic energy can the electron have at a max speed? What is this speed?
    Ee1 + Ek1 = Ee2 + Ek2
    (A1- already solved, Vmax = 5.3*10^3; EK = 1.278*10^-23)

    Q2: Determine w (angular frequency)

    Q3: Determine the period T

    3. How do we determine w using the velocity we calculated? We know Vmax=A root of (k/m) but then we realized that w^2 doesn't = k/m because this isn't simply SHM. We searched up and down but we can't figure out what equations we're supposed to use or how to solve this at all.

    Thank you.
     
  2. jcsd
  3. Nov 16, 2014 #2

    gneill

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    Is there more information about the plates? For example, are their dimensions given? Presumably we will need to be able to find the electric field at points between the plates, but without the dimensions of the plates we can't determine the charge density and hence their fields.
     
  4. Nov 16, 2014 #3
    There is no further information about the plates. This is all the information we were given.
     
  5. Nov 16, 2014 #4

    gneill

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    How then did you go about determining the KE and maximum velocity of the electron?
     
  6. Nov 16, 2014 #5
    using the following: Ee1 +EK1 = Ee2 + EK2
    Ee = (n*k*q1*q2)/r EK = (mv^2)/2
    however, at EK1, the velocity is 0 because the electron is at the turning point and therefore has no velocity so the equation ends up looking like this:
    Ee1 + 0 = Ee2 + EK2

    We were given, n (10000 electron equivalent on plates + 1 electron in the vacuum) k = 9.0*10^9, both q charges are negative so
    (- (1.6*10^-19))*(- (1.6*10^-19)) and then you have all that over r. the r for Ee1 needs to be the sum of Ee1 when r =0.15 and r= 0.35 (since that is at the same point (when it is 0.15m from one plate, it's 0.35m from the other plate))
    For the r of Ee2, we used the r at the center of the tube, that is when the electron is at max velocity. Since the tube was 50cm, the center is at 0.25m and that WAS it's initial position.
    We then plugged all that into the equation Ee1 +0 = Ee2 + EK2 and solved for velocity. At the same time, we were able to get Ek2 which is the kinetic energy at max velocity.
     
  7. Nov 16, 2014 #6
    using the following: Ee1 +EK1 = Ee2 + EK2
    Ee = (n*k*q1*q2)/r EK = (mv^2)/2
    however, at EK1, the velocity is 0 because the electron is at the turning point and therefore has no velocity so the equation ends up looking like this:
    Ee1 + 0 = Ee2 + EK2

    We were given, n (10000 electron equivalent on plates + 1 electron in the vacuum) k = 9.0*10^9, both q charges are negative so
    (- (1.6*10^-19))*(- (1.6*10^-19)) and then you have all that over r. the r for Ee1 needs to be the sum of Ee1 when r =0.15 and r= 0.35 (since that is at the same point (when it is 0.15m from one plate, it's 0.35m from the other plate))
    For the r of Ee2, we used the r at the center of the tube, that is when the electron is at max velocity. Since the tube was 50cm, the center is at 0.25m and that WAS it's initial position.
    We then plugged all that into the equation Ee1 +0 = Ee2 + EK2 and solved for velocity. At the same time, we were able to get Ek2 which is the kinetic energy at max velocity.
     
  8. Nov 16, 2014 #7

    gneill

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    Okay, so you made the assumption that for the electron in the tube, the plates "look like" point charges of 10,000 e- . Given that, I'm seeing a slightly lower maximum velocity than you.

    What numerical values did you get for the electric PE at the initial displacement and at the center?

    I'm not sure how you'll go after the period and angular frequency. The potential function is not trivial so the resulting differential equation of the motion might be tricky to solve.
     
  9. Nov 16, 2014 #8
    for Ee1 (sum of Ee when r =0.15 and r= 0.35, initial displacement ) we got 2.2*10^-23 and for Ee2 (using r=0.25, the center) we got 9.22*10^-24
     
  10. Nov 16, 2014 #9

    gneill

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    I'm seeing a different (larger) value for the PE at the center. It looks like you've only accounted for the effect of one plate.
     
  11. Nov 16, 2014 #10
    Oh! that's right, oops. In that case, our Ee2 is (9.22*10^-24)*2 = 1.84*10^-23 (since it's 0.25m in relation to each of the two plates). Our velocity is then 2.8*10^3. Thank you so much.
    There's still the issue of solving for the angular frequency (w) though.
     
  12. Nov 16, 2014 #11

    gneill

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    You're very welcome.
    Yeah, I'm not sure the best way to go about that. As I mentioned, the differential equation is bound to be tricky due to the form of the PE function. It has the form 1/(1 - z2).

    I suppose it could be solved by numerical approximation (numerical integration of the differential equation), but that's probably not the approach you're expected to use. What's the topic of the unit this comes from in your course or text?
     
  13. Nov 16, 2014 #12
    The whole unit is called energy and momentum. We were doing simple harmonic motion, pendulums, springs, and static electricity as well as energy transformations.
     
  14. Nov 16, 2014 #13

    gneill

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    Okay, so this problem in this form looks to be a bit beyond that, unless I'm missing something obvious.

    Perhaps you might manage an approximation with a quadratic fit to the PE function around the center. That would make the PE function look like that of a spring. I'm not sure how accurate the approximation would be without doing some investigation; maybe a Taylor series expansion to look at the first few terms...

    Sorry I can't think of a path forward at the course's level. I'll have to give it some more thought.
     
  15. Nov 16, 2014 #14
    Knowing our teacher, it's entirely possible that this is beyond curriculum standards. We're going to try using the equations F/m = a = -w^2x and
    F = -(nkq1q2)/r^2 to solve for magnitude of w. Can we do that?
     
  16. Nov 16, 2014 #15

    gneill

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    That would be a crude approximation, but it might be what is intended. I hesitate to just say yes.
     
  17. Nov 16, 2014 #16
    Okay, then I guess that's what we'll be going with. Thank you so much for all your help, we appreciate it very much!
    have a good one.
     
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