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Static electricity and electrons

  1. Feb 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Initially electron moves through a horizontal electric field that accelerates electron. Then electron enters a uniform electric field between two parallel plates. velocity of the electron is horizontal and it lies exactly on the middle line between the plates. This field causes vertical displacement of the electron. A diagram of the electron motion is depicted below. Accelerating potential differences is 100V. Potential differences between parallel plates is 1 V and the length of the plate is 1m. The distance between the plates is 20cm. Electron charge is 1.6 x 10^-19 C; m9e) = 9.11 x 10^-31 kg.

    A. Find the velocity of the electron at the moment it enters the space between the horizontal plates.

    B. Find the time that the electron moves between the plates

    C. Find the vertical displacement of the electrons when it is between the plates

    D. find the magnitude of the velocity at the moment electron leaves electric field

    e. find the magnitude of the velocity at the moment electron leaves electric field

    f. Find the direction of the velocity (the angle with the horizontal).

    please help as soon as possible
     
  2. jcsd
  3. Feb 27, 2008 #2

    Dick

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    Please help us to help you by following the rules of the forum. This means you have to start trying to work the problem out for yourself and show us where you need help. Start with A). What's the energy of the electron when it enters the plate region? How can you turn that number in a velocity?
     
  4. Feb 28, 2008 #3
    Ok for a. You use w= qv to calculate potential work qv =mv^2 .5 so v = 5.93*10^6
     
  5. Feb 28, 2008 #4
    I'm having trouble finding time. Should I use a kinematics formula?
     
  6. Feb 28, 2008 #5

    Dick

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    Ok, you've got the horizontal component of velocity (or you will, as soon as you put units on it). Yes, use kinematics. You know v_horizontal and you know the length.
     
  7. Mar 2, 2008 #6
    I can't get the right answer

    Dick can you help? I seem to be unable to do part B.
     
  8. Mar 2, 2008 #7

    Dick

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    Finding time shouldn't be a problem. You have the horizontal component of velocity and you have the horizontal distance. T=D/V, right?
     
  9. Mar 2, 2008 #8
    I do not think so because the parallel plates voltage is +1 which would change the force which would also means it changes the velocity. unless the x component of velocity is constant which I'm not certain of. also check out that pic I think my reasoning will be clearer if you see it.
     

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  10. Mar 2, 2008 #9

    Dick

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    Exactly. From your problem description the potential difference is perpendicular to x direction. It can only cause the y component of velocity to change, not the x. This is just like a projectile problem. Use kinematics.
     
  11. Mar 2, 2008 #10
    ok then t = 1.686 * 10^-7. now for part c do I have to put the kinematic in 2 dimensions?
     
  12. Mar 2, 2008 #11
    Ok i now am confused? I don't know how to find a I just try two different ways but I think they are both wrong.
    first way: ended up using the Pythagorean theorem to find distance traveled and I plugged it into x = vt +.5at^2 where I use my horizontal velocity as the initial velocity. I think this is wrong because I need to find the vertical force and divide it by mass. that leads me in to version 2

    second way: using the formulas f = q*E, E = k(q/r^2), and V= k(q/r) I got F = V^2/K I then used V =1 and divided by K = 8.99 * 10^9 at that point I divided F/M. after doing such I wonder if what I did is even feasible or if I'm even using the right equations because V=joules per coulumb.

    the answer for the first way is 1.75895 * 10^11
    the answer for the second way is 1.22 * 10 ^20
     
  13. Mar 2, 2008 #12

    Astronuc

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    In the problem treat the horizontal and vertical velocities separately, i.e. the electron has been accelerated horizontally to some energy, for which one solved, and then has a constant velocity.

    With respect to the vertical plates, the electron has no initial vertical velocity, but in the electric field it is accelerated vertically (gravity is neglible), so vertically, one must determine the vertical acceleration and then use the equation of motion for constant acceleration with no initial velocity.

    Remember a = F/m, and in an electric field F = Eq.

    This will help with the relationship between E and V.

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html
     
    Last edited: Mar 2, 2008
  14. Mar 2, 2008 #13
    I was already their my question was if how I found F was correct i applied f = qE, E = k(q/r^2), and V= k(q/r) and derived F = V^2/K. to get the answer you use a= F/m but my question was how to get F.
     
  15. Mar 2, 2008 #14

    Astronuc

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    If one know F = Eq, and E = V/d, where V is the potential difference and d is the plate separation, then one can find F, and then find a.
     
  16. Mar 2, 2008 #15
    so because it is an electron q = 1.6 *10^-19 C and M = 9.11* 10 ^-31. and the Volts in the parallel plates is 1 and the change in vertical distance is .1 M so it should be (q(1/.1))/M. if this is correct then the answer should be a = 1.756 * 10^12. did i do this right if not please tell me where I messed up.
     
  17. Mar 2, 2008 #16
    I mistyped part D the actual queston is
    find the vertical displacement of the electron the moment it leaves the electric field
     
  18. Mar 2, 2008 #17
    for part d i used (at^2)/2 . I already solved a and t. so my answer was .02496
     
  19. Mar 2, 2008 #18

    Astronuc

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    To compute E from V, one needs to use the separation of the plates (0.2 m).

    E = V/d = (1 J/C)/ (0.2 m) = 5 J/C-m.

    The a = F/m = Eq/m = _______________


    One can then find the vertical velocity and displacement as a function of time. Then as the electron emerges at time t, which one determined from the horizontal velocity and distance of 1 m, one can use the vertical and horizontal velocities to determine the resultant velocity.
     
  20. Mar 2, 2008 #19
    for part d I was suppose to find vertical displacement where gravity is negligible so I used the formula x = (at^2)/2 so my answer was .02496

    for part e. i need to find the magnitude of velocity so I use the formula V = V0 + at I used part a's velocity for V0 and I used vertical acceleration from part c and time from part b that gives me the answer of 6.226 * 10^6

    for part f I use tan theta - Vy/VX so theta = tan^-1 vy/vx I use part e's velocity for vy and part a's velocity for VX and my direction equals 46.395 degrees from the horizontal
     
  21. Mar 2, 2008 #20
    ok let me change that is the thought process correct for the rest?
     
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