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Projectile Motion in an Electric Field

  1. Jan 25, 2015 #1

    rlc

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    1. The problem statement, all variables and given/known data
    upload_2015-1-25_15-33-21.png

    In the figure, a uniform, upward-pointing electric field E of magnitude 5.00E3 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 4 cm (0.04 m) and separation d = 2.00 cm (0.02 m). An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v0 of the electron makes an angle θ=45° with the lower plate and has a magnitude of 9.07E6 m/s.

    a) Will the electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates ( m).

    b) The next electron has an initial velocity which has the same angle θ=45° with the lower plate and has a magnitude of 7.28×106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates ( m).

    2. Relevant equations
    qE=ma
    v[initial, y]^2 + (2*a*deltaY) = 0
    acceleration = [qE]/[m]
    v[initial,y] + (a*t) = 0

    3. The attempt at a solution
    find the acceleration
    qE=ma (mass and q are actually constant numbers of 9.11E-31 and 1.6E-19 respectively)
    (1.6E-19)(5.00E3 N/C)=(9.11E-31)a
    a=8.78E14

    then you find the velocity components which is vo(sin theta)
    (9.07E6 m/s)sin(45°)=6.41E6

    solve for time for X and Y
    Tx being: L= (velocity from above)(Tx)
    (0.04 m)=(6.41E6)Tx
    Tx=6.24E-9

    Ty being: d= (.5(acceleration))Ty^2 + (velocity found above)Ty
    0.02 m=((0.5)(8.78E14))Ty^2 + (6.41E6)Ty
    (4.39E14)Ty^2 + (6.41E6)Ty - 0.02 = 0
    Ty=2.64E-9
    or
    Ty=-1.72E-8

    whichever time is smaller, you solve for the height using time*acceleration

    **This question has been eluding me for a while, and that's probably because I can't wrap my head around the question. The attempt I have up there was given to me by one of my classmates and I've attempted to follow along as you can see, but I don't understand it. Is the attempt, in your opinion, good? Did I do those parts correctly? What does it even mean? Tx and Ty? Please, help me.
     
  2. jcsd
  3. Jan 25, 2015 #2

    rlc

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    I found a comment going through this problem. I've posted that entire comment, editing my data into it and doing the calculations. It didn't work. Please help me figure out where I went wrong.

    Attempt 2:
    To determine whether or not the electron strikes one of the plates, we need to determine the time Ty required to travel a vertical distance of y = 0.02 m and the time Tx for a horizontal distance of x = 0.04 m.

    If Ty < Tx, then the electron will strike the negative plate.
    If Ty > Tx, the electron will not strike the plate and we will then determine the vertical distance at which the particle leaves the space between the plates.

    For the most part, this is a kinematics problem, but we need to evaluate the vertical acceleration induced on the electron as it travels through the plates.

    This acceleration is found by equating F = qE = ma --> a = qE/m
    a = qE/m
    a=(1.6E-19)(5E3) / (9.11E-31)
    a=8.78E14 m/s^2

    We also need to isolate the x and y components of the velocity v0.

    Vy = v0sin(45) =(9.07E6)(sin45)= 6.41E6 m/s
    Vx = Vy = 6.41E6 m/s

    Now we find Ty and Tx.

    0.02 = 0+(6.41E6)(Ty)+(0.5)(8.78E14)(Ty)^2 --> Ty = 2.64e-9 s
    0.04 = 0+(6.41E6)(Tx) --> Tx = 6.24e-9 s

    Since Ty < Tx, the electron will in fact strike the plate at a horizontal distance of x = 0+(6.41e6)(2.64e-9) = 0.017 m.
    **I entered 0.017 m into the online homework and it said it was not correct. What did I miss?**

    Part 2

    We repeat the process and find that that Ty < Tx. The electron strikes the plate at a horizontal distance of x = 0+(3.65e6)(4.08e-9) = 0.015 m.
    **This is the data from the person who commented this to another person. I stopped following through after getting the first part wrong, but I'm including this just in case it's kind of right**
     
  4. Jan 25, 2015 #3

    haruspex

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    Umm... no.

    The first stage was right, finding the time to traverse in the x direction. But the questions asks for the y value on exit, so it is more obvious to plug the time for the traversal into the vertical motion equations to find y. If the answer is negative then it hits the lower plate.
    But there is another question that needs to be answered: does it hit the upper plate?
     
  5. Jan 25, 2015 #4

    rlc

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    Since Ty < Tx, the electron will strike the plate...right?
     
  6. Jan 25, 2015 #5

    haruspex

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    But bear in mind it might never reach that height.
    Careful with signs! Which is the positive vertical direction?
     
  7. Jan 25, 2015 #6

    rlc

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    Isn't Ty the positive vertical direction?
     
  8. Jan 25, 2015 #7

    haruspex

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    What are the correct signs for the initial vertical velocity and the vertical acceleration?
     
  9. Jan 25, 2015 #8

    rlc

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    Are they both negative?
     
  10. Jan 25, 2015 #9

    haruspex

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    Up in the diagram is positive, yes?
    Is the initial vertical velocity up or down?
    Is the acceleration up or down?
     
  11. Jan 25, 2015 #10

    rlc

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    The initial vertical velocity is up and to the right. Acceleration...that would be affected by the charges, right? The upper plate is negatively charged while the lower plate is positively charged. An electron is negatively charged.
     
  12. Jan 25, 2015 #11

    haruspex

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    Right, so which way is the acceleration?
     
  13. Jan 25, 2015 #12

    rlc

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    Upward as well? Would it be attracted to the negative charge?
    Or is it the exact opposite?
     
  14. Jan 25, 2015 #13

    rlc

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    Oh wait, it'd be attracted to the positive charge, right?
     
  15. Jan 25, 2015 #14

    haruspex

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    Do like charges attract or repel?
     
  16. Jan 25, 2015 #15

    rlc

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    Like charges repel
     
  17. Jan 25, 2015 #16

    haruspex

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    Yes.
    So what is the sign on the acceleration?
     
  18. Jan 25, 2015 #17

    rlc

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    Negative.
    Would that affect this equation in this way?:
    0.02 = 0+(6.41E6)(Ty) - (0.5)(8.78E14)(Ty)^2
     
  19. Jan 25, 2015 #18

    haruspex

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    Yes.
     
  20. Jan 25, 2015 #19

    rlc

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    0.02 = 0+(6.41E6)(Ty)-(0.5)(8.78E14)(Ty)^2 --> Ty = 4.5183E-9 s
    0.04 = 0+(6.41E6)(Tx) --> Tx = 6.24e-9 s

    Ty < Tx, so x = 0+(6.41e6)(4.5183E-9) = 0.029 m.
     
  21. Jan 25, 2015 #20

    rlc

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    Oh my gosh, thank you! That worked!
     
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