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Magnetic/electric particle deflection

  1. Jun 1, 2008 #1
    1. The problem statement, all variables and given/known data
    a particle having mass of 10^-30 kg and a charge of 10^-18 coulombs and an energy of 5000eV is fired into an electric field produced by two parallel plates. the electric field between the plates is 10V/m and the length of the plates is 0.4 m, how much deflection d, will the charge experience as it traverses the plates?

    d is the distance from the center of the path between the two charged plates

    2. Relevant equations

    electric field E = kq/r^2 where k = 9*10^9, q is the charge, r is the radius

    3. The attempt at a solution

    i am completely lost on this one as i don't know how to implement the given energy of the particle, i am pretty sure the electric field equation will be used but i am sure that there are others. should i assume d = r? in other words, solve for r?

    do i sub in 10V/m for E, 10^-18 for q? what about r?

    i need another equation that factors in energy, mass and the length of the plates, i don't think i have come across such an equation yet? gauss-related???

    any help appreciated, i wish i could give more but that is as much as i have thought of thus far...
  2. jcsd
  3. Jun 1, 2008 #2


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    Find the initial velocity using energy=1/2 mv^2
    Find the acceleration it experiences,using F=EQ and F=ma.
    Kinematic equations
  4. Jun 2, 2008 #3
    using the given values and equation i got 1*10^17 for velocity but i don't know what units to use for velocity, is it still m/s?

    and for the acceleration i used the equation a = (q/m)E where q is charge = 10^-18 coulombs, m is mass = 10^-30kg, E is electric field of 10V/m

    i got 1*10^13 for acceleration but am not sure of the units? m/s^2

    determining deflection:
    how is acceleration and velocity used kinematically to find deflection, d? do i set up a triangle and use pythagorean theorem.
    using 0.4m for one side, d for the vertical... do i use accel and vel components to determine the hypotenuse??
    Last edited: Jun 2, 2008
  5. Jun 2, 2008 #4


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    No, that's not in m/s. I think it's best to convert the energy into Joules first; then redo your procedure to find the speed.
  6. Jun 2, 2008 #5
    i converted eV to joules using 1 eV = 1.6*10^-19 joules and got a velocity of 40,000,000
    i'm still not sure of the units

    what do i do with the velocity and acceleration to find displacement?
  7. Jun 2, 2008 #6


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    That speed sounds right to me.

    As for the units, you converted eV to J, which is (kg m^2 / s^2). You then multiplied that by 2/mass, which cancelled out the kilogram factor, and then you took the square root, leaving m/s.

    Let's orient the problem so that the initial velocity is in the x direction, and the acceleration is in the -y direction. At that point, solving it becomes exactly like a kinematic free fall problem.

    The analogous free fall problem might be: "From the top of a cliff, a ball is thrown horizontally with an initial speed 5 m/s. After it has travelled horizontally 20m, how far has it fallen vertically?" How would you solve that problem? You solve this problem exactly the same way. What do you get?
  8. Jun 2, 2008 #7
    based on your example with the ball-cliff etc problem, do i assume the 'height of the cliff' to be 0.40 m?

    do i use the formula x = vt + 1/2 at^2 where x is deflection, v is initial velocity, a is accel

    how do i find time t? --> should i use time t = distance/velocity = 0.4/40,000,000 = 1*10^-8 sec

    if use that for time i get a deflection d = 0.40 meters --> that can't be right?

    assistance please...
    Last edited: Jun 3, 2008
  9. Jun 3, 2008 #8


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    Remember you have to take into accout two directions and treat them almost independently. If we rewrote this question, we could say that it is really asking: in the time it takes for the particle to travel 0.4 m in the x direction, how far has it fallen in the y direction? (Using the initial velocity to be in the x direction and the acceleration to be in the y direction.)
  10. Jun 3, 2008 #9
    did what i solve for previously, the 0.4m 'deflection', solve anything in terms of what you stated? x or y direction?

    which equations should i use, i'm sure ill be missing things if i treat each direction 'separately,' can i solve for deflection using the v and a that i've found despite them being in different directions?
    Last edited: Jun 3, 2008
  11. Jun 3, 2008 #10


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    The time looks correct, so the particle travels 0.4 meters in 10^-8 seconds.

    However, for the motion in the y-direction, what is the initial velocity in the y-direction? Then use that in the y equation:

    \Delta y = v_{0y} t + \frac{1}{2}a_y t^2[/tex]
  12. Jun 3, 2008 #11
    the initial velocity in the y direction is 0m/s, correct?

    and then using the formula you specified and the acceleration i found, 1*10^13 m/s^2

    i found displacement in the y to be 4.98*10^-4 m, is that what i'm looking for, for deflection?
  13. Jun 3, 2008 #12


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    That looks right to me.
  14. Jun 3, 2008 #13
    thanks so much, cleared up quite a bit for me
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