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Magnetic/electric particle deflection

  • Thread starter scholio
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  • #1
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Homework Statement


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


Homework Equations



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

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...
 

Answers and Replies

  • #2
rock.freak667
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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
 
  • #3
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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??
 
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  • #4
alphysicist
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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?
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.
 
  • #5
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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?
 
  • #6
alphysicist
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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?
 
  • #7
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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...
 
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  • #8
alphysicist
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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.)
 
  • #9
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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?
 
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  • #10
alphysicist
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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:

[tex]
\Delta y = v_{0y} t + \frac{1}{2}a_y t^2[/tex]
 
  • #11
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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?
 
  • #12
alphysicist
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That looks right to me.
 
  • #13
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thanks so much, cleared up quite a bit for me
 

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