Magnetic/electric particle deflection

In summary, a particle with a 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. Using the equations for electric field and kinematics, it can be determined that the particle will experience a deflection of 4.98*10^-4 m as it travels through the plates. This deflection is found by calculating the initial velocity and acceleration of the particle, taking into account the two directions of motion, and using the kinematic equation for displacement in the y-direction.
  • #1
scholio
160
0

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...
 
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  • #2
Find the initial velocity using energy=1/2 mv^2
Find the acceleration it experiences,using F=EQ and F=ma.
Kinematic equations
 
  • #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^2determining 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
scholio said:
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
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
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 canceled 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 traveled 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
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
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
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
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
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
That looks right to me.
 
  • #13
thanks so much, cleared up quite a bit for me
 

FAQ: Magnetic/electric particle deflection

What is magnetic/electric particle deflection?

Magnetic/electric particle deflection is a method used to manipulate the path of charged particles, such as electrons or protons, by using magnetic or electric fields.

How does magnetic/electric particle deflection work?

The process of magnetic/electric particle deflection involves applying a magnetic or electric field perpendicular to the direction of the charged particle's motion. This causes the particle to experience a force, known as the Lorentz force, which deflects it from its original path.

What are some applications of magnetic/electric particle deflection?

Magnetic/electric particle deflection is commonly used in scientific research, particularly in particle accelerators, to study the properties of subatomic particles. It is also used in cathode ray tubes for televisions and computer monitors, as well as in mass spectrometry for analyzing chemical compounds.

What are the differences between magnetic and electric particle deflection?

The main difference between magnetic and electric particle deflection is the type of field used to manipulate the particles. Magnetic deflection uses a magnetic field, while electric deflection uses an electric field. Additionally, magnetic deflection is typically used for charged particles with higher masses, while electric deflection is more suitable for lighter particles.

Are there any limitations to magnetic/electric particle deflection?

Yes, there are some limitations to magnetic/electric particle deflection. One limitation is that it can only be used for charged particles, so neutral particles cannot be deflected. Additionally, magnetic deflection is only effective for particles with a certain range of velocities, while electric deflection can only be used for particles with a specific charge-to-mass ratio.

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