Electron being shot between two plates

In summary: I guess that is the accelerated motion along the x-axis:Vox= Vo * cos45 and x=Vox*tHowever, I do not know the time it takes. From ma= eE + mg I am able to calculate the acceleration.I can calculate Ymax by stating that Vy=0 when the particle reaches its maximum distance.And we know that Vy = Voy - at so Voy= at. This would give me the time that is needed. Calculating this would give me 1.46*10^-8 s. Is this correct?Can i then just plug it in for x=Vox*t= 0.061 m. ?
  • #1
hvthvt
42
0

Homework Statement



In the figure, a uniform, upward-pointing electric field E of magnitude 2.00×103 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 and separation d. Electrons are shot between the plates from the left edge of the lower plate.
The first electron has the initial velocity v0, which makes an angle θ=45° with the lower plate and has a magnitude of 5.86×106 m/s. Suppose the distance d between the plates is large enough so that the electron doesn't hit the upper plate. Determine how far horizontally from the left edge on the lower plate the electron will strike.

Homework Equations



v[initial, y]^2 + (2*a*deltaY) = 0
acceleration = [eE]/[m]
v[initial,y] + (a*t) = 0

The Attempt at a Solution



I do know how to solve this problem when it actually hits the upper plate, but my problem is: how to deal with the fact that it DOESNT hit it? I know that vx is zero when Xmax is reached. I know that the acceleration is 3.5*10^14 m/s^2
Can anybody help me?
 
Physics news on Phys.org
  • #2
hvthvt said:

Homework Statement



In the figure, a uniform, upward-pointing electric field E of magnitude 2.00×103 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 and separation d. Electrons are shot between the plates from the left edge of the lower plate.
The first electron has the initial velocity v0, which makes an angle θ=45° with the lower plate and has a magnitude of 5.86×106 m/s. Suppose the distance d between the plates is large enough so that the electron doesn't hit the upper plate. Determine how far horizontally from the left edge on the lower plate the electron will strike.

Homework Equations



v[initial, y]^2 + (2*a*deltaY) = 0
acceleration = [eE]/[m]
v[initial,y] + (a*t) = 0

The Attempt at a Solution



I do know how to solve this problem when it actually hits the upper plate, but my problem is: how to deal with the fact that it DOESNT hit it? I know that vx is zero when Xmax is reached. I know that the acceleration is 3.5*10^14 m/s^2
Can anybody help me?

Do you see a similarity between this and projectile motion where the downward force is a uniform gravitational force? Can you use the formula relating launch angle, speed and horizontal range for projectile motion for this problem?

AM
 
  • #3
I guess that is the accelerated motion along the x-axis:

Vox= Vo * cos45 and x=Vox*t
However, I do not know the time it takes. From ma= eE + mg I am able to calculate the acceleration.
I can calculate Ymax by stating that Vy=0 when the particle reaches its maximum distance.
And we know that Vy = Voy - at so Voy= at. This would give me the time that is needed. Calculating this would give me 1.46*10^-8 s. Is this correct?
Can i then just plug it in for x=Vox*t= 0.061 m. ?
 
  • #4
hvthvt said:
I guess that is the accelerated motion along the x-axis:

Vox= Vo * cos45 and x=Vox*t
However, I do not know the time it takes. From ma= eE + mg I am able to calculate the acceleration.
I can calculate Ymax by stating that Vy=0 when the particle reaches its maximum distance.
And we know that Vy = Voy - at so Voy= at. This would give me the time that is needed. Calculating this would give me 1.46*10^-8 s. Is this correct?
Can i then just plug it in for x=Vox*t= 0.061 m. ?

I was thinking of R = v02sin2θ/g

See: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra4

How would you adapt this for the electric field in this problem?

AM
 
  • #5


Thank you for reaching out for help with this problem. It seems like you have a good understanding of the equations and concepts involved. In this case, since the electron does not hit the upper plate, we can assume that the acceleration remains constant throughout its motion. Therefore, we can use the equations for constant acceleration to solve for the horizontal distance the electron will travel.

First, we can use the equation v[initial, y] + (a*t) = 0 to find the time it takes for the electron to reach its maximum height. Since the initial vertical velocity is given as 5.86×106 m/s and the acceleration is 3.5×1014 m/s^2, we can solve for t and find that it takes approximately 1.677×10^-8 seconds for the electron to reach its maximum height.

Next, we can use the equation v[initial, y]^2 + (2*a*deltaY) = 0 to find the maximum height reached by the electron. We know that the initial vertical velocity is 5.86×106 m/s and the acceleration is 3.5×1014 m/s^2, so we can solve for deltaY and find that the electron reaches a maximum height of approximately 4.91×10^-7 meters.

Finally, we can use the equation deltaX = v[initial, x] * t to find the horizontal distance the electron will travel before reaching its maximum height. We know that the initial horizontal velocity is given as v0cosθ, which is equal to 5.86×106 m/s * cos(45°). Therefore, the horizontal distance traveled can be calculated as approximately 2.08×10^-7 meters, or 2.08 micrometers.

I hope this helps you solve the problem. Let me know if you have any further questions or need clarification on any of the steps. Good luck!
 

1. What is the concept behind shooting an electron between two plates?

The concept behind shooting an electron between two plates is based on the principles of electricity and magnetism. When an electric field is applied between two plates, it creates a force on any charged particle, such as an electron, causing it to move.

2. How does the speed of the electron affect its trajectory between the plates?

The speed of the electron does not affect its trajectory between the plates, as long as it is within the range of the electric field. The force applied by the electric field will cause the electron to accelerate or decelerate accordingly to maintain its path between the plates.

3. What happens to the electron after it passes through the plates?

After passing through the plates, the electron will continue to move in a straight line until it is acted upon by another force. This force could be another electric field, magnetic field, or collision with another particle.

4. How does the distance between the plates impact the trajectory of the electron?

The distance between the plates does not directly impact the trajectory of the electron. However, a shorter distance between the plates will result in a stronger electric field, which will cause the electron to accelerate at a faster rate.

5. Can the direction of the electric field be reversed to change the trajectory of the electron?

Yes, if the direction of the electric field is reversed, the electron's trajectory will also be reversed. This is because the electron will now experience a force in the opposite direction, causing it to move in the opposite direction as well.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
976
  • Introductory Physics Homework Help
Replies
3
Views
6K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
25
Views
4K
  • Introductory Physics Homework Help
Replies
22
Views
3K
Back
Top