Electron being shot between two plates

  • Thread starter Thread starter hvthvt
  • Start date Start date
  • Tags Tags
    Electron Plates
AI Thread Summary
The discussion revolves around determining the horizontal distance an electron will travel between two charged plates in a uniform electric field. The electric field is set at 2.00×10^3 N/C, and the electron is shot at a 45° angle with an initial speed of 5.86×10^6 m/s. Participants explore how to calculate the time of flight and horizontal distance without the electron hitting the upper plate, drawing parallels to projectile motion. Key equations involve calculating the electron's vertical motion under acceleration due to the electric field and using trigonometric components of the initial velocity. The final calculations suggest a horizontal distance of approximately 0.061 m, contingent on accurate time determination.
hvthvt
Messages
42
Reaction score
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
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
 
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. ?
 
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top