Calculating Horizontal Distance of a Proton in a Parallel-Plate Capacitor

AI Thread Summary
A proton enters a parallel-plate capacitor with a speed of 1.276 x 10-5 m/s and is subjected to a downward electric field of 2.95 x 10-4 N/C. The vertical acceleration of the proton, calculated from the electric force, is 28265 m/s². To determine the horizontal distance traveled before the proton hits the bottom plate, the time taken to fall 0.795 cm vertically must be calculated, using the appropriate kinematic equations. The horizontal distance can then be found by multiplying the time of flight by the initial horizontal velocity. Understanding the relationship between horizontal and vertical motion is crucial for solving this problem.
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Homework Statement



A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.276 x 10-5 m/s. The distance between the two plates is 1.59 cm. The proton enters the capacitor halfway between the top plate and the bottom plate; that is, a distance r = 0.795 cm from each plate. The capacitor has a 2.95 x 10-4 N/C uniform electric field between the plates that points downward from the top plate to the bottom plate. Neglecting gravitational forces, what horizontal distance does the proton traverse before the proton hits the bottom plate?

Homework Equations





The Attempt at a Solution



Since the electric field is 2.95 x 10-4 N/C I used that to get my force by multiplying the charge of an electron 1.6E-19 thus I got F = 4.72E-23 from that I used F = ma to calculated the acceleration which I got 28265 m/s^2. But now I am stuck. The question will have been easier if they gave the width of the capacitor but they gave me the distance between. How can I do this problem? Thanks!
 
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tuggler said:

Homework Statement



A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.276 x 10-5 m/s. The distance between the two plates is 1.59 cm. The proton enters the capacitor halfway between the top plate and the bottom plate; that is, a distance r = 0.795 cm from each plate. The capacitor has a 2.95 x 10-4 N/C uniform electric field between the plates that points downward from the top plate to the bottom plate. Neglecting gravitational forces, what horizontal distance does the proton traverse before the proton hits the bottom plate?

Homework Equations


The Attempt at a Solution



Since the electric field is 2.95 x 10-4 N/C I used that to get my force by multiplying the charge of an electron 1.6E-19 thus I got F = 4.72E-23 from that I used F = ma to calculated the acceleration which I got 28265 m/s^2. But now I am stuck. The question will have been easier if they gave the width of the capacitor but they gave me the distance between. How can I do this problem? Thanks!

What is the direction of the acceleration?
The motion of the proton can be resolved into a horizontal one, parallel with the plates and a vertical one, perpendicular to the plates. How long time does it take for the proton to reach one plate? How far does it travel horizontally during that time?

ehild
 
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The direction of the acceleration will be towards the bottom of the plate, kind of looks like a parabola. But other than that I don't know how to finish this problem.
 
The solution will be similar to projectile motion. The question is almost same as a rock thrown from a cliff with a horizontal velocity.
So what does the horizontal velocity of the proton depend upon? Does it change? What about its vertical velocity?
 
tuggler said:
The direction of the acceleration will be towards the bottom of the plate,

You mean the bottom plate or bottom of the plate?
 
I meant "bottom plate". The velocity will depend on the electric field since they both are perpendicular to each other.
 
Horizontal or Vertical?
 
tuggler said:
I meant "bottom plate". The velocity will depend on the electric field since they both are perpendicular to each other.

Only the initial velocity (horizontal) is perpendicular to the vertical electric field. The proton accelerates vertically towards a plate.
You are right saying that it is like horizontal projectile motion. What are the equations for the projectile?

ehild
 
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Yup, I understand that part now. The velocity is perpendicular to the field where the acceleration is perpendicular to the velocity vector.

But how can continue to get the desired answer?
 
  • #10
How does a projectile move? Do you recall any equations?

ehild
 
  • #11
v_f^2 = v_i^2 + 2ad?
Will this equation work?
 
  • #12
It might work if you use the correct d, vi and vf, but you need the horizontal displacement.

what horizontal distance does the proton traverse before the proton hits the bottom plate?

ehild
 
  • #13
The horizontal displacement is what I am having trouble figuring out. If I knew the horizontal displacement I will know what to do from there. How can I get it?
 
  • #14
The horizontal displacement is the question in the problem.

As for projectile motion, this link may help.

http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Vectors/ProjectilesMotion.html

ehild
 
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  • #15
I am still confused
 
  • #16
Well, just calculate the time taken for proton to fall in the plate and multiply it to horizontal velocity. Remember vertical velocity is zero.
 
  • #17
The proton accelerates vertically with a=28265 m/s2. It travels Δy=0.795 cm distance vertically. How long time does it travel till it reaches the bottom plate?
You need the formula between displacement and time for uniformly accelerating motion.
Horizontally, the proton travels with constant speed. How far does it reach (horizontally) during its time of flight?

ehild
 
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