Vertical Deflection of Electron: 1.1×107 m/s & 3.2×10-16 N

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SUMMARY

The discussion focuses on the vertical deflection of an electron moving horizontally at a speed of 1.1 × 107 m/s under a constant vertical force of 3.2 × 10-16 N. Participants emphasize the importance of treating horizontal and vertical motions separately, as the vertical force does not affect the horizontal speed. The acceleration of the electron is calculated using the formula a = F/m, resulting in 2.5125 × 1014 m/s2. The final vertical distance deflected is determined to be approximately 1.678 × 10-3 m.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with kinematic equations for constant acceleration
  • Basic knowledge of forces and mass (F=ma)
  • Concept of separating vector components in motion
NEXT STEPS
  • Study the kinematic equations for vertical motion under constant acceleration
  • Learn about the effects of forces on particle trajectories in physics
  • Explore the concept of vector decomposition in two-dimensional motion
  • Investigate the relationship between force, mass, and acceleration in different contexts
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of charged particles in electric fields will benefit from this discussion.

  • #31
If we instead consider the position of a particle I mentioned earlier. If we keep the x co-ordinate constant, but changed the y co-ordinate, the direction we would need to travel in order to reach the particle has changed. It doesn't mean the x co-ordinate has.

Do you see how we can relate this to the velocities?
 
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  • #32
Stephen Hodgson said:
If we instead consider the position of a particle I mentioned earlier. If we keep the x co-ordinate constant, but changed the y co-ordinate, the direction we would need to travel in order to reach the particle has changed. It doesn't mean the x co-ordinate has.

Do you see how we can relate this to the velocities?
i see know
 
  • #33
J-dizzal said:
maybe not, if there are no forces acting in the -x direction, then the x component of its velocity doesn't change. its only the y component increasing right?
Exactly! :smile: The y component is increasing, thus changing the direction. The x component does not need to change for this to happen
 
  • #34
So the x velocity is constant. What equation can we therefore use to calculate the time the particle is in motion for?
 
  • #35
Stephen Hodgson said:
If we instead consider the position of a particle I mentioned earlier. If we keep the x co-ordinate constant, but changed the y co-ordinate, the direction we would need to travel in order to reach the particle has changed. It doesn't mean the x co-ordinate has.

Do you see how we can relate this to the velocities?
so
Stephen Hodgson said:
Exactly! :smile: The y component is increasing, thus changing the direction. The x component does not need to change for this to happen
Stephen Hodgson said:
So the x velocity is constant. What equation can we therefore use to calculate the time the particle is in motion for?
t=dist/v = 3.091x10-9s
 
  • #36
That seems correct to me.

We now need to use that time to calculate the vertical distance travelled.
 
  • #37
Stephen Hodgson said:
That seems correct to me.

We now need to use that time to calculate the vertical distance travelled.
yes just distance now. so Δy = v0t - 1/2 at2 this look like a good formula of motion to use?
 
  • #38
This looks good.

What values of v, a and t are you going to use?
 
  • #39
Stephen Hodgson said:
This looks good.

What values of v, a and t are you going to use?
v0=0, a= 3.5126x1014, and t=3.091x10-9
 
  • #40
Great! v0 should be 0 because in the y direction it is initially at rest :smile:.

You may want to check your equation again though. Are you sure about the - sign?

Let me know what you get as your final answer.
 
  • #41
Stephen Hodgson said:
Great! v0 should be 0 because in the y direction it is initially at rest :smile:.

You may want to check your equation again though. Are you sure about the - sign?

Let me know what you get as your final answer.
542872.33 m that seems huge. yea i rechecked formula thanks its + because v is the unknown not v0
 
  • #42
Stephen Hodgson said:
Great! v0 should be 0 because in the y direction it is initially at rest :smile:.

You may want to check your equation again though. Are you sure about the - sign?

Let me know what you get as your final answer.
542872.33 m seems large
 
  • #43
J-dizzal said:
542872.33 m seems large
forgot to sqare the time one sec
 
  • #44
1.678 m
 
  • #45
Almost. Check your units.
 
  • #46
Stephen Hodgson said:
Almost. Check your units.
1.67810^-3
 
  • #47
lol thanks
 
  • #48
Great! Glad I could help. It seems like you learned a lot doing that question :smile:
 
  • #50
Stephen Hodgson said:
Great! Glad I could help. It seems like you learned a lot doing that question :smile:
yea i learned some fundamental concepts that i know will help on the exam
 
  • #51
main thing being that force changes the velocity of a particles seems so obvious now
 
  • #52
Yeah. The other important thing is to know that you can simply treat the velocities separately. It is often easier to do this than try to work with vectors.
 
  • #53
I just had a look at the circular motion problem. Which part are you having problems with?
 
  • #54
Stephen Hodgson said:
I just had a look at the circular motion problem. Which part are you having problems with?
i can't seem to find any unknowns. i know that 3/4 of the distance of the circle is traveled by the particle in 9.90s-3.30s but when i try to derive an equation for total time(T) i cant. to solve for T i have so far 6.6T = 3pi/2 but plugging that solution into r =vt/2pi its to get the radius its not working
 
  • #55
So is the y co-ordinate obvious from your diagram?

To calculate T, a bit of common sense needs to be used. If 3/4 of the circle is completed in 6.6s, how long does it take to complete a full circle?
 
  • #56
Stephen Hodgson said:
So is the y co-ordinate obvious from your diagram?

To calculate T, a bit of common sense needs to be used. If 3/4 of the circle is completed in 6.6s, how long does it take to complete a full circle?
8.8s would be the whole right?
the y coordinate of?
 
  • #57
Stephen Hodgson said:
So is the y co-ordinate obvious from your diagram?

To calculate T, a bit of common sense needs to be used. If 3/4 of the circle is completed in 6.6s, how long does it take to complete a full circle?
ok 8.8 works, it must of made a mistake in the math last time i tried it.
 
  • #58
So that question is now solved?
 
  • #59
Stephen Hodgson said:
So that question is now solved?
just working on the y coordinate now
 
  • #60
Take a look at your diagram.
 

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