# Electron traveling between charged plates

• jisbon
In summary, the problem involves finding the minimum velocity of an electron between two plates with a length of 50cm, spacing of 1cm, and a potential difference of 20V. The equation F=ma is used, with the vertical component of the final velocity being solved for. The direction that does not involve acceleration also needs to be taken into account. The equation for this is Sx=uxt+1/2axt^2. The maximum time for the electron to travel from one plate to the other without touching either plate is also a factor to consider. The equation used for this is Sy=uyt+1/2ayt^2, with ay being equal to Eq/m. The gravitational force can
jisbon
Homework Statement
Plates have length 50cm, spacing 1cm, and P.D of 20V. Find minimum velocity of electron such that it just emerge from plates.
Relevant Equations
##F_{e}=qe##
##E=\frac{\triangle V}{\triangle d}##
F=ma

So I can understand how to find out velocity of electron moving between these 2 plates, by using:
##F=ma##
##a=\frac{F}{m} =\frac{qE}{m}##
##v^2=2as = \frac{2q}{m}Ex##
##v = \sqrt{\frac{2q}{m}Ex}##
where x = 0.01 , E =20 etc. Answer seems to be way too off, which is expected.
How am I supposed to find the 'minimum' velocity though?

jisbon said:
Homework Statement: Plates have length 50cm, spacing 1cm, and P.D of 20V. Find minimum velocity of electron such that it just emerge from plates.
Homework Equations: ##F_{e}=qe##
##E=\frac{\triangle V}{\triangle d}##
F=ma

View attachment 249240
So I can understand how to find out velocity of electron moving between these 2 plates, by using:
##F=ma##
##a=\frac{F}{m} =\frac{qE}{m}##
##v^2=2as = \frac{2q}{m}Ex##
##v = \sqrt{\frac{2q}{m}Ex}##
where x = 0.01 , E =20 etc. Answer seems to be way too off, which is expected.
How am I supposed to find the 'minimum' velocity though?
You seem to have solved for the vertical component of the final velocity.

How long does it take (i.e., how much time does it take) for the electron to travel from one plate to the other if the initial, vertical component of the velocity is 0?

collinsmark said:
You seem to have solved for the vertical component of the final velocity.

How long does it take (i.e., how much time does it take) for the electron to travel from one plate to the other if the initial, vertical component of the velocity is 0?
##S_{x}=u_{x}t+1/2a_{x}t^2 =0 +1/2at^2##
May I ask why is that the vertical component that I found? Isn't it supposed to be the 'overall' final velocity?

And if I only have final vertical velocity, can I consult if I am still able to use ##S_{y}=u_{y}t+1/2a_{y}t^2## where a= 9.8?

jisbon said:
##S_{x}=u_{x}t+1/2a_{x}t^2 =0 +1/2at^2##
May I ask why is that the vertical component that I found? Isn't it supposed to be the 'overall' final velocity?
For the problem to be solved, you need to solve for both the vertical component and the horizontal component of the final velocity.

In the work you showed in the original post, you have solved for the component in the direction that involves acceleration. So now what about the direction which does not involve acceleration?

What's the maximum amount of time it will take an electron to get from the left side to the right side without touching either plate?
And if I only have final vertical velocity, can I consult if I am still able to use ##S_{y}=u_{y}t+1/2a_{y}t^2## where a= 9.8?

The equation is valid, but you'll likely find that the gravitational force is completely negligible compared to the electric force. So although the equation is fine, no, you wouldn't use a=9.8.

----

By the way, just for clarity, is there anything left out of the problem statement such as the initial direction of the electron? Does it specify that the electron is initially traveling in the horizontal direction or can it be shot in at a slight angle?

you can study the problem exactly like that of a projectile with initial velocity v (which seems to be only horizontal in our case) inside a gravitational field but instead of ##a_y=g## you take ##a_y=\frac{Eq}{m}##.

jisbon said:
Homework Statement: Plates have length 50cm, spacing 1cm, and P.D of 20V. Find minimum velocity of electron such that it just emerge from plates.
Homework Equations: ##F_{e}=qe##
##E=\frac{\triangle V}{\triangle d}##
F=ma

View attachment 249240
So I can understand how to find out velocity of electron moving between these 2 plates, by using:
##F=ma##
##a=\frac{F}{m} =\frac{qE}{m}##
##v^2=2as = \frac{2q}{m}Ex##
##v = \sqrt{\frac{2q}{m}Ex}##
where x = 0.01 , E =20 etc. Answer seems to be way too off, which is expected.
How am I supposed to find the 'minimum' velocity though?
1) if the electron has very small initial horizontal velocity, it will be deflected into one of the plates.

2) if the electron has a very high initial horizontal velocity, it will shoot between the plates with mininal deflection.

Your task is to find the horizontal velocity at the transition between being too slow and fast enough to escape.

Why do you use spoilers on your homework posts?

PeroK said:
1) if the electron has very small initial horizontal velocity, it will be deflected into one of the plates.

2) if the electron has a very high initial horizontal velocity, it will shoot between the plates with mininal deflection.

Your task is to find the horizontal velocity at the transition between being too slow and fast enough to escape.

Why do you use spoilers on your homework posts?
Delta2 said:
you can study the problem exactly like that of a projectile with initial velocity v (which seems to be only horizontal in our case) inside a gravitational field but instead of ##a_y=g## you take ##a_y=\frac{Eq}{m}##.
collinsmark said:
For the problem to be solved, you need to solve for both the vertical component and the horizontal component of the final velocity.

In the work you showed in the original post, you have solved for the component in the direction that involves acceleration. So now what about the direction which does not involve acceleration?

What's the maximum amount of time it will take an electron to get from the left side to the right side without touching either plate?The equation is valid, but you'll likely find that the gravitational force is completely negligible compared to the electric force. So although the equation is fine, no, you wouldn't use a=9.8.

----

By the way, just for clarity, is there anything left out of the problem statement such as the initial direction of the electron? Does it specify that the electron is initially traveling in the horizontal direction or can it be shot in at a slight angle?

To all:
Electron is assumed as initially traveling in the horizontal direction.
I used spoilers because I thought my attempt is not related to the solution.
So as far as I'm concerned, there is only acceleration due to the electric field in the vertical direction and no acceleration in the horizontal direction.
So my job is to find out the velocity for just the electron to squeeze out from the plates, so something like this? (Pardon my drawing)

Alright so to find the time taken, since initial velocity of the vertical component is 0 (since it is initially traveling in the horizontal direction),
##S_{y}=u_{y}t+1/2a_{y}t^2 ##
where ##u=0##, so,
##S_{y}=1/2a_{y}t^2 ##
where ##a_y=\frac{Eq}{m} ##
So can I assume:
##0.05=\frac{1}{2}\frac{Eq}{m}t^2=\frac{1}{2}\frac{(20)(1.6*10^{-19})}{9.11*10^{-31}}t^2##
So ##t=1.69*10^{-7}s##?
Am I on the right track here?

Delta2
yes you are definitely on the right track. Now all you need to do is to find the horizontal velocity, such that in the same time t, it travels horizontal distance d=length of plate.

jisbon said:
To all:
Electron is assumed as initially traveling in the horizontal direction.
I used spoilers because I thought my attempt is not related to the solution.
So as far as I'm concerned, there is only acceleration due to the electric field in the vertical direction and no acceleration in the horizontal direction.
So my job is to find out the velocity for just the electron to squeeze out from the plates, so something like this? (Pardon my drawing)
View attachment 249252
Alright so to find the time taken, since initial velocity of the vertical component is 0 (since it is initially traveling in the horizontal direction),
##S_{y}=u_{y}t+1/2a_{y}t^2 ##
where ##u=0##, so,
##S_{y}=1/2a_{y}t^2 ##
where ##a_y=\frac{Eq}{m} ##
So can I assume:
##0.05=\frac{1}{2}\frac{Eq}{m}t^2=\frac{1}{2}\frac{(20)(1.6*10^{-19})}{9.11*10^{-31}}t^2##
So ##t=1.69*10^{-7}s##?
Am I on the right track here?
Yes, although why calculate ##t##?

There are only three variables: acceleration, length and width. Time is bound to cancel out if you stick with the algebra.

PeroK said:
Yes, although why calculate ##t##?

There are only three variables: acceleration, length and width. Time is bound to cancel out if you stick with the algebra.
Won't I need to find t to find out the horizontal velocity? Since I'm determining the velocity needed.

Delta2 said:
yes you are definitely on the right track. Now all you need to do is to find the horizontal velocity, such that in the same time t, it travels horizontal distance d=length of plate.
Hmm okay :)
So since ##t=1.69*10^{-7}s ##, and horziontal distance traveled is 0.1m,
##S_{x}=u_{x}t+1/2a_{x}t^2## where ##a_{x}##=0,
##0.1=u_{x} (1.69*10^{-7}s)##,
##u_{x} = 591715.9763m/s##
Since I have ##u_{x}## and ##v_{y}## (##v_{y}=u_{y}t+at= 0 + \frac{Eq}{m} (1.69*10^{-7}s) =594375.4116m/s##
So technically, shouldn't the velocity be ##\sqrt{u_{x}^2+v_{y}^2}## = 838695.3717m/s?

jisbon said:
Won't I need to find t to find out the horizontal velocity? Since I'm determining the velocity needed.Hmm okay :)
So since ##t=1.69*10^{-7}s ##, and horziontal distance traveled is 0.1m,
##S_{x}=u_{x}t+1/2a_{x}t^2## where ##a_{x}##=0,
##0.1=u_{x} (1.69*10^{-7}s)##,
##u_{x} = 591715.9763m/s##
Since I have ##u_{x}## and ##v_{y}## (##v_{y}=u_{y}t+at= 0 + \frac{Eq}{m} (1.69*10^{-7}s) =594375.4116m/s##
So technically, shouldn't the velocity be ##\sqrt{u_{x}^2+v_{y}^2}## = 838695.3717m/s?
It's the initial horizontal velocity you are looking for.

At this level of physics, a fear of algebra is a serious if not fatal handicap.

jisbon said:
Won't I need to find t to find out the horizontal velocity? Since I'm determining the velocity needed.Hmm okay :)
So since ##t=1.69*10^{-7}s ##, and horziontal distance traveled is 0.1m,
##S_{x}=u_{x}t+1/2a_{x}t^2## where ##a_{x}##=0,
##0.1=u_{x} (1.69*10^{-7}s)##,
##u_{x} = 591715.9763m/s##
Since I have ##u_{x}## and ##v_{y}## (##v_{y}=u_{y}t+at= 0 + \frac{Eq}{m} (1.69*10^{-7}s) =594375.4116m/s##
So technically, shouldn't the velocity be ##\sqrt{u_{x}^2+v_{y}^2}## = 838695.3717m/s?
Also, you need to decide what the horizontal distance actually is. You have it as variously 10, 20 and 50cm in your posts.

PeroK said:
It's the initial horizontal velocity you are looking for.

At this level of physics, a fear of algebra is a serious if not fatal handicap.
Correct me if I'm wrong, but won't the initial and final horizontal velocity be the same?

PeroK said:
Also, you need to decide what the horizontal distance actually is. You have it as variously 10, 20 and 50cm in your posts.

##S_{x}=u_{x}t+1/2a_{x}t^2##
Where ##S_{x}= 0.5 ,a_{x}=0##
After doing the same steps as above, I will still get resultant velocity as ##3.02*10^7 m/s##

jisbon said:

##S_{x}=u_{x}t+1/2a_{x}t^2##
Where ##S_{x}= 0.5 ,a_{x}=0##
After doing the same steps as above, I will still get resultant velocity as ##3.02*10^7 m/s##
There is always lots of room for error when you work with complicated numbers with lots and lots of significant digits.

Why not try an an algebraic approach?

The irony is that I've done it algebraically and can see that the length must be ##50cm##. If I'd used your approach maybe I would have got something wrong as well.

jisbon said:
To all:
Electron is assumed as initially traveling in the horizontal direction.
I used spoilers because I thought my attempt is not related to the solution.
So as far as I'm concerned, there is only acceleration due to the electric field in the vertical direction and no acceleration in the horizontal direction.
So my job is to find out the velocity for just the electron to squeeze out from the plates, so something like this? (Pardon my drawing)
View attachment 249252
Alright so to find the time taken, since initial velocity of the vertical component is 0 (since it is initially traveling in the horizontal direction),
##S_{y}=u_{y}t+1/2a_{y}t^2 ##
where ##u=0##, so,
##S_{y}=1/2a_{y}t^2 ##
where ##a_y=\frac{Eq}{m} ##
So can I assume:
##0.05=\frac{1}{2}\frac{Eq}{m}t^2=\frac{1}{2}\frac{(20)(1.6*10^{-19})}{9.11*10^{-31}}t^2##
So ##t=1.69*10^{-7}s##?
Am I on the right track here?
This value of ##t## is wrong.

PeroK said:
This value of ##t## is wrong.
Oh... Let me check and get back :)

Hmm, so assuming this is correct:
##0.05=\frac{1}{2}\frac{Eq}{m}t^2##

##t^2 = 0.05/0.5/ \frac {(20)(1.6*10^{-19})}{9.11*10^{-31}}##
##t^2 = 2.84*10^{-14}##
##t = 1.69 *10^{-7}## ?
Not sure which part I messed up

jisbon said:
Oh... Let me check and get back :)
I never calculated ##t## in the first place. All I did was divide the length by the answer for velocity you gave and it was different from the ##t## you got. You could have done that yourself.

PeroK said:
I never calculated ##t## in the first place. All I did was divide the length by the answer for velocity you gave and it was different from the ##t## you got. You could have done that yourself.
But this equation is correct, right? ##0.05=\frac{1}{2}\frac{Eq}{m}t^2##

jisbon said:
But this equation is correct, right? ##0.05=\frac{1}{2}\frac{Eq}{m}t^2##
Leaving aside the issue of units, yes.

PeroK said:
Leaving aside the issue of units, yes.
And :
##E= 20V##
##q = 1.6*10^{-19}##
##m = 9.11*10^{-31}##?

jisbon said:
And :
##E= 20V##

For a capacitor:

##E = V/d##

An Electric field is dimensionally different from a voltage.

jisbon said:
But this equation is correct, right? ##0.05=\frac{1}{2}\frac{Eq}{m}t^2##
0.5 cm ≠ .05 m. This might be the "issue of units" that @PeroK referred to.

PeroK
PeroK said:
Leaving aside the issue of units, yes.
TSny said:
0.5 cm ≠ .05 m. This might be the "issue of units" that @PeroK referred to.
Hi all, thanks for your help :) The value seems to be alright, but the power is off by ##10^1##, was wondering if anyone of y'all could help figure it out :)

##0.5=\frac{1}{2}\frac{Eq}{m}t^2 = \frac{1}{2}\frac{\frac{V}{d}q}{m}t^2 = \frac{1}{2}\frac{\frac{20}{0.
01}(1.6*10^{-19})}{9.11*10^{-31}}t^2##
## t = 5.33*10^{-8}##
##u_{x}=s_{x}/t = l/t = 0.5/ (5.33*10^{-8})= 9380863.03939..##
----------------------
##v_{y}=0.5a_{y}t^2 = 0.5 (\frac{\frac{V}{d}q}{m})t^2= 0.5 (\frac{\frac{20}{0.01}1.6*10^{-19}}{9.11*10^{-31}})(5.33*10^{-8})^2 = 0.49894##

So ##v=\sqrt{u_{x}^2+v_{y}^2}=\sqrt{(9380863.03939)^2+( 0.49894)^2} =9380863.03939m/s =9.38*10^6##

Final answer's supposed to be ##9.38*10^7## :/

jisbon said:
##0.5=\frac{1}{2}\frac{Eq}{m}t^2 =##….
Explain how you got the value of 0.5 on the left side. What does this number represent and what are its units?

Delta2
First of all I don't understand why you calculate ##v_y## you are not asked for that (you are not asked for the magnitude of the vertical component at exit or the magnitude of the velocity at exit), you are asked only to calculate ##v_x##, that is the horizontal velocity with which the electron enters the electric field.

Other than that @TSny was just ahead of me in time :D.

TSny
TSny said:
Explain how you got the value of 0.5 on the left side. What does this number represent and what are its units?
50cm = 0.5m

jisbon said:
50cm = 0.5m
But this equation you wrote ##0.5=\frac{1}{2}a_yt^2##is for the vertical motion. 0.5m is the length of the plates not the spacing between the plates which is 1cm. Shouldn it be ##\frac{1}{2}cm=\frac{0.01}{2}m=0.005m=\frac{1}{2}a_yt^2##??

Last edited:
Delta2 said:
But this equation you wrote ##0.5=\frac{1}{2}a_yt^2##is for the vertical motion. 0.5m is the length of the plates not the spacing between the plates which is 1cm. Shouldn it be ##\frac{1}{2}cm=\frac{0.01}{2}m=0.005m=\frac{1}{2}a_yt^2##??
Oh gosh. Yep I realize my mistake. Kept messing up between the length and width

Delta2

## 1. How does an electron travel between charged plates?

The movement of an electron between charged plates is governed by the electric field created by the charged plates. When a potential difference is applied between the plates, the electric field exerts a force on the electron, causing it to move from the negatively charged plate to the positively charged plate.

## 2. What happens to the electron's speed as it travels between charged plates?

The electron's speed increases as it moves from the negatively charged plate to the positively charged plate. This is because the electric field accelerates the electron, increasing its kinetic energy.

## 3. How does the distance between the charged plates affect the electron's travel time?

The distance between the charged plates has a direct impact on the electron's travel time. The greater the distance between the plates, the longer it will take for the electron to travel between them. This is because the electric field weakens as the distance increases, resulting in a weaker force on the electron and slower acceleration.

## 4. Can an electron travel between charged plates in a vacuum?

Yes, an electron can travel between charged plates in a vacuum. In fact, a vacuum is often used in experiments involving electron travel between charged plates to eliminate the effects of air resistance and other external forces.

## 5. What is the significance of electron travel between charged plates in physics?

The movement of electrons between charged plates is a fundamental concept in physics, known as the flow of electric current. It is essential for understanding the behavior of electricity and plays a crucial role in various technologies, including electronics, power generation, and communication systems.

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