# Electron traveling between charged plates

#### 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.
Homework 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?

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#### collinsmark

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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?

#### jisbon

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?

#### collinsmark

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Gold Member
$S_{x}=u_{x}t+1/2a_{x}t^2 =0 +1/2at^2$
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?

#### Delta2

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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 $a_y=\frac{Eq}{m}$.

#### PeroK

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2018 Award
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?

#### jisbon

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?
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 $a_y=\frac{Eq}{m}$.
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

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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.

#### PeroK

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2018 Award
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.

#### jisbon

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.

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 travelled 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?
Answer seems to be $9.38*10^7$ instead :/

#### PeroK

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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 travelled 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?
Answer seems to be $9.38*10^7$ instead :/
It's the initial horizontal velocity you are looking for.

You do like your numbers!

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

#### PeroK

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2018 Award
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 travelled 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?
Answer seems to be $9.38*10^7$ instead :/
Also, you need to decide what the horizontal distance actually is. You have it as variously 10, 20 and 50cm in your posts.

#### jisbon

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

You do like your numbers!

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?

#### jisbon

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

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

#### PeroK

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My bad,

$S_{x}=u_{x}t+1/2a_{x}t^2$
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.

#### PeroK

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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.

#### jisbon

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

#### PeroK

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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.

#### jisbon

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? $0.05=\frac{1}{2}\frac{Eq}{m}t^2$

#### PeroK

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

#### jisbon

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

#### PeroK

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And :
$E= 20V$
For a capacitor:

$E = V/d$

An Electric field is dimensionally different from a voltage.

#### TSny

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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.

#### jisbon

Leaving aside the issue of units, yes.
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 any one 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 $9.38*10^7$ :/

#### TSny

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

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