Electric field between two parallel plates

[meher4real]

Let me explain to you my situation here.
I used to study physics in french (my last course was four years ago), this is my first time doing it in English, plus i used to work for three years on marketing, this year I've decided to complete my education and quit my job so i can get a master degree in Geology as a foreign student.
It's a challenging stage for me and I'm motivated to pass it.
(Hope you understand why I'm struggling here)

 

[Delta2]

Ok I see now, in French and 4 years ago, that explains it :D.

Unfortunately I don't know French (except a few words hehe) so I don't know how Work and Work-Energy theorem is in French.

We ll try to do this with the kinematic approach. Do you recall the equations $d=v_0t-\frac{1}{2}at^2$, $v=v_0-at$?

 

[meher4real]

Thank you for understanding
Of course, but how it can work with Electromagnetism ?

 

[Delta2]

For this problem we don't use a lot from Classical Electromagnetism, just that $F=Eq$ and that $E=\frac{\sigma}{\epsilon_0}$, the rest is classical mechanics stuff.
Also for the second part you have to use the formula $V=E\cdot l$ where l is the spacing of the two plates.

 

[Delta2]

If the force is $F=Eq$ what is the acceleration $a$ from the electric field on the electron?

 

[meher4real]

For this problem we don't use a lot from Classical Electromagnetism, just that $F=Eq$ and that $E=\frac{\sigma}{\epsilon_0}$, the rest is classical mechanics stuff.
Also for the second part you have to use the formula $V=E\cdot l$ where l is the spacing of the two plates.

Thank you for you time mate it means a lot
I'll take my time working on that problem, I'm watching DALLAS game rn hehe, your answers helps alot.
Can you expain the "stop" thing, what does it mean ?

 

[Delta2]

It means that the velocity of all the electrons that constitute the beam , from $0.6x10^8m/s$ becomes 0 as the electrons travel a distance of d=10mm inside the electric field of the parallel plates. The electric field E exerts a force $F=Eq$ to each electron which decelerates the electrons with constant deceleration $a$ (sorry if I said acceleration), and their velocity $v$ is reduced according to the equation $v=v_0-at$ until the time $t$ is such that $v_0-at=0$, i.e. the time that the electron stops. Of course here you are given the distance and not the time but you can work out the details via the other equation $d=v_0t-\frac{1}{2}at^2$.

 

[Delta2]

You should focus solving the problem btw, not watching games that distract you...

 

[meher4real]

It means that the velocity of all the electrons that constitute the beam , from $0.6x10^8m/s$ becomes 0 as the electrons travel a distance of d=10mm inside the electric field of the parallel plates. The electric field E exerts a force $F=Eq$ to each electron which decelerates the electrons with constant deceleration $a$ (sorry if I said acceleration), and their velocity $v$ is reduced according to the equation $v=v_0-at$ until the time $t$ is such that $v_0-at=0$, i.e. the time that the electron stops. Of course here you are given the distance and not the time but you can work out the details via the other equation $d=v_0t-\frac{1}{2}at^2$.

I literally was lost, thank you so much.

 

[meher4real]

You should focus solving the problem btw, not watching games that distract you...

I'm a diehard DALLAS fan, i'll be back studying after the game is finished.

 

[Delta2]

Ok

 

[meher4real]

Hi !
Is it normal to find Voltage = 111.046x10^3 V

 

[Delta2]

Hi.

How did you find this number? Step by step please.

 

[meher4real]

d=V0xt => 10x10^-3 = 0.6x10^8 x t => t = 1.66x10^-10 s
d = V0t+1/2xaxt^2 (initial velocity is 0) (a=?) then 10x10^-3 m = 1/2 a (1.66x10^-10)^2
a = 7.81x10^17 m/s^-2
Also a= F/m = qE/me then E = (a x me)/q = 444.18x10^4 NC^-1
Voltage = ? l = 25x10^-3 m
V= El = 111.046x10^3 V

 

[Delta2]

You just can't say that $d=v_0t$ and also that $d=v_0t+1/2at^2$ , its like saying that the electron does move with constant velocity for first and that it does move with constant acceleration for the second. You just can't have both, if it moves with constant acceleration, then the velocity isn't constant.

However there is some logic in the general procedure that you follow, specifically your aim to find first $a$ and then $E$. You are correct in finding $E$ from $a$ but you aren't correct in finding $a$.

So in order to find the correct value for acceleration $a$ (which is actually deceleration):
The initial velocity of the electron is not 0 but it it is 0.6x10^8m/s.
The final velocity is zero (0).

The equation for d is really $d=v_0t-1/2at^2$. In this equation you have two unknowns the deceleration a and the time t it takes to reach zero velocity. Which other equation can you use, so that you have two equations with the same two unknowns so that you can solve the system for a and t?

 

[meher4real]

Thank you for your detailed response.
But i still confused about the difference between d=v0t-1/2at^2 and d=v0+1/2at^2 ??
How to solve "t" if you use it in that equation when "a" still unknown ?
Is the value of "t" correct ? if yes so it's about the "d" equation !

 

[Delta2]

No the value of t you calculated is not correct.

You have to use a system of two equations with two unknowns to solve for both a and t. I will tell you that the two equations are :
$d=v_0t-1/2at^2$
$0=v_0-at$.

 

[meher4real]

I found t= 3.33x10^-10 s and a = 1.80x10^17
E= 102.47x10^4
V= 25.6x10^3 (Still big)
What do you think ?

 

[Delta2]

Again you should tell me the intermediate steps and which formulas you used and how you solved that system of two equations, I haven't done the arithmetic calculations my self to know the arithmetic values for t and a.

 

[Delta2]

Your value of t seems correct to me now that i did some arithmetic calculations myself.
Your value for a seems correct to me too.

 

[Delta2]

A few tens of KV for voltage (10-100KV) is typical for such experiments btw.

 

[meher4real]

Is it normal a big Voltage ?
This problem tooks me two days to solve it now i have to face 7 other problems before next week hehe.
All this equations for just two questions.
Gad may be with us.
Anyways, Thank you so much

 

[Delta2]

25 KV (25 Kilo Volts) is typical for this kind of experiments as I said in earlier post.

Not so many equations if you ask me, just 4, the two at my earlier post and $F=Eq=ma$ and $V=El$.
You should write a Phd or Master thesis, or even worst decide to write a book, there you would have to deal with hundreds if not thousands of equations.

 

[meher4real]

25 KV (25 Kilo Volts) is typical for this kind of experiments as I said in earlier post.

Not so many equations if you ask me, just 4, the two at my earlier post and $F=Eq=ma$ and $V=El$.
You should write a Phd or Master thesis, or even worst decide to write a book, there you would have to deal with hundreds if not thousands of equations.

Good Luck with that fam !
I feel you.

 

[kuruman]

d=V0xt => 10x10^-3 = 0.6x10^8 x t => t = 1.66x10^-10 s
d = V0t+1/2xaxt^2 (initial velocity is 0) (a=?) then 10x10^-3 m = 1/2 a (1.66x10^-10)^2
a = 7.81x10^17 m/s^-2
Also a= F/m = qE/me then E = (a x me)/q = 444.18x10^4 NC^-1
Voltage = ? l = 25x10^-3 m
V= El = 111.046x10^3 V

If you use the shortcut equation 2ad = v_f² - v₀² with displacement d = 10 mm, initial velocity v₀ = 0.6×10⁸ m/s and final velocity v_f = 0, you will not get the same value for the acceleration. You probably made a mistake somewhere in calculating the time. Of course, if there is an error in the acceleration, there will be an error in the electric field value.

This shortcut not only gets the answer faster but also reduces the number of calculations which makes it less likely to make a mistake. It is very useful for relating acceleration, displacement, and speed if time is of no importance as is the case here.

You can find this and other useful kinematic equations here
https://www.physicsforums.com/threa...ductory-physics-formulary.110015/#post-905663

 

[Delta2]

Ehm @kuruman I did it as you said and I got the same value. It is $$(0.6x10^8)^2=2a1010^{-3}\Rightarrow 0.36\times 10^{16}=2a10^{-2}\Rightarrow a=0.18\times 10^{18}$$ which is the same as $1.8\times 10^{17}$.

 

[Delta2]

Ah i see you referring to the old post of his, check post #48 for the new values.

 

[kuruman]

Yes, I was referring to the older post #44 and missed the updated value in #48. Sorry about the confusion.

 

[meher4real]

Thank you all