Electric field between two parallel plates

[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