Proton release inside a parallel capacitor Find velocity

AI Thread Summary
The discussion revolves around calculating the speed of a proton released from rest in a parallel-plate capacitor with a given electric field strength and plate spacing. The key equation mentioned is ΔU = -0.5eEd, which relates to the change in potential energy. Participants express confusion about the derivation of this equation and its application to the problem. Clarification is sought regarding the use of the factor of 0.5 in the equation and the overall approach to solving for the proton's final speed. The conversation highlights the challenges of applying theoretical concepts to practical problems in college-level physics.
conov3
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Homework Statement



The electric field strength is 47100 N/C inside a parallel-plate capacitor with 8.60 mm spacing. A proton is released from rest at the positive plate. What is the speed of the proton (in m/s) when it reaches the negative plate?

Homework Equations


Vf=√(-2U/m) I think?
ΔU=-.5eEd

The Attempt at a Solution



I tried using that equation to find ΔU=-3.768*10^-15
then plugged it in seem to have the wrong final answer though
 
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welcome to pf!

hi conov3! welcome to pf! :wink:
conov3 said:
ΔU=-.5eEd

why .5 ?
 
Thank you! I figured since I am in a college level physics, I might as well try to get some help if I need it! ha

Using my Physics book.. it says ΔU=Uf-Ui= (Uo+ (-e)Ed)-(Uo+(-e)E(d/2)
Then under that it says =-1/2eEd

I am confused about it also and am really unsure.
 
conov3 said:
Using my Physics book.. it says ΔU=Uf-Ui= (Uo+ (-e)Ed)-(Uo+(-e)E(d/2)

i don't understand that … it doesn't seem to fit the question :confused:
 

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