Electrical Potential Questions

[seamstander]

Homework Statement

The questions is as follows:

Two isolated infinite parallel conducting sheets are .05 m apart and are initially uncharged. Electrons are then removed from one sheet and placed on the other sheet resulting in a voltage difference of 200V between the sheets. A proton is released from rest halfway between the sheets. Find the speed of the proton just before it hits one of the sheets.

Homework Equations

V being voltage and d being distance between the sheets

Using Coloumb's law to swap out F for E.

Using KE to determine that max velocity.

The Attempt at a Solution

Once I get to the last equation I simple plug in the charge of a proton for q and the mass of a proton for m, along with the given 200V different for V. Is this the correct assumption for this problem?

When I plug it all in it comes to 1.95x10^5 m/s

 

[Delphi51]

Welcome to PF, Seamstander.
Why not do it the high school way:
v² = 2a(d/2) = Fd/m = qEd/m = qV/m
This gives a smaller answer. A factor of 2 smaller in the v².

 

[seamstander]

Thank you for your reply Delphi.

Often, my problem when learning new physics material is that I make strange substitutions that may or may not hold up to criticism. I don't mean to ask too much, but can you see where my substitutions failed?

 

[Delphi51]

I had trouble understanding your first line. I would have said the integral of E*dr is E*d.

In the second line it appears you are using a = F/m, for which I would get a = q*E/m.

In the third line, I think there should be a 1/2 somewhere because the particle only accelerates through half of the potential difference.

The hardest thing for me to understand in your work is how the 3 lines relate to each other - I so no relationship whatsoever! Of course, I'm just an old high school teacher.

 

[seamstander]

Delphi, thank you for your analysis. And I think you should give yourself more credit in your Physics ability. I don't know you, but the fact that you're helping me is something I am grateful for.

You are right, my formulas are quite incoherent. I was trying to take it from integration, which is something I think I should leave until I have a firmer grasp on the subject.

I should start with KE = qV = .5 * m * v^2

(from the equation KE = (1/2)mv^2) I'm not sure where the 2 is getting left in.

 

[Delphi51]

Yes, KE = qV = .5 * m * v^2 is a nice starting point all right.
But the V should be half the 200 Volts because the proton is starting in the middle of the electric field between the plates, not at the top or bottom. It only moves through half the 200 V potential difference.

 

[seamstander]

Ah, I do believe I see what you're saying now Delphi. Thank you so much for your time.

If the electron would've been released from some unspecified location would we be able to say that the max KE would be (1/2)mv^2 but since we know that it was released half way through we can say that it will only be half? (This may seem obvious but I just want to make sure I know where the flaw in my thinking was.)

 

[SammyS]

Ah, I do believe I see what you're saying now Delphi. Thank you so much for your time.

If the electron would've been released from some unspecified location would we be able to say that the max KE would be (1/2)mv^2 but since we know that it was released half way through we can say that it will only be half? (This may seem obvious but I just want to make sure I know where the flaw in my thinking was.)

It's perhaps more proper to say the proton is accelerated through an electric potential difference of 100V, since it's midway between the plates.

Therefore, [tex]e(\Delta V)=(1/2)m_pv^2,[/tex] where ΔV = 100 V.