Determining speed of electron in a parallel plate capacitor

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1. Dec 2, 2017

mitchy16

1. The problem statement, all variables and given/known data
A proton is released from rest at the positive plate of a parallel plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 50,000 m/s. What will be the final speed of an electron released from rest at the negative plate?

2. Relevant equations
KE= 1/2 mv^2
U= qV

3. The attempt at a solution
I tried using conservation of energy, the left side is the proton and right side is the electron:
KE1 + U1 = KE2 + U2
1/2mv^2 + qV = 1/2mv^2 + qV
I do not understand what to plug in for V? I have tried searching for similar threads but I can't find a similar one. Any guidance would be appreciated, thank you.

2. Dec 2, 2017

rude man

Your equation works for ANY value of V! Tell you anything?

3. Dec 2, 2017

Delta²

First your equation has a small (or is it big, cant tell if it is a typo(or not familiar with $TeX$) or you misunderstood the whole thing) error, in the left hand side in the KE term you should have put the mass of proton $m_p$ while in the respective term in the right hand side the mass of electron $m_e$. Also the same holds for the velocities in the KE terms.

Second and most important, though your equation might be true (at the ...very end of the day) I don't think you got it by applying conservation of energy in a proper manner.

You have to apply conservation of energy for the proton first, at initial position in the positive plate and at final position in the negative plate and get one equation (1).
Then apply it again for the electron with initial position at the negative plate and final position at the positive plate and get a second equation (2).

By combining equation (1) and (2) you ll get the equation you write at your initial post.

so your final equation should be

$\frac{1}{2}m_pv_p^2=\frac{1}{2}m_ev_e^2$

(we can omit the qV terms since $q_p=-q_e$ and because the particles start in vice versa positions regarding the plates of capacitor).

You have been given $v_p=50km/s$. Have you been given the values for the masses or the ratio of masses $\frac{m_p}{m_e}$?

Last edited: Dec 2, 2017
4. Dec 2, 2017

mitchy16

Yes! The mass is given in a formula sheet. I understand my mistake, thank you! I re-did the question and got 2.1 x 106 which is the correct answer.