Electron/proton released from charged parallel plates

[PolarBee]

Homework Statement

Two parallel plates 1.40 cm apart are equally and oppositely charged. An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate. How far from the negative plate is the point at which the electron and proton pass each other?

Known:
Q_e = -1.6 x 10^-19 C
Q_p = 1.6 x 10^-19 C
m_e = 9.11 x 10^-31 kg
m_p = 1.6 x 10^-27 kg

Homework Equations

a = qE/m
d = 1/2*a*t²[/B]

The Attempt at a Solution

[/B]
Acceleration of electron
a_e = q_eE / m_e

Acceleration of proton
a_p = q_pE / m_p

Distance of electron
d_e = a_et² / 2

Distance of proton
d_p = a_pt² / 2

Since I don't know time, I tried to eliminate it by
t² = 2d_e/a_e
and plugging it into distance formula but from there, I'm stuck on what to do.

 

[TSny]

Hello, and welcome to PF!

You are given the distance D between the plates. Can you make use of this information?

 

[PolarBee]

I'm not really sure how distance D relates to the problem, but my guess would be if electron is traveling from distance 0 to 1.4 cm and the electron is traveling from 1.4 cm to 0, if looking at it on an x-axis... then the distance the electron travels is S, while the distance the proton travel is 1.4 cm - S.

Rewriting the equations would be...

S = 1/2 * a_e * t²

and

0.014 - S = 1/2 a_p * t²

How would I get the acceleration without knowing the electric field E?

 

[TSny]

I'm not really sure how distance D relates to the problem, but my guess would be if electron is traveling from distance 0 to 1.4 cm and the electron is traveling from 1.4 cm to 0, if looking at it on an x-axis... then the distance the electron travels is S, while the distance the proton travel is 1.4 cm - S.

Rewriting the equations would be...

S = 1/2 * a_e * t²

and

0.014 - S = 1/2 a_p * t²

Good.

How would I get the acceleration without knowing the electric field E?

Maybe you won't need actual values for the accelerations. Just keep working with symbols. What can you learn about the time from your equations above? For now, use D for the distance between the plates rather than use the number .014 m.

 

[PolarBee]

Ah I see.
I know that t would be the same for both because that's when they meet with each other.

Would I solve for t and make them equal to each and solve for S?

t² = 2S / a_e and t² = 2(D-S) / a_p

And solving for S would be...

S = D * a_e / a_p + a_e

Unless my my algebra is wrong.

 

[TSny]

S = D * a_e / a_p + a_e

Did you mean S = D * a_e / (a_p + a_e) so that both a_p and a_e are in the denominator?

Can you write your result in terms of the ratio of a_e to a_p?

 

[PolarBee]

Yes they are both in the denominator.

Using the formula a = qE/m , I know that E should be the same for both the proton and electron, so I can set them equal.
I'm not sure if this is what you were asking but I did
(a_e * m_e ) / q_e = (a_p * m_p ) / q_p

Then solving for a_e I got
a_e = a_p * 1756.31

 

[TSny]

Yes they are both in the denominator.

Using the formula a = qE/m , I know that E should be the same for both the proton and electron, so I can set them equal.
I'm not sure if this is what you were asking but I did
(a_e * m_e ) / q_e = (a_p * m_p ) / q_p

Then solving for a_e I got
a_e = a_p * 1756.31

OK. What do you get when you use this information in your result for S?

 

[PolarBee]

Ah! I see.

I plugging in everything I know, the accelerations got canceled resulting in S = 0.01399 m. This makes sense because electrons move faster than protons.

 

[TSny]

Looks good.