Charged parallel plates, proton and electron released simultaneously

[dewert]

[SOLVED] Charged parallel plates, proton and electron released simultaneously

Homework Statement

Two parallel plates 0.700 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?

Homework Equations

U = qEs
v^2 = (v_0)^2 + 2ad
a = qE/m
*Taking q to mean +1.60E-19 ie. charge of proton

The Attempt at a Solution

Attempt to use energy for entire system:

Define s-axis as running from negative plate on left to positive plate on right.

U_i = U_f + K_f

qE*s_p_i (for proton) - qE*s_e_i (for electron, =0) = -qEs_e + qEs_p + (1/2)m_p*(v_p)^2 + (1/2)m_e*(v_e)^2

Subbed in qE/m for a in v^2 = 2ad (v_0 = 0) to get (1/2)(2*qE/m_p*(7E-3-s_e)) for proton, (1/2)(2*qE/m_e*(s_e)) for electron.

Solved all of this for s (E's and q's cancelling all over the place) and got something... wrong. I was trying a lot of other stuff before, involving x = x_0 + v_0*t + 1/2*at^2, but it ended up in tautologies... sigh. Can anyone help, by at least telling me if I'm on the right track at all?

 

[Google_Spider]

I would solve it like this:

Find acceleration of electron (a_{e})


a_{e}=\frac{eE}{m_{e}}




Find acceleration of proton (a_{p})




Apply S_{e}=\frac{a_{e}t^2}{2}......(1)



and S_{p}=\frac{a_{p}t^2}{2}.......(2)


Eliminate 't' from equation 1 and 2 to get eq(3).


S_{e}+S_{p}=0.007....(4)



solve (3) and (4) to get S_{e}.


_______________________________
I may be wrong...

 

[dewert]

Thanks google_spider!

Ah... lovely. And so easy! I swear I would have thought of it eventually :P

It makes sense, because the proton is several orders of magnitude more massive than the electron. No way it would move nearly as fast. Should've thought of that one right off the bat. Good to have mathematical proof though.

Thanks!