# Charged parallel plates, proton and electron released simultaneously

1. Feb 13, 2008

### dewert

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

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

U = qEs
a = qE/m
*Taking q to mean +1.60E-19 ie. charge of proton

3. 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?

2. Feb 13, 2008

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.......

3. Feb 13, 2008