# Uniform electric field moving a proton and electron

## Homework Statement

A uniform electric field of magnitude 640 N/C exists between two parallel plates that are 4.00 cm apart. A proton is released from the positive plate at the same instand that an electron is released from the negative plate. Determine the distance from the positive plate that the two pass each other. (Ignore the electrostatic attraction between the proton and electron.)

r = 0.04 m
E = 640 N/C
qp = 1.6 * 10-19 C
qe = -1.6 * 10-19 C
mp = 1.67 * 10-27 kg
me = 1/1830 * mp

## Homework Equations

 x = x0 + v0 + 1/2*a*t2
 F = ma = qE

## The Attempt at a Solution

PROTON:
using , a = qp*E/mp
subbing into , xp = 1/2(qp*E/mp)t2

ELECTRON:
xe = r + 1/2(qe*E/me)t2

see when xp = xe
t = $$\sqrt{2*r/(E(qp/mp - qe/me}$$
i get t = 2.67 * 10-8 seconds

but when i plug it in for xp i get 2.18*10-5 m, whereas xe is 4.00 * 10-2 m

they should be equal =S

where did i go wrong? thanks! ## Answers and Replies

Should they be equal? Does equal force imply equal acceleration in this case?

I thought they would be different because a proton has 1830 times the mass of an electron, and as F = m * a... Am I wrong?

No, no, you're right in that. They won't meet in the middle, though, because they've been accelerated unequally. You want not xp = xe, but xp + xe = 4.

Oh, I understand what you mean now. However, they are released from opposite ends, so I think when they cross, their x values should be equal. That is why my x equation for the electron starts at r = 4.

If one particle starts at 0, and the other at r = 4, then one will accelerate in the positive r direction, and the other will accelerate in the negative x direction.