- #1

gsquare567

- 15

- 0

## 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

q

_{p}= 1.6 * 10

^{-19}C

q

_{e}= -1.6 * 10

^{-19}C

m

_{p}= 1.67 * 10

^{-27}kg

m

_{e}= 1/1830 * m

_{p}

## Homework Equations

[1] x = x

_{0}+ v

_{0}+ 1/2*a*t

^{2}

[2] F = ma = qE

## The Attempt at a Solution

PROTON:

using [2], a = q

_{p}*E/m

_{p}

subbing into [1], x

_{p}= 1/2(q

_{p}*E/m

_{p})t

^{2}

ELECTRON:

x

_{e}= r + 1/2(q

_{e}*E/m

_{e})t

^{2}

see when x

_{p}= x

_{e}

t = [tex]\sqrt{2*r/(E(q

_{p}/m

_{p}- q

_{e}/m

_{e}}[/tex]

i get t = 2.67 * 10

^{-8}seconds

but when i plug it in for x

_{p}i get 2.18*10

^{-5}m, whereas x

_{e}is 4.00 * 10

^{-2}m

they should be equal =S

where did i go wrong? thanks!