# Charged Capacitor with Electron and Proton

1. Sep 16, 2008

### SundayST

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

The plate separation of a charged capacitor is 0.0800 m. A proton and an electron are released from rest at the midpoint between the plates. Ignore the attraction between the two particles, and determine how far the proton has traveled by the time the electron strikes the positive plate.

2. Relevant equations

ending velocity of proton = - ((mass of electron)(ending velocity of electron)/mass of proton)

(I got to the equation using conservation of energy and assuming starting velocity equalled zero for both, so two pieces of the conservation of energy equation cancelled to begin with.)

3. The attempt at a solution

My attempt is really rather scatter-shot. I tried several ways of coming at the problem, but I can't find anything that gets me past a point where I'm stumped or need more information or a formula I just don't know to apply.

Ve= 0 m/s, and Vp= 0 m/s. (Particles start at rest.)

Force on the two particles are equal, since they have an equal magnitude of charge.

I set:
ap = Proton's acceleration
ae = Electron's accelertion
Fp = Force of the proton
mp = mass of the proton
Fe = Force on the elctron
me = mass of the electron

Fp=mp*ap
Fe=me*ae

Forces are equal, so:

mp*ap = me*ae

tp = time it takes the proton to travel half the distance
te = time it takes the electron to travel half the distance

te = sqrt( (9.1x10^-31kg)(0.0400m)/Fe )
(Substituted in mass of electron and half the distance in the problem.)

After that, there doesn't seem to be a point to solving any further, as if I substitute in any of the above formulas and solve for force, I'm back where I started.

I am completely out of ideas on this.

If you can answer this, can you answer it as to a novice who has not had anything except high school physics, has had calculus (but not since over 10 years ago) and now needs to understand this problem enough to help someone else understand it on their homework. (I told him I'd try to help figure it out. But it doesn't do either of us any good if all I can do is copy the steps and an answer.)

Please explain as if explaining to someone with next to NO idea what they are doing.

Thanks!

2. Sep 16, 2008

### edziura

Assuming that the electric field between the plates is uniform, the force acting on each particle has the same magnitude (they are of equal charge magnitude) and is constant, so each has a constant acceleration.

Writing the kinematics equation (constant a): d= 0.5 a t^2 for each and F = m a for each, and combining simplifying

dp = $$\frac{m_{e}}{m_{p}}$$ de

3. Sep 16, 2008

### SundayST

Thank you SO much!

Once I applied the kinematics, it worked out. (At least, I think it's right. Will find out after it's graded. If I did the math right, it should be right.)

I had not even thought of using that. But I understand it well enough to explain how to do it now.