# Charged parallel plates, proton and electron released simultaneously

• dewert
In summary, the problem involves two parallel plates with equal and opposite charges, an electron and a proton being released simultaneously from rest on the surfaces of the plates. Using the equations for energy and acceleration, the distance between the negative plate and the point at which the electron and proton pass each other can be calculated. The solution involves finding the acceleration of each particle and using it to find their respective distances traveled.
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
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?

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...

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!

## 1. What are charged parallel plates?

Charged parallel plates refer to two flat plates that are placed parallel to each other and carry equal and opposite charges. These plates create an electric field between them.

## 2. What is the significance of releasing a proton and electron simultaneously between charged parallel plates?

Releasing a proton and electron simultaneously between charged parallel plates allows for the study of the effects of an electric field on these particles. It also helps in understanding the principles of electrostatics and the behavior of charged particles in an electric field.

## 3. How do the charges on the parallel plates affect the motion of the proton and electron?

The charges on the parallel plates create an electric field, which exerts a force on the proton and electron. The direction and magnitude of this force depend on the charges and separation distance of the plates.

## 4. What happens to the proton and electron as they move between the charged parallel plates?

As the proton and electron move between the charged parallel plates, they experience a force due to the electric field. This force causes them to accelerate and change their direction of motion. The magnitude of their acceleration depends on the strength of the electric field and the mass of the particles.

## 5. How does the motion of the proton and electron between the charged parallel plates demonstrate the principle of conservation of energy?

The motion of the proton and electron between the charged parallel plates demonstrates the principle of conservation of energy because the electric field does work on these particles, converting their potential energy into kinetic energy. This means that the total energy of the particles remains constant throughout their motion in the electric field.

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