Question

Two parallel plates 0.800 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?

Express your answer with the appropriate units.

Answer 1

Answer:

0.79 cm

Explanation:

The computation is shown below:-

Particle acceleration is

$a = \frac{qE}{m}$

We will take d which indicates distance as from the negative plate, so the travel by proton is 0.800 cm - d at the same time

$d = \frac{1}{2} a_et^2\\\\0.800 cm - d = \frac{1}{2} a_pt^2\\\\\frac{d}{0.800 cm - d} = \frac{a_e}{a_p} \\\\\frac{d}{0.800 cm - d} = \frac{m_p}{m_e} \\\\\frac{d}{0.800 cm - d} = \frac{1836m_e}{m_e}$

After solving the equation we will get 0.79 cm from the negative plate.

Therefore it is 0.79 cm far from the negative pate i.e the point at which the electron and proton pass each other

Answer 2

The point at which the electron and proton pass each other will be 0.79 cm.

What is the charge?

When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.

The given data in the problem is;

d' is the distance between the two parallel plates= 0.800 cm

The acceleration is given as;

$\rm a= \frac{qE}{m}$

The distance from Newton's law is found as;

$d = ut+\frac{1}{2} at^2 \\\\ u=0 \\\\ d= \frac{1}{2} at^2 \\\\ d-d' = \frac{1}{2} a_pt^2 \\\\ 0.800-d= \frac{1}{2} a_pt^2 \\\\\ \frac{d}{0.800-d} =\frac{a}{a_p} \\\\ \frac{d}{0.800-d} =\frac{m_p}{m} \\\\ \frac{d}{0.800-d} =\frac{1836m_e}{m_e} \\\\ d=0.79 \ cm$

Hence the point at which the electron and proton pass each other will be 0.79 cm.

To learn more about the charge refer to the link;

https://brainly.com/question/24391667