The electric field between two parallel plates is uniform, with magnitude 646 N/C. A proton is held stationary at the positive plate, and an electron is held stationary at the negative plate. The plate separation is 4.26 cm. At the same moment, both particles are released. Required:
Determine the distance from the positive plate at which the two pass each other.

The distance from the positive plate at which the two pass each other is 0.0023 cm.

Explanation:

We need to find the acceleration of each particle first. Let's use the electric force equation.

For the proton

For the electron

Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.

Both of them have an initial speed equal to zero. So we have:

With this time we can find the distance from the positive plate (x(p)).

Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.

I hope it helps you!

the answer would be a

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!