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**1. The plates of a parallel plate capacitor are separated by a distance (4cm). A point charge of magnitude (q=2e) (3.2*10^-19 C) is fired horizontally through a small hole in the positive plate of the capacitor. The charge is fired with speed (8.5 *10^5 m/s) and, upon reaching point M (located @ 2cm between each plate), is observed to have speed (4*10^5 m/s). The point M is located from the negative plate. Assume the charge has mass 2proton (3.34 *10^-27 kg).**

**2. Will the charge reach the negative plate? If so, what will its speed be when it reaches**

the negative plate? If not, what is its closest point of approach to the negative plate?

the negative plate? If not, what is its closest point of approach to the negative plate?

**3. the answer is: The charge will be momentarily brought to rest at a point ~1.33 cm**

from the right hand plate.

I do not understand how they came up with the above answer.

When I calculate the Vf using the Vi as the Velocity at point M I get a very large velocity (not zero). What equation should I use to get their answer?

I currently use:

Vf= sqrt (2q deltaV)/(m)

from the right hand plate.

I do not understand how they came up with the above answer.

When I calculate the Vf using the Vi as the Velocity at point M I get a very large velocity (not zero). What equation should I use to get their answer?

I currently use:

Vf= sqrt (2q deltaV)/(m)