The electric field strength is 4.60×104 v/m inside a parallel-plate capacitor with a 2.1 mm spacing. a proton is released from rest at the positive plate. what is the proton's speed when it reaches the negative plate? give answer in m/s and show all work!

Explanation:

It is given that,

Electric field strength inside a parallel plate capacitor,

Spacing between plates, d = 2.1 mm = 0.0021 m

A proton is released from rest at the positive plate, u = 0

We need to find the speed of proton when it reaches the negative plate of the capacitor. Let it is given by v.

Using third equation of motion as :

Where,

q is the charge on proton

m is the mass of the proton

v = 136052.12 m/s

or

So, the speed of proton when it reaches the negative plate is . Hence, this is the required solution.

the answer is b

explanation: