physics question any help please???

Question

The following diagram shows a proton entering the region between two oppositely charged plates. The plates are 35.0 mm long and are separated by 13.5 mm.  A potential difference of 0.408V generates an electric field between the plates.  The proton is given an initial velocity of 1.25 x 104 m/s, in the positive x direction. Use this information to determine if the proton will make it through the plates and exit the other side.

Answer 1

Potential Difference between the plates V = 0.408

Distance between the plates d = 13.5mm = 13.5 * 10^-3 m

 

Initial Velocity vi = 1.25*10^4 m/sec

Given that Initial Velocity = Distance/ time

time = Distance / Velocity

t = 13.5 * 10^-3 / 1.25*10^4

t = 1.08 * 10^-6 sec

Time taken for proton to travel a distance of 13.5mm is 1.08 * 10^-6 sec.

 

We know that Charge of the proton q = 1.602 * 10^-19

Mass of the Proton m = 1.6762 * 10^-27 kg

Let us draw Free body diagram of the proton is

Force acting on the Proton is given by

F = Fe - mg

Where Fe is the Electric Field = qE

E is the Electric Field = Potential Difference V/ distance between the plates d.

Fe = qV/d = (1.602 * 10^-19 * 0.408) / (13.5 * 10^-3 m) = 4.8416 * 10^-18

mg is the Gravitational Force acting on the proton

 

From the Newton's 2nd law, F = ma

ma = Fe - mg

a = Fe/m - g

a = (4.8416 * 10^-18)/(1.6762 * 10^-27 kg) - 9.8

a = 2.88 * 10^9 m/sec²

 

We know that Acceleration a = V/t

vf/t = 2.88 * 10^9 m/sec²

vf = 2.88 * 10^9 * t

vf = 2.88 * 10^9 * 1.08 * 10^-6

vf = 3.1104 * 10^3 m/sec

 

Resultant Velocity v = √(v_f² + v_i²) = √(9.674 * 10^6 + 1.56*10^8 ) = 1.288 * 10^4 m/sec

Therefore the Resultant Velocity is 1.288 * 10^4 m/sec.