Question

Part A

An electron is moving east in a uniform electric field of 1.52 N/C directed to the west. At point A, the velocity of the electron is 4.54×10⁵ m/s pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.360 m east of point A?

Part B

A proton is moving in the uniform electric field of part A. At point A, the velocity of the proton is 1.89×10⁴ m/s , again pointed towards the east. What is the speed of the proton at point B?

Answer 1

An electron is moving east in a uniform electric field of magnitude 1.52 N/C directed to the west.
Force on electron will be directed to east and acceleration of electron will be positive
Initial speed = u = 4.54×10^5 m/s
as
Acceleration =
Ae = F/m = qE/m
Acceleration will be

Ae = (1.6*10^-19*1.52) /(9.11*10^-31)

Acceleration of electron is = Ae = 2.66*10^11 m/s^2

as

distance = s = 0.360 m
Thereby using Third eqn of motion we get
Vb^2 - u^2 = 2as

Vb = sq rt (u^2 +2* a* s)

the speed of the electron when it reaches point Vb = sq rt (u^2 +2as)

Vb = sq rt ((4.54*10^5)^2 + 2* 2.66*10^11* 0.360 )

Vb = 6.30*10^5 m/s

The speed of the electron when it reaches point B is 6.30*10^5 m/s

PART B:

A proton is moving in the uniform electric field of part A. At point A, the velocity of the proton is 1.89×104 m/s , again pointed towards the east. What is the speed of the proton at point B?

Initial velocity of proton = u =1.89×10^4 m/s
The proton is retarted because force of field is west and initial velocity is east.

the speed of the proton at point B
Vb = sq rt ( u^2 +2as )
now,

acceleration of proton =
Ap = -F/m
Ap =-qE/m
Ap =-(1.6*10^-19 * 1.54) / (1.67*10^-27 )

acceleration of proton is Ap = -1.47*10^8 m/s^2
Therefore Speed of the proton at point B

Vb = sqrt ( ( 1.89 * 10^4 )^2 -(2* 1.47 * 10^8 * 0.395) )

the speed of the proton at point B = Vb = 1.53 * 10^4 m/s