Question

a.) An electron is moving east in a uniform electric field of 1.50 N/C directed to the west. At point A, the velocity of the electron is 4.48×105 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? b.) A proton is moving in the uniform electric field of part A. At point A, the velocity of the proton is 1.92×104 m/s, again pointed towards the east. What is the speed of the proton at point B?

Answer 1

Answer:
2.007 * 10^11 m/s, 19100.52m/s

Step-by-step explanation:

Given data:

Electric field = 1.50 N/C

velocity of electron is 4.48 * 10^5 m/s

distance of point B from point A is 0.360 m

we know that acceleration of particle is given as

a) for electron: a = qE/m
= 1.6 * 10^-19 *1.50/9.1 * 10-31
= 4 * 10^-21
from equation of motion we have

v^2 = u^2 +2as
= (4.48 * 10^5)^2 + 2(4 * 10^-21 * 0.360)
≈ 2.007 * 10^11 m/s
Please verify my calculations lol

d=V/E=1.91×10^4/1.48

d=12905.4m

At point B

Distance=12905.4+0.360=12905.770

V=Ed=12905.760*1.48=19100.52m/s