Course Home Chapter 21-Electric Charge and Electric Field Exercise 21.25-Enhance
ID: 1878095 • Letter: C
Question
Course Home Chapter 21-Electric Charge and Electric Field Exercise 21.25-Enhanced - with Solution 12 of 17 > Constants Part A A proton is traveling horizontally to the right at Find the magnitude of the weakest 4.10 106 m/s You may want to review (Pages 695- 699) For related problem-solving tips and strategies, you electric field that can bring the proton uniformly to rest over a distance of 3.20 cm may want to view a Video Tutor Solution of Electron E3.30 10 Submit Previous Answers Request Answer O Type here to searchExplanation / Answer
A)
Use law of conservation of energy
Initial kinetic energy of proton = final potential energy of proton in electric field
KEi = Uf
½*mv2= q*V
½*mv2= q*E*x
E = ½*mv2*q*x
Plugging values,
E = [½*1.67*10^-27*(4.10*10^6)^2]/(1.6*10^-19*0.0320)
E = -2.74*10^6 N/C ………..directed opposite to the initial velocity
C)
For on proton = F= q*E
Acceleration of proton = a= (q*E)/m
Plugging values,
a = (1.6*10^-19*2.74*10^6)/(1.67*10^-27)
a = -2.62*10^14 m/s^2 ……….directed opposite to the initial velocity
No use kinematic eqn,
vf=vi+at
plugging values,
0 = 4.10*10^6 - 2.62*10^14*t
t = 1.56*10^-8 s
D)
m=9.1*10^-31 kg
E = ½*mv2*q*x
Plugging values,
E = [½*9.1*10^-31*(4.10*10^6)^2]/(1.6*10^-19*0.0320)
E = 1493.866 N/C ………..directed in the direction of the initial velocity
E)
Since E is directed in the direction of the initial velocity
= angle between E and vi = 0o
= 0o