Part a) A n electron is to be accelerated from a velocity of 4.50×10 6 m / s to
ID: 1491381 • Letter: P
Question
Part a) An electron is to be accelerated from a velocity of 4.50×106m/s to a velocity of 9.50×106m/s . Through what potential difference must the electron pass to accomplish this? The answer for that is V1-V2= -199 VPart b) Through what potential difference must the electron pass if it is to be slowed from 9.50×106 m/s to a halt? V1-V2=? Part a) An electron is to be accelerated from a velocity of 4.50×106m/s to a velocity of 9.50×106m/s . Through what potential difference must the electron pass to accomplish this? The answer for that is V1-V2= -199 V
Part b) Through what potential difference must the electron pass if it is to be slowed from 9.50×106 m/s to a halt? V1-V2=?
Part b) Through what potential difference must the electron pass if it is to be slowed from 9.50×106 m/s to a halt? V1-V2=?
Explanation / Answer
a) V*q = 0.5*m*(vf^2-vi^2) ====> V = -199.28 V (due to -ve charge)
b) Same equation, vf = 0, vi = 9.5*10^6 m/s
V*q = 0.5*m*(vf^2-vi^2) ====> V = +256.93 V