Question

A electron moves a distance of 6 cm when accelerated from rest by an electric field of strength
2 x 10-4NC-1 Calculate the time of travel. The mass and charge of electron are 9 x 10-31kg and
1.6 x 10-19C respectively.​

Answer 1

Explanation:

Given E=2×10^-4

q=1.6×10^-19

F=Eq

F=3.2×10^-23

m=9×10^-31

a=F/m

=3.2×10^-23/9×10^-31=0.355×10^8=35.5×10^7

v²=u²+2as

initial velocity is u=0

v²=2(35.5×10^7)0.06

v=6.52×10^7

a=v/t

t=v/a=6.52×10^7/35.5×10^7=0.18sec