An electron with a speed `5.00xx10^6` m/s enters an electric field of magnitude `10^3` N/C, travelling along the field lies in the direction that retards its motion.
a. How far will the electron travel in the field before stopping momentarily?
b. How much time will have elapsed?
c. If the regin with the electric field is only 8.00 mm long (too short from the electronn to stop with in it), what fraction of the elctron's initial kinetic energy wil be lost in that region.

Answer

A, B, C, D

Explanation

a. `s=u^2/(2a)=u^2/(2(qE//m))=(mu^2)/(2qE)`
`=((9.1xx10^-31)(5xx10^6)^2)/(2xx1.6xx10^-19xx10^3)`
`=7.1xx10^-2m=7.1cm`
b. `t=u/a=u/(qE//m)=(mu)/(qE)`
`((9.1xx10^-31)(5xx10^6))/((1.6xx10^-19)(10^3))`
`=2.84xx10^-8s`
c Loss of energy (in fraction)
`(1/2"mu"^2-1/2"mu"^2)/(1/2"mu"^2)=1-u^2/u^2`
`=1-(u^2-2as)/u^2=(2as)/u^2=(2qEs)/("mu")^2
`=(2xx1.6xx10^-19xx10^3xx8xx10^-3)/(9.xx10^-31xx(5xx10^6)^2)`
`=0.11`