Question

An electron is released from rest in a weak electric field given by Ể--1.60 x 10-10 N/Cj. After the electron has traveled a vertical distance of 1.2 μm, what is its speed? (Do not neglect the gravitational force on the electron.) 7.65 mm/s eBook

Answer 1

answer) we know the formula for acceleration of the electron is given by

a=qE/m=-1.6*10^-19*(-1.6*10^-10)jm/s²)/9.1*10^-31=28.1m/s² j

and the gravitational acceleration=-9.8m/s² ( downwards)

so total acceleration=28.1-9.8=18.3m/s²

now we know the formula

v²=u²+2as=0+2*18.3*1.2*10^-6m=43.92*10^-6m/s

v=$\sqrt{}$43.92*10^-6=6.63*10^-3m/s=6.63mm/s

so the answer is 6.63mm/s or 6.6mm/s or 7mm/s