Question

a uniform field is oriented toward the right and has a magnitude 54_N/C in which direction will the electron move once it is released at rest? Find the speed of the electron after it has moved .24m?

Answer 1

direction of motion of electron must be opposite to the direction of field, so to the left

acceleration acts on electron

a = eE / m = 1.6* 10^-19* 54 / ( 9.11* 10^-31) = 9.484* 10^12 m/s^2

now using 3rd equation of motion

v^2 = 2 a d = 2* 9.484* 10^12* 0.24

v = 2.133* 10^6 m/s

=======

comment in case any doubt, will reply for sure.. goodluck