An electron moving to the right at 4.0% the speed of light enters a uniform electric field parallel to its direction of motion. If the electron is to be brought to rest in the space of 5.0 cm, determine what the strength of the field is.

Answer 1

At only 4% of the speed of light, you can neglect relativity effects and assume the electron kinetic energy starts out as
(1/2)mV^2 = (1/1250)m c^2
where I used V = c/25.
m is the electron mass.

To stop the electron in a distance X = 0.05 m,

E*X = (1/1250)mc^2

Solve for E, the required field strength.