Question

A particle (mass = 6.7 × 10^−27 kg, charge = 3.2 × 10^−19 C) moves along the positive x axis with a speed of 4.8 × 10^5 m/s. It enters a region of uniform electric field parallel to its motion and comes to rest after moving 2.0 m into the field. What is the magnitude of the electric field?

Please explain and give formulas. I can do the math/work.

Answer 1

Kinetic energy stored in the particle = work done against the particle by the field
$$\frac{1}{2} mv^2=F_ed=(eE)d\implies E=\frac{mv^2}{ed} $$

Answer 2

2ed....

Answer 3

E=mv^2 / 2ed

Answer 4

yeah, sorry for the typo

Answer 5

Thank you very much!