Magnitude of electric field?

Question

A proton initially moves left to right long the x axis at a speed of 2.00 x10^3 m/s. It moves into an electric field, which points in the negative x direction, and travels a distance of 0.200 m before coming to rest. If the proton's mass and charge are 1.67 x 10^-27 kg and 1.60 x 10^-19 C respectively, what is the magnitude of the electric field?

a. 13.9 N/C

b. 0.017 N/C

c. 0.104 N/C

d. 28.3 N/C

e. 0.038 N/C

Thank you for your help!!

Answer 1

The acceleration of the proton is a = 0.5*V^2/s,  The force on the proton to cause this is F = m*a.  This comes from the electric field F = E*q;  E = F/q, E = 0.5*m*V^2 / (q*s)

I get c) 0.104 N/C