Electric potential and potential energy

[dpogre]

A proton is initially at rest (in a vacuum) in a uniform electric field of magnitude 1.30 10^3 N/C.
(a) Calculate the force applied to the proton by the field. ANSWER: 2.08E-16 N

(b) Apply Newton's second law to calculate the acceleration of the proton. (Don't be surprised if the acceleration is large. It isn't applied for a very long time.) ANSWER: 1.24E11 m/s^2

(c) Calculate the time it takes for the proton to move 1 cm in this field, starting from rest. (Hint: The acceleration is uniform.)
in part c i cannot figure out how to find the uniform acceleration.

the equation given to find the time it takes for the proton to move 1cm in this field is
x = 1/2at^2 and I'm solving for t.
0.01 m = 1/2 (Vf - Vi) t^2
0.01 m = 1/2 (1.30x10^3 - 0) t^2
but the answer i got was wrong. what am i doing wrong to find the time it takes for the proton to move 1 cm?

 

[hage567]

(c) Calculate the time it takes for the proton to move 1 cm in this field, starting from rest. (Hint: The acceleration is uniform.)
in part c i cannot figure out how to find the uniform acceleration.

I would think you just use the acceleration you found in (b).

the equation given to find the time it takes for the proton to move 1cm in this field is
x = 1/2at^2 and I'm solving for t.

Yes, this equation will work. Now you know what the acceleration is and don't need to be substituting for it in your calculations. Just solve.

0.01 m = 1/2 (Vf - Vi) t^2
0.01 m = 1/2 (1.30x10^3 - 0) t^2

Why did you but 1.30x10^3 in for Vf? That number is the strength of the electric field, it has nothing to do with velocity. Also Remember that a = Δv/Δt

 

[dpogre]

thank you