How Does a Proton's Velocity Change in a Uniform Electric Field?

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A proton enters a uniform electric field of 10 N/C traveling at 10^5 m/s and covers a distance of 4 cm. The key question is how the proton's velocity changes due to the electric field. The work done on the proton can be calculated using the formula W = q*ΔV, where the distance and electric field strength are used to determine the work done on the proton. The confusion arises around the application of the distance in the context of the electric field's effect on the proton's kinetic energy. Understanding these concepts is crucial for determining the proton's final velocity as it exits the field.
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Homework Statement


1. A uniform electric field is directed towards the east with a magnitude of 10 N/C. A proton, moving with a speed of 10^5 m /s, enters the field traveling parallel to it. The proton covers a distance of 4 cm while in the field. Find the proton’s final velocity as it leaves the field.


Homework Equations


F=(k)(q1)(q2)/r, F=mv^2/r, Ac=v^2/r


The Attempt at a Solution


I don't understand how to attempt this problem. Confused about the 4cm, not sure if where to use it. How does the proton's velocity change when it enters an electric field?
 
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1 Newton / coulomb = 1 volts / meter

And W = q*ΔV

If you know how far it went in the field (.04 m) * 10 v/m that's how much work was exerted on increasing its kinetic energy isn't it since it was || to its direction of motion?
 
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