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

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SUMMARY

The discussion centers on calculating the final velocity of a proton moving in a uniform electric field of 10 N/C. The proton, initially traveling at 10^5 m/s, covers a distance of 4 cm (0.04 m) while in the field. The work done on the proton can be calculated using the equation W = q * ΔV, where ΔV is derived from the electric field strength. The final velocity can be determined by applying the work-energy principle, which states that the work done on the proton increases its kinetic energy.

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Haniszmi
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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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