Proton Distance Help: Find After .5s

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A proton released in a uniform electric field of 2.5e-8 N/C travels a certain distance after 0.5 seconds. The correct approach involves calculating the proton's acceleration using the formula Fe = q*E, where q is the charge of the proton and E is the electric field strength. The kinematic equation x = vot + 1/2at^2 is applicable, but the initial assumption of using gravity (9.81 m/s²) is incorrect since the proton is influenced by the electric field. The correct distance traveled after 0.5 seconds is 0.3 meters, not 1.2 meters as initially calculated. Understanding the context of electric fields is crucial for solving this problem accurately.
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Homework Statement


A proton is released from rest at t=o in a uniform electric field whose magnitude is 2.5e-8 N/C. How far has the proton traveled after .5 s?


Homework Equations



x = vot + 1/2at^2

The Attempt at a Solution


x = 1/2(9.81)(.5)^2 = 1.2 but it says the answer is .3m. I'm prety sure I'm using the correct equation so I don't know what is wrong.
 
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matt72lsu said:

Homework Statement


A proton is released from rest at t=o in a uniform electric field whose magnitude is 2.5e-8 N/C. How far has the proton traveled after .5 s?


Homework Equations



x = vot + 1/2at^2

The Attempt at a Solution


x = 1/2(9.81)(.5)^2 = 1.2 but it says the answer is .3m. I'm prety sure I'm using the correct equation so I don't know what is wrong.

Fe = q*E = ma Solve for acceleration, a, first. Then do your kinematics. The proton is accelerating in an electric field, I am assuming you don't have to worry about g... Have you not been studying E fields in your class? Remember what you are studying in class and then apply to the "old" ways.
 


thanks!
 


matt72lsu said:
thanks!

happy day
 
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