What is the maximum distance and velocity of a charge in an electric field?

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SUMMARY

The maximum distance a charge of -0.061 C can move in an electric field of 888 N/C is determined by applying the force exerted on the charge, calculated using the formula F = qE. The charge, with a mass of 0.72 kg and an initial velocity of 54 m/s, can be analyzed using conservation of energy principles to find its return velocity to the origin and its velocity at x=5.75 m. The problem is straightforward once the correct approach is recognized, emphasizing the importance of understanding the relationship between electric force and motion.

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Homework Statement



A charge of -0.061 C is intiialy moving to the right with a velocity of 54m/s. It is moving in an electric field of 888N/C to the right, and has a mass of 0.72 kg. What is the maximum distance the charge moves to the right? What velocity does it have when it returns to the origin? What velocity does it have at x=5.75m?


Homework Equations



I don't know how to start.



The Attempt at a Solution



I tried using the electric field E=kq/r2but I can't get the solution.

I been looking at this problem blankly for a good two hours now and I could not find anything with google. Here is the link to the problem and solution.

http://books.google.com/books?id=vN...the origin has a magnitude of 845 N/C&f=false

Page 96 question 3.46
 
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This problem might sound a little harder than it actually is.

All you need to notice that an electric field exerts a force on a charged particle given by

\vec{F}=q\vec{E}

Once you consider that force, the problem can be solved with the usual high school mechanics techniques. (conservation of energy, anyone?)
 
Ty, I got the same solution as the book. I guess I was over thinking such a simple problem.
 

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