Change in elect. PE given work, KE, charge

  • Thread starter kevnm67
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  • #1
kevnm67
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Homework Statement


The work done by an external force to move a -8.4*10^-6C charge from point a to point b is 1.9*10^3J. If the charge was started at rest and has 4.81*10^4J of KE when it reach pt. B, whats the change in PE of the charge?


Homework Equations


Not sure
W=change in U
PE=KE


The Attempt at a Solution


I tried a few things but im just not sure how to approach this problem. I attempted V=Uelect/q and W= change in U, U=qV.... just looking for some direction, thanks!
 

Answers and Replies

  • #2
Fightfish
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Not sure
W=change in U
PE=KE
Two points of contention:
i) The work done on a system increases the total energy of the system, not just the potential energy
ii) Why must PE = KE ?? PE = KE is actually a statement demonstrating the case when there is purely conversion of energy from kinetic to potential or vice versa without any additional energy input into the system.
 
  • #3
tiny-tim
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hi kevnm67! :smile:

(try using the X2 icon just above the Reply box :wink:)
The work done by an external force to move a -8.4*10^-6C charge from point a to point b is 1.9*10^3J. If the charge was started at rest and has 4.81*10^4J of KE when it reach pt. B, whats the change in PE of the charge?

i don't think the amount of the charge, or the fact that the field is electric, has anything to do with the question :rolleyes:

you're given ∆KE and W, and you're asked for ∆PE :wink:
 
  • #4
kevnm67
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hi kevnm67! :smile:

(try using the X2 icon just above the Reply box :wink:)


i don't think the amount of the charge, or the fact that the field is electric, has anything to do with the question :rolleyes:

you're given ∆KE and W, and you're asked for ∆PE :wink:

Thanks for the responses.

OK, so I am still not sure. Work= change in Uelect. and Uelect.= q'V?

So what exactly is going on here? You have a charge that moves in a straight line from its initial position (where KE=0) to point B where it has 4.81x104J. The question asks for change in PE. So it has more PE at point A? and looses it to KE as it moves to B? How do you determine which equations to use and why? I am confused and have not seen a question like this so I completely lost. I appreciate the help
 
  • #5
tiny-tim
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hikevnm67! :smile:

I don't think it matters whether it goes in a straight line or not …

try using just W = ∆KE + ∆PE :wink:
 
  • #6
kevnm67
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hikevnm67! :smile:

I don't think it matters whether it goes in a straight line or not …

try using just W = ∆KE + ∆PE :wink:

OK, I figured out why I started to try to come up with a nonsense explanation....Thanks Tim, your correct and this was the first thing I tried and could not figure out why I did not get the correct answer. This question is from an old exam my professor posted and my mac deleted the negative sign in front of the exponents so my answer was a little off :)

Thanks for your help!
 

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