Calculating Charge Interactions: Half-Speed & Beyond

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SUMMARY

The discussion focuses on calculating the speed of a charge q= +5.11 µC released from rest in the electric field of a fixed charge Q= +3.65 µC located at the origin. The solution for part (a) reveals that the speed of charge q when it moves infinitely far from the origin is 9.07202 m/s. For part (b), the user struggles to determine the distance from the origin where charge q reaches half of its final speed, indicating a misunderstanding of energy conservation principles in electrostatics.

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Homework Statement



A fixed charge Q= +3.65 µC is held fixed at the origin of an xy plane. A second charge q= +5.11 µC is released from rest at the xy coordinate of (+1.15 m, +0.810 m).

a) (a) If the mass of q is 2.90 g, what is its speed when it moves infinitely far from the origin?

(b) At what distance from the origin does q attain half the speed it will have at infinity*?
(*Please note where most of the energy exchange occurs!)

--------

Homework Equations



Ei=Ef
KQq/r=Fe=qV
1/2mv^2


The Attempt at a Solution



I solved for a) and got 9.07202 m/s
Can't seem to figure out way to do b. Initially I just took 1/2 the velocity I got in a), plugged it into the equation KQq/r=1/2mv^2 to solve for a n R, but that's wrong. Any help is appreciated.
 
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