(adsbygoogle = window.adsbygoogle || []).push({}); A point charge Q = +85.5 µC is held fixed at the origin. A second point charge, with mass m = 0.0526 kg and charge q = -2.73 µC, is placed at the location (0.323 m, 0).

(a) Find the electric potential energy of this system of charges.

(b) If the second charge is released from rest, what is its speed when it

reaches the point (0.125 m, 0)?

ok so i found part (a) -6.496597059 J but how do i calculate (b)?? i've tried subtracting the electric potential energy of the distance traveled from .323 to .125 (.198 m) from the electric potential energy of the system to find the change in electric potential energy. Then i divided that by my second charge (-2.73E-6) to get my change in potential so i could plug that into [tex]v= \sqrt{2q\Delta V/m}[/tex] and am getting wrong answers... any guess on where i'm going wrong??

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# Homework Help: Speed of a charge

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