It seems like this should be really easy but I've been stuck for awhile...(adsbygoogle = window.adsbygoogle || []).push({});

A small particle has charge -5.00 uC and mass 2.00E-4 kg. It moves from point a, where the electric potential is Va=+200V, to point b, where the electric potential is Vb=+800V. The electric force is the only force acting on the particle. The particle has speed 5.00 m/s at point a. What is its speed at point b?

Here's what I've tried. I did Ka + Ua = Kb + Ub

so that's (.5)(.200g)(5.00m/s)^2 + (-5.00E-6C)(200V) = (.5)(.200g)(vb)^2 + (-5.00E-6C)(800V)

And that gave me vb of 5.00 m/s so I figure that lead me in a circle bc I know that's not the answer bc the answer is in the back of the book... So then I tried:

V=kq/r and did -600V[Va-Vb]=9E9Nm^2/C^2 x -5.00E-6C / r and got r=75m so I assumed that's the distance between a and b. So then I used that with E=V(ab)/d and got E=-8N/C and so I tried to do something like E=kq/r^2 to find q but obviously, that only will give me q=-5.00E-6C so I'm making lots of little circles but I can't seem to figure out the charge of these points.. Any suggestions on what to do with those potentials (+200 and +800) I don't know where to go from here..

Kristi-Lynn

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# Homework Help: Electric force/velocity

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