Asked by Anonymous

Good afternoon! I'm stumped with this question here.

<b>Q: An electron starts from rest 32.5 cm from a fixed point charge with Q=-0.125μC. How fast will the electron be moving when it is very far away?</b>

I got an answer of <i>5.54 * 10^-16 J</i>, but apparently it is incorrect. Please help! Thanks!
Answered by bobpursley
The PE available is : kQ/r=9e9 (.125e-6)/.325=3462 joules
1/2 m v^2=PE
v= sqrt(2*3462/(.325*9.1e-31)) = I dont get your answer, or even near it.

Paste this in your google search window sqrt(2*3462/(.325*9.1e-31)) =
Answered by Anonymous
Oh. So we're first using the Voltage formula to find the voltage given k is a constant, and we have Q and r. Then, we plug in the voltage into the 1/2mv^2 formula. Now I'm confused though -- what is the PE?

P.S. Isn't there also another formula: 1/2 m v^2 = KE? I don't have the PE one but I have a KE one on my notebook here.
Answered by bobpursley
when you move a charge to a new location, it has potential energy, the same amount of work that was done to get it that. In the case of a charge , that work moving is kQ/r, and that is its potential energy. If it moves from there, back into new position, the work done converts to KEnergy.
if you move a box up in a gravitational field you do work, mgh. At the end, it has PE mgh. If you release it, 1/2 mv^2. If you start a repelling charge, at some r, the work done on it by the electric field gives it KE. Work is energy, energy is work. Anyway, PE comes for potential theory (which comes in many forms: electrical, gravitational, nuclear) which can be converted to KE.
Answered by Damon
Voltage is potential energy per unit charge:
see:
http://www.studyphysics.ca/2007/30/06_forces_fields/12_voltage.pdf
Answered by Anonymous
Ohhhhh... I see.

Thanks! :)
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