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Change in potential energy/work formula for electric charges. 
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#1
Mar812, 01:27 AM

P: 368

I have the following formula on my formula sheet:
ΔU = U_a  U_b = q(V_a  V_b) I was wondering if 'a' is final and 'b' is initial or is it the other way around? Also when I plug in my charge q into the formula, if it was a negative charge do I plug the negative sign into the formula? I realize that some formulas assume that you plug in the magnitude of the charge so I am not so sure about this formula... Thanks! 


#2
Mar812, 02:24 AM

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Thanks
P: 10,380

The change of potential energy from a to b is UbUa=q(VbVa), but the work done by the electric field on a charge q when it moves from a to b is W(ab)=UaUb. If the charge moves in free space the work done by the field increases its kinetic energy: UbUA=KE(b)KE(a), which means that U+KE=const, conservation of energy.
Imagine that the (positive) charge moves across a resistor from a to b. The moving charge constitutes current; the current flows from positive to negative, in the direction of decreasing potential: I=(VaVb)/R, and the work done (and dissipated on the resistor) while q charge moves from a to b is q(VaVb). VaVb sometimes is called "voltage" or "potential drop". Ohm's Law means that Current=potential drop divided by resistance. The charge q can be either positive or negative, but you can omit the sign when you are interested only in the magnitude of work . ehild 


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