Find potential difference of negative charged particle?

AI Thread Summary
To find the potential difference between points A and B for a -8.5μC charge, the work done by an external force is 15x10^-4 J, and the kinetic energy at point B is 4.82x10^-4 J. The calculation shows that the potential energy at point A is -0.001 J, leading to a potential difference of 119.7 volts. However, since the charge is negative, the potential difference is expressed as -119.7 volts, indicating that point A is at a lower potential than point B. The discussion clarifies the definition of potential difference as the work needed to move a unit charge from point B to point A.
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Homework Statement


The work done by an external force to move a -8.5μC charge from point A to point b is 15x10^-4 J. If the charge was started from rest and had 4.82x10^-4 J of kinetic energy when it reached point b, what must be the potential difference between a and b


Homework Equations


W(nonconservative)=ΔKE+ΔPE
V=ΔPE/Q


The Attempt at a Solution


15x10^-4 - 4.82x10^-4 J= .001 J (Work = ke f - ke i (0) + PE f (0) - PE i) So PE i would equal -.001 J.
Then v= -.001J/-8.5x10^-6 C = 119.7 Volts.
But the answer is supposed to be -119.7 volts?
 
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My first thoughts are that the question is badly worded! A Potential DIFFERENCE is surely just a value ...119.7V
The negative charge starts from rest and gains KE which tells me that point A is a negative charge and the external force of the charge at point A repels the negative charge to point b
So point A is -119.7V with respect to point b
 
When we say potential difference what do we mean? My textbook defines potential difference as the work required to move a unit of electric charge from point b to point a.
 
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