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Electric Potential Sign Check

  • #1
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Homework Statement



The work done by an external force to move a -8.50 mu C charge from point a to point b is 1.90×10^-3 J.

If the charge was started from rest and had 4.86×10^-4 J of kinetic energy when it reached point b, what must be the potential difference between a and b?


Homework Equations



V = PE / q
Vba = - W/q

The Attempt at a Solution



If the charge started from rest, all the energy it had was potential energy which got converted to kinetic energy at point b.

V = 4.86 x 10^-4 / -8.50 x 10^-6 C
V at point a = -57.176

Vb - (-51.176 V) = -1.90 X 10^-3 J / -8.5 X 10^-6 C

Vb + 51.716 V = 223.5 V
Vb = 165.76 (166 V)

I use MasteringPhysics and it told me the answer should have been negative but I don't understand why. My book states "When the electric force does positive work on a charge, the kinetic energy increases and the potential energy decreases. The difference in potential energy, PEb - PEb, is equal to the negative of the work, Wba, done by electric field to move the charge from a to b."
 

Answers and Replies

  • #2
Simon Bridge
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Don't see how you can know the potential energy at a particular point - what are you using for your zero of PE?

From the looks of the numbers - the applied force did much more work than the resultant kinetic energy ... so the charge must have been encountering an opposing force, which means the motion was against an opposing potential. Thus Vb < Va (since negative charges roll uphill).
 
  • #3
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Simon Bridge - Thank you, that makes sense. Initially, I wasn't entirely sure how to approach the problem. I thought I the zero point was at "B" when all the energy was kinetic.
 
  • #4
Simon Bridge
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I thought I the zero point was at "B" when all the energy was kinetic.
However, you also found a distinct value for Vb.

It looks like you found Vb from Vb - Va = 223.5V (and plug in your value for Va). But Vb-Va is the voltage difference you have been asked to find.

Have you tried expressing everything in terms of change in energy?
 
  • #5
191
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No, but the next time I come across a problem like this I will express things in terms of ΔU.
 
  • #6
Simon Bridge
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OK - have fun ;)
 

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