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Homework Help: Find potential difference of negative charged particle?

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data
    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


    2. Relevant equations
    W(nonconservative)=ΔKE+ΔPE
    V=ΔPE/Q


    3. 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?
     
  2. jcsd
  3. Jan 22, 2012 #2
    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
     
  4. Jan 22, 2012 #3
    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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