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Electrostatic potential

  1. Oct 6, 2005 #1
    This is more of a general question and not a homework question, just to make it clear. Say two test charges are brought separately, one after the other, into the vicinity of a charge +Q. First test charge +q is brought to point B a distance r from +Q. This charge is removed and a test charge -q is brought to the same point. Now do we say that the electrostatic potential of +q is greater because it has a positive sign compared to -q? Or do we say their potentials are the equal because their magnetudes are the same? I assume it's the latter, isn't it?
     
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  3. Oct 6, 2005 #2

    Physics Monkey

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    The electrostatic potential of the system is always
    [tex]
    \phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}
    [/tex]
    assuming the test charges are truly infinitesimal (i.e. we neglect their contribution).

    The electrostatic potential energy (or just energy) is
    [tex]
    E_{q}= \frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}
    [/tex]
    in the case of a positive test charge and
    [tex]
    E_{-q}= -\frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}
    [/tex]
    in the case of a negative test charge. The energy or electrostatic potential energy is different in each case, but the electrostatic potential of the charge Q is the same in both cases. It's mostly a matter of terminology.
     
  4. Oct 6, 2005 #3
    So to clear this up, the electrostatic potential energy of the positive test charge is greater?
     
  5. Oct 6, 2005 #4

    Physics Monkey

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    Yes. It has positive energy while the negative charge has negative energy.
     
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