Comparing Electrostatic Potentials of Positive and Negative Charges

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The electrostatic potential of a charge +Q remains constant regardless of the test charge brought into its vicinity, as described by the formula φ(r) = (1/4πε₀)(Q/r). While the electrostatic potential energy differs between a positive test charge (+q) and a negative test charge (-q), with the former having positive energy and the latter negative, the potential itself is unaffected by the sign of the test charge. Thus, the electrostatic potential is the same for both charges at the same distance from +Q. The distinction lies in the potential energy, which is greater for the positive charge. Overall, the key takeaway is that the electrostatic potential is independent of the test charge's sign.
Tony11235
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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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The electrostatic potential of the system is always
<br /> \phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}<br />
assuming the test charges are truly infinitesimal (i.e. we neglect their contribution).

The electrostatic potential energy (or just energy) is
<br /> E_{q}= \frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}<br />
in the case of a positive test charge and
<br /> E_{-q}= -\frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}<br />
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.
 
Physics Monkey said:
The electrostatic potential of the system is always
<br /> \phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}<br />
assuming the test charges are truly infinitesimal (i.e. we neglect their contribution).

The electrostatic potential energy (or just energy) is
<br /> E_{q}= \frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}<br />
in the case of a positive test charge and
<br /> E_{-q}= -\frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}<br />
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.

So to clear this up, the electrostatic potential energy of the positive test charge is greater?
 
Yes. It has positive energy while the negative charge has negative energy.
 
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