# Electrostatic potential

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?

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

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

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

Physics Monkey