Is electric potential at a point defined due to an "isolated" point charge?

Question

If there is a fixed point charge then it has its fields in the surrounding and it doesn't matter whether there is another charge near or not.

But for potential, it is defined as electric potential energy per unit charge and for potential energy we need two charges at least (in order to define a system).

But what if we just have a single fixed point charge. Will there be any potential near the charge ? Is it even meaningful to define potential in this case since there is nothing to have potential energy with ?

Answer 1

There's a difference between electric potential and electric potential energy. The potential energy is a single number, and tells you how much energy you need to construct the system starting from point charges that are infinitely far away. The potential is a function of space, and tells you how much energy you would need to bring a unit charge from infinity to a given spatial point, holding all the other charges fixed.

For an isolated charge, (ignoring the self-energy, that is the energy needed to make the charge in the first place), the potential energy is zero. Since there are no other charges around, no energy is needed to move this point charge from infinity.

However, the potential is not zero, in fact it is $kq/r$, where $k$ is Coulomb's constant, $q$ is the charge, and $r$ is the distance from a given spatial point to the point charge. The potential tells us how much energy we would need to give to another point charge with charge $Q=1\ {\rm C}$ (assuming we are working in SI units) to bring it from infinity to a given point in space, while holding the original point charged fixed.

Answer 2

Consider $N$ point charges $q_i$ at fixed positions $\mathbf{r}_i'$. The electric potential of these source charges is (taking the standard reference point of $V \rightarrow 0$ as $r\rightarrow\infty$) $$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N}\frac{q_i}{||\mathbf{r}-\mathbf{r}_i'||} \tag{1}.$$ Notice that this formula only makes reference to the positions and charges of the source particles. It is independent of the presence (or absence) of a "test charge." In the case of a single charge $q$ at position $\mathbf{r}'$, the sum in Eq. (1) reduces to a single term, the well-known Coulomb potential: $$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\frac{q}{||\mathbf{r}-\mathbf{r}'||}.$$ The negative of the gradient of this potential is the electric field of the charge, a meaningful physical quantity. Electric potential, therefore, has meaning even in the case of an isolated point charge.

Answer 3

Just as the electric field (in N/C) is defined in terms of the force on a “test” charge, the potential is defined (in J/C) in terms of the potential energy of a test charge at a point relative to a chosen reference point. One or more charges produce the field and potential at a point and another charge is needed to measure them.

Answer 4

The potential difference $V$ between two points is the work per unit charge required to move the charge between the two points. The potential difference between points 1 and 2 due to the electric field produced by a point charge $Q_1$ at point 1 does not depend on the existence of a unit charge at point 2. It equals the work required to move a unit charge $Q_2$ hypothetically located at point 2 to point 1, sometimes referred to as a "test charge".

The potential at point 1 due to the field generated by $Q_1$ requires one to define a point where the potential is zero (zero reference potential). While a theoretically arbitrary decision, for a point charge the zero reference is generally taken for $r=\infty.$ Then the potential difference is the work required to move a unit test charge from $\infty$ to point 1.

Hope this helps.