Which is the difference between electrostatic potential energy and electrostatic potential stored energy?

Question

here it's written that:

The electrostatic potential energy, $U_E$, of one point charge $q$ at position $r$ in the presence of a point charge $Q$, taking an infinite separation between the charges as the reference position, is: $$U_E(r) = \frac{1}{4\pi\epsilon_0r} qQ.$$

I see it's a function of position between the observer (where is put the test charge $q$) and the source charge $Q$.

Then, it says, about the energy stored in such a system:

The electrostatic potential energy of a system containing only one point charge is zero, as there are no other sources of electrostatic force against which an external agent must do work in moving the point charge from infinity to its final location.

Well, I can't understand the difference between energy stored and electrostatic potential energy. My doubt regards also other situations. For instance, always in that page, it is written that for a system of two point charges:

• Electrostatic potential energy of $N$ point charges

$$U_E(r) = \frac{q}{4\pi\epsilon_0} \sum_{i=1}^{N} \frac{Q_i}{r_i}$$

It depends on the distance between the observer (test charge $q$) and each source $Q_i$.

• Electrostatic potential energy stored on a system of $N$ point charges

$U_E(r) = \frac{1}{2}\frac{1}{4\pi\epsilon_0} \sum_{i=1}^{N} Q_i \sum_{j=1}^{N,j\neq i} \frac{Q_j}{r_{ij}}$

It doesn't depend on the distance between the observer (test charge $q$) and each source $Q_i$, but it depends on the distance between two charges of each possible couple.

So, which is (physically) the difference between such quantities? They are both indicated with $U_E$, but they have different expressions and one depends on the position and the other one not. If the electrostatic potential energy is $qV$, how is defined the stored energy?

Answer 1

Energy is always the ability to do work and there is no difference between "energy stored" and general "energy". The difference between your two formulas depends on the physical system: in the first case, the charges $Q_i$ are fixed in space and therefore do not do any work onto each other by electrical forces. In the second case, they are free to move and affect each other electrically.

The first example $$ U(r) = \frac{q}{4\pi \epsilon_0} \sum_{i=1}^N \frac{Q_i}{r_i} $$ is the amount of work that would be done on a small test charge $q$ if we were to place it at $r$, assuming that the positions of all charges $Q_i$ remain fixed. We can interpret this as "stored energy" in the following way: If $U(r)$ is positive, it would be equal to the kinetic energy that the small test charge $q$ would have after it is pushed very far away (to "infinity") by the other charges. (If $U(r)$ is negative, it is the kinetic energy needed to escape from the other charges, like the how the gravitational potential energy is the amount of energy needed to escape a planet's attraction.) This is also the case where the formula $U = qV$ is applicable, and in a sense it can be viewed as a definition of the electric potential $V$ at a given point in space. Here we assume that all the charges $Q_i$ remain (for whatever reason) fixed in space, and therefore cannot do any work onto each other. Thus we do not need to think about any forces between the charges.

In the second example, we do not hold the position of the other charges fixed any more, so that they can also do work onto each other! The amount of work that they can do onto each other is given by your second formula, which takes into account the forces between the electric charges $Q_i$.

So the formula that you use depends on the physical situation. Do you have one charge $q$, which is much smaller than the other charges, so that moving it around does not change the position of the other charges? And do you only want to know about the work that would be done on $q$? Then use the first formula. Do you have multiple charges that interact with each other? And do you want to know how they interact with each other (for example when writing down the system's Lagrangian $L = T-U$, kinetic energy minus potential energy)? Then use the second formula.