What is the energy of a single charge system?

Question

I will try to limit the question in the case of the electric fields, but is something that applies also to the magnetic ones. There are two ways to express the energy in a capacitor:

• By Voltage : $U = 1/2 CV^2 $
• And by Field : $U = 1/2 \varepsilon E^2Ad$, With Energy Density: $u = 1/2 \varepsilon E^2 $

Unless i understood everything wrong and these two are NOT the same quantity, i have the following question. When we have two charges placed at points A and B, then in order to calculate the energy of the system, we will take the first charge, place it at point A WITHOUT doing any work, and then we will calculate the work needed to place the second charge at point B.

The weird thing to me here, is that while we have placed the first charge, without generating any work the system will still have the energy held in the field of the charge!

There is obviously something that i miss, but what?

Answer 1

Let's suppose you have two point charges $q_1$ and $q_2$. Each charge has its own electric field $\vec{E}_1$ and $\vec{E}_2$, and the total electric field is $\vec{E} = \vec{E}_1 + \vec{E}_2$. If we look at the total potential energy "stored in the fields" in this configuration, we can split it up into three pieces: \begin{align*} U = \iiint u \, dV &= \frac{\epsilon_0}{2} \iiint \left( \vec{E}_1 + \vec{E}_2 \right)^2 \, dV \\ &= \frac{\epsilon_0}{2} \iiint \left( \vec{E}^2_1 + \vec{E}^2_2 + 2 \vec{E}_1 \cdot \vec{E}_2 \right) \, dV \\ &= \underbrace{\frac{\epsilon_0}{2} \iiint \vec{E}^2_1 \, dV}_{U_1} + \underbrace{\frac{\epsilon_0}{2} \iiint \vec{E}^2_2 \, dV}_{U_2} + \underbrace{ \epsilon_0 \iiint \vec{E}_1 \cdot \vec{E}_2 \, dV}_{U_\text{int}} \end{align*}

Now, what do each of these three pieces mean? $U_\text{int}$ turns out to be the easiest to interpret: if you calculate this integral over all over space assuming that the charges are a distance $r$ apart, you get $$ U_\text{int} = \frac{q_1 q_2}{4 \pi\epsilon_0 r}; $$ in other words, this is the potential energy that we know & love. (Doing this exercise is a fun way to test your calculational mettle.) This term can therefore be thought of as the energy due to the interaction between the two charges $q_1$ and $q_2$. But what about $U_1$ and $U_2$? It's not too hard to see that these two quantities are, in fact, infinite; writing out the integral in spherical coordinates, we get $$ U_1 = \frac{\epsilon_0}{2} \iiint_\text{all space} \left( \frac{q_1}{4 \pi \epsilon_0 r^2} \right)^2 r^2 \sin \theta \, dr \, d\theta \, d \phi = \frac{q_1^2}{8 \pi \epsilon_0} \int_0^\infty \frac{1}{r^2} dr = \frac{q_1^2}{8 \pi \epsilon_0} \left[ \frac{1}{r} \right]_0^\infty, $$ which diverges at its lower limit. Uh-oh.

The most common interpretation of this divergence is to note that we never actually care about the absolute value of the potential energy; we only care about the differences between potential energies of various configurations. The quantities $U_1$ and $U_2$ don't depend on the location of the other charge; so we can view them as a fixed amount of energy that each charge carries around with it somehow. We are free to reset our "zero" for potential energy so that the potential energy of the system goes to zero as their separation $r \to \infty$, by subtracting the "constant" $U_1 + U_2$ from our definition for $U$ above; and then our new potential energy is $U_\text{int}$ by itself and everything is hunky-dory.

For most people, that resolves that, and if you're OK with the above explanation, you don't need to read the next paragraph. That said: it's still a bit odd & unsatisfying that we have these infinities running around. Really, what this calculation is telling us is the physics version of GIGO. Point charges have infinite charge density, and so we shouldn't be surprised when other important quantities (like energy) also end up being infinite when we use such ill-behaved charge distributions. If we model the charges as uniform balls with radius $R$, this whole problem never arises (though the integral $U_\text{int}$ becomes much harder to calculate exactly.) Classical electrodynamics is full of problems that arise from taking the idea of a "point charge" too seriously (I'm looking at you, Abraham-Lorentz force), and it's better to keep in the back of your mind that "point charges" are an idealization that can occasionally bite you.

Answer 2

The weird thing to me here, is that while we have placed the first charge, without generating any work the system will still have the energy held in the field of the charge!

It is not so weird, all bodies have some energy. Charged particles have energy even if far from other charged particles, provided they are composed of smaller charged parts. This energy indicates there is work one could extract if all the elementary charges the particle contains were separated from each other.

If the particle is elementary, i.e. it has no charged parts, then one cannot extract the work, so there is no need to associate energy with it.

For example, if the particle is an isolated charged point, there is no need to associate any EM energy with it. You may think that the energy density $\frac{1}{2}\epsilon_0 E^2$ implies there is EM energy around the point particle, but that is actually not necessary. The expression for energy density $\frac{1}{2}\epsilon_0 E^2$ was derived by Poynting under assumption the charge density is not singular. If it is as singular as point charge, the derivation breaks down, so the formula is not valid.

Answer 3

Potential energy is always defined for multi particle systems but can be defined for a single particle(in this case it is charge and hence electrostatic potential energy) if we assume the other charge to be present at infinity(a point far away from the vicinity of the charge will do) then you can calculate the potential energy for that system by taking the potential at infinity to be zero which is theoretically the potential energy of the system but can be labeled as the potential energy of the single particle.So what you actually did when you placed the charge at 'A' was,you assumed another particle to be present at infinity and wrote down the potential energy of that system.

Answer 4

The capacitor stores its energy in form of electric field.

Initially both the plates were neutral. After applying the potential difference, electrons move form one plate to another.

So to calculate the energy, you would first take a charge $dq$ from one plate to another. By this action, you just created an opposite charge $-dq$ on the other plate the moment you took the charge from it. This will set up an E. field. You set up an electric field that wasn't there before, so some work has to be done.

We will keep taking charge from one plate and deposit it on the other plate till the charge on the plate becomes $Q$.

Let's assume that at a point of time, charge on a plate is $Q'$. Therefore at the same time, the electric field between the plates will be $E=\frac{Q'}{A\epsilon_{\circ}}$.

Now, we will move charge $dq$ and the work done will be given by :

$dW=Ed.dq$

$W=\frac{d}{A\epsilon_{\circ}}\int_0^Q Q'dq$

$W=\frac{1}{C}.\frac{Q^2}{2}$

On rearranging,

$W=\frac{CV^2}{2}$

When you place a $1^{st}$ charge at point A (as per your question) you bring it to such a point where it interacts will no other charge. Thus, we come to the conclusion that work done here is $0$. The work will only be done in bringing the $2^{nd}$ charge at point B.

Answer 5

Besides the things mentioned in the other answers, here is one more thing to consider:

The energy of the single charge at some point in space can be thought of as arrived at by starting with an infinitesimal charge, and adding (from infinity) a little bit of charge. That will take a little bit of work. The next bit of charge will take more work, and so it continues. In other words - a single (finite) charge (assumed distributed over a small region of space - you can't have a finite charge in an infinitesimal volume) must have some energy associated with it.