# Why does electrostatic energy not obey superposition principle? What does the extra energy term account for?

Energy stored in an electrostatic field is given by $\frac{\epsilon_0}{2}\int\text{E}^2\text{d}\tau$ where the integration is over all space.
If $\vec{\text{E}_0}=\vec{\text{E}_1}+\vec{\text{E}_2}$ and find the energy stored in $\vec{\text{E}_0}$ then, we get this energy term $\frac{\epsilon_0}{2}\int2\vec{\text{E}_1}.\vec{\text{E}_2}\text d\tau$ in addition to the energy terms of the individual fields. Lets call this energy $U.$ What does $U$ account for? Does it account for the energy associated with the establishment of the electrostatic field $\vec{\text{E}_0}$ using the individual fields? If this is correct then, while establishing the field, what causes $U$ to be subtracted when the two individual fields are in opposite direction as $\vec{\text{E}_1}.\vec{\text{E}_2}=-\text{E}_1\text{E}_2$ while it is to be added when they are in the same direction? If I am wrong then what does it account for?
One may tell that the work done in bringing a test charge from infinity to a point in the field against the field will be less in the first case hence lesser energy stored than the latter case where the two fields add up in the same direction and hence create a stronger field. But I can't understand how $U$ amount of extra energy is stored when the latter configuration is established but not in the first case? It seems $U$ amount of energy is released while establishing the field in the first case.

The expression for the total potential energy stored in the fields is given by

$$\frac{\epsilon_0}{2} \int \left| \mathbf{E}_1 + \mathbf{E}_2 \right|^2 d\tau = \frac{\epsilon_0}{2}\left( \int \left| \mathbf{E}_1 \right|^2 d\tau + \int \left| \mathbf{E}_2 \right|^2 d\tau + 2 \int \mathbf{E}_1 \cdot \mathbf{E}_2 d\tau \right)$$

Notice that the first and second terms of the right-hand side are the potential energies stored in, respectively, the first field (produced independently by some charge configuration $\rho_1$) and the second field (produced independently by some charge configuration $\rho_2$). However, when the two fields are superimposed on one another, there is an interference between them. The third term accounts for the interference, which may be constructive to the total energy ($\mathbf{E}_1$ is in the same direction as $\mathbf{E}_2$, so they constructively add which results in a higher net field and therefore a higher energy) or destructive to the total energy ($\mathbf{E}_1$ is in the opposite direction as $\mathbf{E}_2$, so they destructively interfere which results in a lower net field and therefore a lower energy). Notice also that if $\mathbf{E}_1 \cdot \mathbf{E}_2 = 0$, then $\mathbf{E}_1$ and $\mathbf{E}_2$ are orthogonal to one another, so their superposition results in no interference (one could be oriented along $\hat{x}$ and the other along $\hat{y}$). In this very special case, the energy happens to obey a superposition principle.

• If we integrate over all space there should always be an increase in energy by $U$ amount for all cases right? because if somewhere destructive is taking place then in some other place constructive is taking place in order to conserve energy and $U$ amount of energy is stored which was work done to establish the system. – Jolie Aug 28 '15 at 6:24

It's a bit unfortunate that you called it $J$ because energy is stored in electromagnetic fields and so it moves around but sometimes the net flow of electromagnetic field energy into a region is not completely balanced by an increase in the field energy at that point and one time that happens is when the fields exchange energy with electric charges. And the rate that happens is $\vec J \cdot \vec E.$

Because that is the rate that work is done on the charges. Then, since $\vec \nabla \times \vec B =\mu_0\vec J+\mu_0\epsilon_0\frac{\partial \vec E}{\partial t},$ so then we get $\vec J \cdot \vec E=$ $$\left(\frac{1}{\mu_0}\vec \nabla \times \vec B -\epsilon_0\frac{\partial \vec E}{\partial t}\right)\cdot \vec E.$$

Which equals $$\frac{1}{\mu_0}\vec E \cdot (\vec \nabla \times \vec B) - \epsilon_0\vec E \cdot \frac{\partial \vec E}{\partial t}.$$

Which also equals $$\frac{1}{\mu_0}\vec E \cdot (\vec \nabla \times \vec B ) -\frac{\partial}{\partial t}\left(\frac{\epsilon_0}{2}\vec E \cdot \vec E\right).$$

So when charges gain or lose energy from work being done on them, the fields lose or gain energy at an equal rate. So if the current and the electric field point in the same direction the charges gain energy and the fields lose it hence $-\frac{\partial}{\partial t}\left(\frac{\epsilon_0}{2}\vec E \cdot \vec E\right)$ is energy created right there.

If instead the current pointed in the opposite direction as the field then the charges lose energy and hence the fields gain it.

So really the energy came from the charges. For instance if you have two like charges then they have their own energy but then when you bring them closer together the fields have more energy but that is because you had to push them to bring them together.

You can imagine a situation where two charges are rushing towards each other. They start out with kinetic energy. Work is done by the fields and they slow down. That energy had to go somewhere. The energy goes into the fields right then and there at the exact location of the charge as work is done on it.