Electric field energy deduce, why one of the term goes zero?

Question

When we derive equation for energy of electric field, we end up below during the process $$U=\frac{\epsilon_0}{2}\left[\int (\nabla\phi)\cdot(\nabla\phi)dV-\int\nabla\cdot(\phi\nabla\phi)dV\right]$$ and then, $$U=\frac{\epsilon_0}{2}\left[\int (\nabla\phi)\cdot(\nabla\phi)dV-\oint(\phi\nabla\phi)\cdot d\mathbf a\right]$$ I think I can quiet grasp the idea that when we integrate over the entire space, the second term will be calculated to have $r^{-1}$ term which leads to zero, thus negligible.

But my question is, do we always pretend that second term is zero? Is there any situation where we need to account the second term?

Such as when we are dealing with parallel plate condenser, especially the infinite sized case, second term doesn't seem like to go zero at all in that case.

And what about the case of energy density? though we write $$u=\frac{\epsilon_0 E^2}{2}$$ when electric field is close to the source or where electric field is strong enough, the second term seemingly no longer obsolete thus I thought maybe there's a case where we calculate energy density of electric filed by $$u=\frac{\epsilon_0 E^2}{2}+\frac{\epsilon_0}{2}\nabla\cdot(\phi \mathbf E)$$

If we never use the second term, why is that? Is there any way to prove that second term is always negligible?

Answer 1

$U$ in your first formula is the total Coulomb energy of a continuous distribution of charge in space. This energy is a number assigned to the system as a whole, and is not necessarily distributed in space.

But it can be thought of as distributed to space and to its boundary the way you wrote it. However, the boundary is arbitrary, so we can push it to infinity and then it turns out (for static field that decays fast enough in infinity) contribution of the boundary is zero. Thus we can think of the Coulomb energy as distributed in infinite space, no boundaries needed.

There is still an infinity of different ways to distribute the same energy in infinite space, including the Poynting density $u = \frac{1}{2}\epsilon_0E^2$ and the density you suggest $u' = u+\frac{1}{2}\epsilon_0 \nabla\cdot(\phi\mathbf E)$.

There are reasons for preferring the Poynting density as the definition of energy density of electric field, in the general case. It is the simplest and symmetric in terms of fields. However, remember this is just a definition of density that we chose, a very useful one, but we could work instead with $u'$ and with any finite system, we get the same total energy, because contribution of the divergence term is zero when taken over the whole space. It changes distribution of energy in space, but not the total energy. There is no way to decide which density expression is the right one in physical sense: it is a matter of practicality and aesthetics, and most often, the Poynting expression $u$ is preferred.

Energy of an infinite parallel plate capacitor is infinite, so both densities will get that right.