Energy Density with a Linear Dielectric Inconsistency The energy density of an electromagnetic field with a linear dielectric is often expressed as $0.5 E \cdot D$. It is also known that energy can be found by $ 0.5\int_{V} \rho V dV $. Using the latter, the energy density is found to be $0.5 E^2 \varepsilon_0$, as is well known. If you integrate the latter only over free charge and ignore bound charge, you write $\epsilon \nabla \cdot E=     \rho$, use integration by parts, and obtain the first result. Does the first result neglect the energy from bound charge? If not, why does $0.5 E^2 \varepsilon_0$ break down (I.e. why can’t one find the energy with a dielectric by treating the bound charge as its own independent charge arrangement and using formulae for a vacuum?)
Here is a concrete example:
Consider again a parallel plate capacitor with charge $Q$, area $A$, separation $d$, and ignore edge effects. Fill it with dielectric of constant $k$. Now, we know that there will be a surface charge density of $-\frac{(k-1)Q}{kA}$ on the side of the dielectric next to the positively charged plate, and the opposite density on the side of the dielectric next to the negatively charged plate. This reduces the electric field by a factor of $k$, as is well known. If you consider this charge arrangement in a vacuum, it has energy $\frac{AdE^2 \varepsilon_0}{2k^2}$. But with a dielectric, apparently you must replace $ \varepsilon_0 $ with $ \varepsilon $, and the energy is now suddenly $\frac{AdE^2 \varepsilon_0}{2k} $ for the same charge distribution!
 A: Yes. The first result neglects the energy from bound charge. 
Given you consider that the E is net E(=E due to free charge +E due to bound charge) . 
In the example of a capacitor with a dielectric between its plates the first result represents the energy that is retrievable from the capacitor(by discharging the capacitor). The first result , neglects the energy in dielectric , but, it keeps track of the effect of dielectric on the energy retrievable from capacitor. If you write $Q*V$ for all charges(those on capacitor plates and those in dielectric) in this example. You will find that the total energy in the system(capacitor plates + dielectric) is $EQd/2$ + some other term. (Here E is net E in dielectric , d is distance between the plates and Q is the charge on capacitor plates) . This $EQd/2$ is equal to  $0.5E.D$ .  
$V$ is potential in above discussion
A: Griffith’s gives a good exposition. Essentially the vacuum formula does not take into account “spring energy” in dipoles (which could be electric in nature, but is still neglected because we use an averaged, macroscopic E field). 
