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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!

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  • $\begingroup$ Agreed on the first point. As for the second point, I’m unsure why it should take more energy in a dielectric. Why not the same amount of energy? Qualitative, nonrigorous reasoning for same energy: If you consider a parallel plate capacitor with a dielectric for example, I agree you need more free charge for the same field, raising energy, but doesn’t the interaction of the free charge with the bound surface charge on the dielectric lower the energy? $\endgroup$ – Arjun Puri Mar 10 at 6:50
  • $\begingroup$ In $\epsilon \nabla \cdot E= \rho$ the E is the net E.. due to both free charge and bound charge. But $\rho$ consists only of free charge. It would take more energy to establish the same amount of electric field E in dielectric than it takes to establish in vaccum. The first result doesn't neglect the energy from bound charge(if E is net E in the dielectric). $\endgroup$ – Jeevesh Juneja Mar 10 at 6:54
  • $\begingroup$ I recognize $\rho$ is only the free charge in the macroscopic Gauss’s law, but it is all charge in $ 0.5\int_{V} \rho V dV $. Substituting using Gauss’s law into this integral integrates over only free charge, which is part of my concern $\endgroup$ – Arjun Puri Mar 10 at 7:00
  • $\begingroup$ Sorry , I was wrong.. do tell what you think of the answer below $\endgroup$ – Jeevesh Juneja Mar 10 at 8:33
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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

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  • $\begingroup$ Would discharging the capacitor not get rid of the polarization in the dielectric as well, allowing that energy to be used too? $\endgroup$ – Arjun Puri Mar 10 at 18:43
  • $\begingroup$ Discharging of capacitor would get rid of polarization in the dielectric .The energy of polarization would be gone. But that energy would not appear in the circuit through which the capacitor discharged $\endgroup$ – Jeevesh Juneja Mar 11 at 12:39
  • $\begingroup$ The dielectric will be hotter after the capacitor has discharged through the outside circuit. $\endgroup$ – Jeevesh Juneja Mar 11 at 13:24
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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).

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