AP36020

But why should the energy in a dielectric differ from the same charge distribution in a vacuum? For 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 on the side of the dielectric next to the negatively charged plate. This reduces the electric field. 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!

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 \(\int_{V} \rho V dV\). 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?)