Confusion About Energy Density in Electric Field

[AP36020]

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?)

Edited March 10, 2019 by AP36020

[studiot]

5 hours ago, AP36020 said:

The energy density of an electromagnetic field with a linear dielectric is often expressed as 0.5E⋅D . It is also known that energy can be found by ∫VρVdV . Using the latter, the energy density is found to be 0.5E2ε0 , as is well known.

 

Not quite.

You are mixing up a vacuum and a material dielectric.

The volume integral refers to a vacuum (since you have used [math]{\varepsilon _0}[/math] )

There is no polarisation in a vacuum!

 

Remember also that the electric field comes from somewhere.

In the case you have described it comes from a parallel plate capacitor.

Now conservation of energy tells us that we can either consider the energy due to the separation of charge on the plates.

Or we can consider the energy distributed in the field.

If you have a material dielectric between the plates then

[math]D = {\varepsilon _0}E + P[/math]

or you can use susceptibility.

 

Edited March 10, 2019 by studiot

[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!

Edited March 10, 2019 by AP36020