Dielectric constant in the energy density function

[RoyalCat]

The energy density function for the electric field in vacuum is
$$u=\dfrac{\epsilon_0}{2}E^2$$

And the cited textbook result for the energy density inside a dielectric is:
$$u=\dfrac{\epsilon_0 \epsilon_r}{2}E^2$$

Now, one way to reach the upper formula is to look at the energy as $$\tfrac{1}{2}\int_\text{over all space} \rho \phi dV$$ and then go through some vector identities to reach the proper representation.

Now, going through the same derivation inside a dielectric, I can't quite find where the dielectric constant comes in.
I mean, if we're dealing with the total field, E, and the total charge density, $$\rho$$, why should it differ?

I've seen one derivation where it's built ground up from the infinitesimal amount of work required to add a small amount of free charge, but that still doesn't quite click for me.

I'd really appreciate an explanation of how that dielectric constant got into the formula. :)

With thanks, Anatoli.

 

[tiny-tim]

Hi Anatoli!

… going through the same derivation inside a dielectric, I can't quite find where the dielectric constant comes in.

Using dimensional analysis …

energy = work done = force x distance = E x charge x distance

energy density = E x (charge x distance per volume)

= E x (charge per area) = E x coulombs per square metre

= E x D​

(the electric displacement field D is surface charge density, or coulombs per square metre)

So energy density is proportional to E times D (the "1/2", we have to find some other way).

Since D/E is the the dielectric constant (permittivity, "capacitivity"), ε, we have to write it either εE² or D²/ε.

I mean, if we're dealing with the total field, E, and the total charge density, $$\rho$$, why should it differ?

No, we're not dealing with the total charge density, only with the free charge density …

the bound charge doesn't have to go across the whole field, it only has to go a very short distance (it's bound! ) …

we still have to take account of it, of course, and we do so by recognising that the work done in polarising the bound charges has the effect of decreasing the E field (by ε_r = ε/ε₀) …

sooo … why does a dielectric appear to increase the energy density?

Because we must compare like with like … if we introduce a dielectric into an existing situation, then D will stay the same (not E), so the energy density of D²/2ε₀ will be reduced (the field "pulls" the dielectric in, and if we later remove the dielectric, we must do work against the field, thus increasing the field's energy) …

it only looks as if the energy density in increased because we usually write it εE²/2, and we forget that increasing ε decreases E² even faster!

Finally, why is the free field always greater than the total field? Because the bound field always opposes the free field! …

(imagine a bound pair of charges between the plates of a capacitor … the charges go + - + -)

which is why we conventionally write the bound field as minus P, to "help" explain why the D field is larger!

 

[RoyalCat]

Thanks for the reply, but that wasn't quite what I was asking.
The problem for me is understanding why we need only take account of the free charge when calculating the integral.

We know that in a vacuum, the total energy is given by $$\tfrac{1}{2}\int \rho \phi dV$$

If we interpret the dielectric simply as additional charge density (Positive and negative close together), why should it not come into our calculation? Why do we only need to run over the free charge?

Now, the dimensional analysis is all fine and good when you're working in SI, but in CGS, you find yourself with D and E in the same units, so that argument falls short of satisfactory.

Now, my understanding of the integral $$\tfrac{1}{2}\int \rho \phi dV$$ is that the potential and total charge density at every point are known functions, and we go around sampling them, putting the factor one half to avoid counting twice (For every infinitesimal charge, we're counting its contribution once by seeing what its energy is in the potential due to all other charges, and once more, when we're counting the contributions of all other charges, while taking into account the potential due to the first charge).

The major block I'm experiencing here is that the energy density involves the microscopic E field, so why does the expression change in the first place? The one thing close to an explanation friends and I came up with, is that the dielectric medium has forever been in its place, and that the energy required to construct the dipoles (The total potential energy of the dielectric minus the energy required to align them) is something we didn't have to pay for, but that the original integral does include.

EDIT:
I finally found a place dealing with the subject properly in Jackson 4.7, and it elucidated why that formula isn't valid inside a dielectric. It was derived assuming you brought the charge in bit by bit against the then-existing electric field. But the work we do in constructing the system within a dielectric includes bringing in the charges, as well as polarizing the material.

 

[ideasrule]

You get from $$\tfrac{1}{2}\int \rho \phi dV$$ to $$u=\dfrac{\epsilon_0}{2}E^2$$ by using Gauss' law, which is that the divergence of E is ρ/epsilon_0. ρ is the total charge density, which is the sum of free charge density and polarization density. To rewrite this equation in terms of free charge density, we replace epsilon_0 by k*epsilon_0, because the dielectric constant represents how much total charge density exceeds free charge density.

In this case, and in most other cases we care about, we don't want to include polarization density. For calculating the potential energy, we certainly don't want to include it because induced dipoles contribute very little potential energy. That's why we get factors of k everywhere when talking about dielectrics.

 

[RoyalCat]

You get from $$\tfrac{1}{2}\int \rho \phi dV$$ to $$u=\dfrac{\epsilon_0}{2}E^2$$ by using Gauss' law, which is that the divergence of E is ρ/epsilon_0. ρ is the total charge density, which is the sum of free charge density and polarization density. To rewrite this equation in terms of free charge density, we replace epsilon_0 by k*epsilon_0, because the dielectric constant represents how much total charge density exceeds free charge density.

In this case, and in most other cases we care about, we don't want to include polarization density. For calculating the potential energy, we certainly don't want to include it because induced dipoles contribute very little potential energy. That's why we get factors of k everywhere when talking about dielectrics.

I find it a stretch that we're simply ignoring the contribution of the dipoles to the potential energy. There must be some other explanation as to why we're only considering free charge in the integration.

The matter isn't how to rewrite the equation in terms of the free charge, but why it must be written in terms of the free charge, and why writing it in terms of the total charge is a mistake. This is what I cannot understand.
All the textbooks and derivations I could find (Jackson's derivation and one that follows it in another place) take it for granted that the charge density added to the system, $$\delta \rho$$ is one of free charge, and don't care to explain why it must be so, and why when adding charge to the system we do not consider the work done in changing the bound charge (Be it its position and thus the potential affecting it, or the orientation of dipoles relative to the electric field), nor do they care to explain why this work in constructing the system is ignored.

It is clearly evident that the energy stored in a capacitor with a dielectric is different than that without, for a given voltage. And the energy density functions there differ by exactly the factor of relative permittivity. I find it hard to believe that it is because the energy went some place we can neglect it, such an explanation makes little sense to me.

 

[tiny-tim]

It's never been clear to me where the bound charges are in a polarised dielectric.

If a conductor is placed between a parallel-plate capacitor whose plates have charge ±Q, then free charges inside the conductor move to the "top" and "bottom" surfaces to give them a charge of ± Q also, and the capacitor effectively becomes two capacitors in series, and the capacitance increases.

But if instead an insulator (a dielectric) is placed there, then there are no free charges to move, and instead the bound charges inside the insulator, ie the dipole moments, re-orient themselves to oppose the field.

Am I right in thinking that in a linear isotropic material, the dipole moment density inside the material is zero (because divP is proportional to divE which is 0), and the only dipole moments are on the "top" and "bottom" surfaces of the material?

And that if the the material has, say, susceptibility 9 and relative permititivity 10 (and fills the whole gap), then that surface charge is ±0.9Q?

Suppose a capacitor, with a fixed applied voltage, has charge ±1, and energy 1 "unit".

Now insert a dielectric material with relative permititivity 10 … that multiplies the capacitance by 10, and the charge becomes ±10, and the energy becomes 10 units (energy = QV/2).

But doesn't the dielectric material have an opposing surface charge of ±9?

Then why isn't the energy situation the same as if the total surface charge was still ±(10 - 9) = ±1?

For example, with the dielectric material in position but with the bound charges somehow "pinned down" so that they can't move, increase the plate charge to ±10 (with no fixed voltage): that increases the energy (Q²d/2ε₀A) to 100 times what it would be with Q = ±1 ("100 units");

then keep the plate charge at ±10 (presumably by disconnecting them), and "unpin" the bound charge so that its charges move the opposite way, so reducing the energy …

obviously it should reduce it by 90 units, so that the total charge is 10 units …

but (looking only at this second part) how do we get 90 out of the maths?? ​

EDIT: I forgot to mention: I found useful online lecture notes by Richard Fitzpatrick of TexasU at http://farside.ph.utexas.edu/teaching/em/lectures/node69.html#e6.5" etc