Is capacitor energy the electrostatic energy of the whole system?

[chingel]

It is known that the energy required to charge a parallel plate capacitor (thickness $x$, surface area $S$) with a dielectric (with dielectric constant $\epsilon$) is
$$E=\frac{xQ^2}{2S\epsilon\epsilon_0}.$$

I would think that the work done goes only into the electrostatic energy between all the charges. There is a charge
$$-Q\left( 1 - \frac{1}{\epsilon} \right)$$
on the dielectric surface right next to the charge $Q$ (and opposite sign charges on the other side, on the plate and dielectric).

When I calculate the total potential energy by summing up over all charges (factor 1/2 is because we have to include each pair only once),
$$E = \frac{1}{2} \sum_{i,j} \frac{k q_iq_j}{r_{ij}},$$
then the contribution from the charge on the dielectric surface cancels out the contribution for an equivalent charge on the plate next to it, leaving a charge
$$Q_2= Q -Q\left( 1 - \frac{1}{\epsilon} \right) = \frac{Q}{\epsilon}$$
on both plates with a remaining contribution. So it seems to be like a capacitor with this new amount of charge on the plates (with no dielectric), which would have total energy
$$E_2 = \frac{xQ^2}{2S\epsilon^2\epsilon_0}.$$

However, this is different from the known answer. What is wrong with this reasoning?

 

[vanhees71]

The energy in the charged capacitor is
$$E=\frac{C}{2} U^2.$$
The capacity is $C=\epsilon_r C_{\text{vac}}$. To charge the capacitor to a given voltage you need more energy with the dielectricum included, because you need not only energy to transport charges in the vacuum from one plate to the other but also to polarize the medium. This polarization energy is
$$E_{\text{pol}}=\frac{1}{2}C_{\text{vac}}(\epsilon-1)U^2.$$

 

[chingel]

If I think in terms of the charge on the plate, then the energy is smaller with the dielectric inside
$$E = \frac{Q^2}{2C} = \frac{Q^2}{2\epsilon C_\text{vac}}. $$ It seems to me that some of the energy of the charge on the dielectric surface cancels out the energy of the charge on the capacitor plate, because they are right next to each other and with opposite sign.

But if I think of the total charge (dielectric + plate) on one side, which would be $$Q_2=Q/\epsilon,$$ then the total energy would seem to me to be
$$E = \frac{Q_2^2}{2 C_\text{vac}} = \frac{Q^2}{2\epsilon^2 C_\text{vac}},$$ because it is the total potential energy between this net charge $Q_2$ on both sides and so it is like for a parallel plate capacitor in vacuum. However this is not the right answer, but why?

 

[anuttarasammyak]

When I calculate the total potential energy by summing up over all charges (factor 1/2 is because we have to include each pair only once),

Then we should consider potential energy between charges in the same plate which have same sign as well as between charges in the opposite plate which have different sign. It seems complicated to sum up.

[EDIT]I am not certain how you treat energy minus sign for interaction of plus charges on one plate and minus charges on the other plate.

 

[vanhees71]

For fixed charge, $Q$, you need a smaller potential difference between the plates with the dielectricum compared to the same capacitor without it. That's why you need less energy to charge the capacitor to the given amount of charge $Q$ with the dielectricum than without it.

 

[chingel]

I can calculate the potential energy of a charge on the capacitor by calculating the work done by the electric field when I take it to a plane which goes through the middle of the plates that is parallel to them (because on this plane the electric field is normal to the plane by symmetry and then taking the charge to infinity on this plane requires no work).

Let's say the top plate of the capacitor is at a potential $U$ compared to the bottom one. Then any charge on the top plate has compared to infinity potential $U/2$ and on the bottom plate the potential is $-U/2$ (because that is the potential difference to the parallel plane through the middle of the plates). So if the total net charge on the top plate and dielectric surface is $Q_2=Q/\epsilon,$ where $Q$ was the charge on the conducting plate, then the total electrostatic energy would be $$E=\frac{1}{2} \left( Q_2 \frac{U}{2}+ (-Q_2)\frac{(-U)}{2} \right) = \frac{Q_2U}{2} = \frac{QU}{2\epsilon}.$$
Since this is not equal to $QU/2$, which is the typical capacitor energy formula with a dielectric, it seems that the energy formula of the capacitor doesn't include all the potential energy of all the charges (including the polarization charges), but why?

 

[vanhees71]

The energetics is that if the capacitor is charged to a voltage $U$ and you want to transport a further "infinitesimal charge" $\mathrm{d} Q$ you need an energy
$$\mathrm{d} E=\mathrm{d} Q U = \mathrm{d} Q \frac{Q}{C} \; \Rightarrow \; \frac{\mathrm{d} E}{\mathrm{d} Q} = \frac{1}{C} Q,$$
Defining the energy to be $0$ for an uncharged capacitor this is integrated to
$$E=\frac{1}{2C} Q^2=\frac{C}{2} U^2.$$

 

[hutchphd]

Since this is not equal to QU/2, which is the typical capacitor energy formula with a dielectric, it seems that the energy formula of the capacitor doesn't include all the potential energy of all the charges (including the polarization charges), but why?

Because the relationship between free charge Q and V (i.e. its capacity) is changed by the presence of the polarizability. You cannot mix the two systems using C values selectively at your whim. With the dielectric, the Capacity to hold free charge at a given voltage increases.
One also needs to define whether the process of "adding" polarizeability is for fixed (free) charge or fixed voltage as mentioned.

 

[TSny]

Consider these diagrams which show the same parallel plates under different conditions.

In diagram (A) there is no dielectric and the plates are given free charge $\pm Q_f$.

In diagram (B) there is a dielectric material between the plates with dielectric constant $K$. The plates are charged with the same free charge $\pm Q_f$ as in (A). Polarization of the dielectric material will cause some bound charge $\mp Q_b$ to appear on its upper and lower surfaces as shown.

In diagram (C) there is no dielectric. The plates have again been given the same free charge $\pm Q_f$ as in (A) and (B). In addition, charge $\pm Q_b$ has somehow been “stuck” to the inner surfaces of the plates as shown. $Q_b$ in (C) is selected to be the same as in (B). So the distribution of free and bound charge is the same in (B) and (C).

The electric field $E$ between the plates is indicated in each diagram. (B) and (C) have the same field, which is less than the field in (A) by a factor $1/K$.

The energy stored in the system in each case is shown. I let $W_0$ represent the energy stored in (A). The energy $W_0$ is the work required by an external agent to bring the free charge $\pm Q_f$ bit by bit from infinity and add it to the plates. (Or, if you prefer, the external agent could transfer $Q_f$ from the lower plate to the upper plate. I will think of the agent as bringing in charge from infinity.)

The energy stored in (B) is less than (A) by a factor $1/K$. So, the external agent does less work in bringing in $\pm Q_f$ in (B) than in (A). Note that the agent does not need to bring in the bound charge. The bound charge appears “on its own” due to polarization of the dielectric. Nevertheless, as the agent brings in bits of free charge, the work required is affected by the presence of the bound charge. So, the total work to bring in $\pm Q_f$ is different for (A) and (B).

The energy is least in scenario (C). In this case the external agent not only has to bring in the charge $\pm Q_f$ from infinity, but also must bring in the charge $\pm Q_b$. The total work to bring in both $\pm Q_f$ and $\pm Q_b$ turns out to be less than the work in (A) by the factor $1/K^2$.

The distributions of the charges $\pm Q_f$ and $\pm Q_b$ are the same in (B) and (C). So, the electrostatic energy associated with these charges is the same in (B) and (C) and is equal to $W_0/K^2$, which is the total energy of (C). The total energy of (B) is $W_0/K$ and is, therefore, larger than the electrostatic energy of the system of charges $\pm Q_f$ and $\pm Q_b$. As @vanhees71 pointed out in post #2, the extra energy in (B) is energy associated with the polarized molecules in the dielectric.

A very nice treatment of energy in dielectrics is given in section 4.4.3 of Griffiths’ Introduction to Electrodynamics.

 

[chingel]

Thanks for the responses.

I understand that the actual total potential energy of all the charges would be the total work done on the system, but we use the average field inside the dielectric, which removes some of the field (and so potential) that a charge of every dipole feels. Actually every charge of every dipole should feel a zero electric field to not move, but the average field we use is clearly not zero.

So in this case this force keeping the charges from moving is like an extra force and to move the charges we have to do work against this force.

If the charge density is $\rho$ near some small volume $dV$ (let's say density of the positive charges of the dipoles), then they would feel a force $$\vec{F}=\rho \vec{E}\,dV,$$ due to the averaged out electric field $\vec{E}$, which is balanced out by the so to say extra force, because the charges aren't moving. The work done against this force is $$W = \int \vec{F}\,d\vec{d}= dV \int \vec{E}\rho\,d\vec{d},$$ where $\vec{d}$ is the displacement of the charges in a dipole. If the dielectric is linear, $$\vec{P}=\epsilon_0\chi \vec{E}=\rho \vec{d},$$ which we can use to replace $$\rho d\vec{d} = \epsilon_0\chi d\vec{E}$$ in the work relation: $$W = \epsilon_0\chi dV\int \vec{E}d\vec{E} = \frac{\epsilon_0\chi E^2}{2}dV.$$ This we would have to integrate over the whole dielectric volume. So then the total work to charge the capacitor would be (using $\chi=\epsilon-1$) $$W_T = \frac{xQ^2}{2S\epsilon^2\epsilon_0}+ \int \frac{\epsilon_0(\epsilon-1) E^2}{2} dV,$$ which would give the correct result (using $E=Q/(S\epsilon\epsilon_0)$ and $V=Sx$) $$ W_T = \frac{xQ^2}{2S\epsilon\epsilon_0} $$ and explain in detail what I was missing in my calculation before.