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The following text is from the book Concepts of Physics by Dr. H.C.Verma, from the chapter on "Capacitors", page 154, under the topic "Energy in the Electric Field in a Dielectric":

Consider a parallel-plate capacitor with a dielectric of dielectric constant $K$. The energy stored in the capacitor is $U=\frac 1 2 CV^2$. The energy density in the volume between the plates is

$$u=\frac{U}{Ad}=\frac{\frac 1 2 \left(\frac{K\epsilon_0A}{d}\right)V^2}{Ad}=\frac 1 2 K\epsilon_0\left(\frac V d\right)^2=\frac 1 2 K\epsilon_0E^2$$

where $E=V/d$ is the electric field between the plates.

We see that the energy density in the dielectrics is greater than that in vacuum for the same electric field. The dipole moments interact with each other so as to give this additional energy.

I understood the mathematical part of the above text, according to which, energy density in a dielectric ($u=\frac 1 2 K\epsilon_0E^2$) is greater than that in vacuum ($u=\frac 1 2 \epsilon_0E^2$) by a factor of $K$. A unit volume of dielectric stores an excess energy equal to $\frac 1 2 (K-1)\epsilon_0E^2$ other than what is stored by vacuum.

How is this excess energy in the same amount of volume stored? According to the author the reason is "The dipole moments interact with each other so as to give this additional energy", but I'm unable to understand this statement. I know dipole moment vectors in a dielectric try to align themselves with the external electric field applied. Thermal forces tend to destroy this kind of uniformity. But, how do dipole moments interact to increase the energy density and where is the energy coming from? I believe in the Law of Conservation of Energy and so I think some other form of energy is converted to electrical potential energy. But why does this happen?

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In a comment on another answer

If dielectrics tend to diminish the field due to the presence of induced dipole moments, how can field lines propagate more effectively than in vacuum?

This is true if we took an isolated capacitor and then put in the dielectric. i.e. if there was a fixed amount of charges on the plates, then yes the dielectric would thus decrease the electric field between the plates, lower the potential difference between the plates, and therefore increase the capacitance of the capacitor, as I'm sure you are familiar with.

However, if we are holding $Q$ constant then we really should be using the equation $U=\frac12\frac{Q^2}{C}$ which then predicts a decrease in energy as $C$ increases. What is going on? The issue is in what is being held constant. In your book's explanation $V$, and hence $E$ is being held constant. So then when you put in your dielectric more charge ends up on the plates. Loosely speaking, field lines start on positive charges and end on negative charges. Therefore, since we have the same electric field magnitude associated with more charges, we have a larger "field density", and hence a larger energy density.

Note that this is why the book says "for the same field" rather than "for any capacitor". The claim only holds true for comparing capacitors at the same potential difference. I think the given explanation about the dipoles is suspicious though, because the dipoles are present even in the scenario where we hold the charge constant and the energy density decreases. I would say in the constant potential case the additional energy comes from the battery, as more work is done to put more charges across the same potential difference.

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    $\begingroup$ Thank you. If my understanding is correct, electric energy density must depend on the field in the medium rather than the cause. So, I've some trouble understanding this statement - "Therefore, since we have the same electric field magnitude associated with more charges, we have a larger "field density", and hence a larger energy density." At first, even I thought bringing the dipole concept is irrelevant, then I realised there is no other factor which could contribute to the increased energy density in dielectrics. $\endgroup$ – user14250 Dec 18 '19 at 7:21
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    $\begingroup$ Further, the author wants us to conclude the same for a general case rather than limiting our understanding to that of capacitors. Because, he introduced the concept of energy density due to electric field for capacitors and at last said the same applies to electric field due to any source which may or may not be capacitors. I understood rest of your answer. $\endgroup$ – user14250 Dec 18 '19 at 7:23
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    $\begingroup$ Aaron, thank you for your clarifications, I have edited my original answer and credited you in it. -NN $\endgroup$ – niels nielsen Dec 18 '19 at 17:45
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    $\begingroup$ Thank you very much. I understood your answer and comments completely. I understood why energy density increases when dielectric is placed between the plates of a capacitor maintained at constant potential difference when compared to a capacitor without dielectric maintained at the same potential difference. I understood my question on the lines of a capacitor. But still the main part of the question seems to be unclear to me. Let me condense my question : Let us forget about capacitors. We know that electrical energy in electrostatics is stored in the electric field. $\endgroup$ – user14250 Dec 20 '19 at 16:02
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    $\begingroup$ Suppose we have a certain amount of electric field $\vec E$ in vacuum. Energy density is given by $\frac 1 2 \epsilon_0E^2$. Now in a dielectric of dielectric constant $K$, the same amount of electric field $\vec E$ is maintained by some external mechanism (which we aren't interested in), the energy density is given by $\frac1 2 K \epsilon_0E^2$. Generally $K>1$ and so energy density in dielectric is higher than that in vacuum for the same amount of field intensity. Electric field lines look the same in both vacuum and in dielectric in our case since the field intensity is maintained constant. $\endgroup$ – user14250 Dec 20 '19 at 16:02
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The amount of charge that a capacitor can store at a given voltage is increased by filling the gap between its plates with dielectric material. This means that if you have two identical capacitors in parallel, connected to the same source of voltage, the one with its interplate gap filled with a dielectric will contain more stored energy than the other that has its gap filled with a vacuum.

One way to visualize this is to see that the effect of the dielectric is to trick the plates into thinking they are closer together than they really are. Moving the plates closer together puts the plates into closer communication, increasing the capacitance of the assembly. If you disconnect the plates from a voltage source and then move the plates closer together, the voltage across them will decrease. If you then reconnect them to the voltage source, current will flow into the capacitor until the original voltage is restored. But now the capacitor is holding more charge and hence is storing more energy.

My original answer seems to be causing confusion; I have hence edited it and I recommend you read Aaron Stevens' excellent clarifications in his answer, posted below.

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    $\begingroup$ Thank you. I'm unable to understand this statement "...the effect of the dipoles in the dielectric is to propagate electric field lines between the plates more effectively than a vacuum can..." because earlier, I've learnt that electric field in a dielectric is diminished by a factor $K$ dielectric constant ($\vec E =\vec{E_0}/K$ where $\vec{E_0}$ is the electric field in vacuum, $\vec E$ is the electric field in dielectric in electrostatic). If dielectrics tend to diminish the field due to the presence of induced dipole moments, how can field lines propagate more effectively than in vacuum? $\endgroup$ – user14250 Dec 18 '19 at 2:58
  • $\begingroup$ remember this: the effect of the dielectric is to trick the plates into thinking they are closer together than they really are. moving the plates closer together puts the plates into closer communication, increasing the capacitance of the assembly. If you disconnect the plates from a voltage source and then move the plates closer together, the voltage across them will decrease. If you then reconnect them to the voltage source, current will flow into the capacitor until the original voltage is restored. But now the capacitor is holding more charge and hence is storing more energy. $\endgroup$ – niels nielsen Dec 18 '19 at 17:35

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