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Thermodynamics of Fluids in Electric Fields

by J. M. Haile

When a fluid is subjected to an electric field, many properties of the fluid change relative to their values when no field is present; in general, the changes become more dramatic with increasing field strength. The purpose of this article is to explain those changes. To keep the discussion simple, we discuss the phenomena within the context of an ideal, parallel-plate capacitor. An external electric field can do work on a fluid, and we show how that work appears in the first law of thermodynamics.

1. Parallel-Plate Capacitors

Consider two metallic plates, each of area A, aligned parallel to one another and separated by distance r, which is small relative to the width of one plate. Evacuate the space between the plates and attach a battery to each, charging one plate with a positive charge of magnitude Q, the other with a negative charge of the same magnitude. Such a device is called a capacitor (or a condenser in older literature); see Figure 1. The charges create an electric field between the plates and the plates exert attractive forces on one another.

Figure 1

Figure 1. Schematic of a parallel-plate capacitor with a vacuum between the plates. In this drawing, the distance between plates, r, has been exaggerated for clarity. We assume that the field Eo is uniform between the plates; that is, we neglect end effects on the field.

The field Eo between the plates is the surface charge density, that is, it is the charge per unit area on one plate,

Eo = Q/A

(1.1)

We use subscript "o" to indicate properties for the situation in which a vacuum exists between the plates. The field Eo has dimensions of force per unit charge. We use the electrostatic system of units (esu), so in (1.1), Q has units of (esu of charge), A is in square cm, and Eo is in dyne/(esu of charge). In SI units, the permittivity of the vacuum would appear in the denominator of (1.1); then Q would be in Coulomb (C), A in square meters, and Eo in Newtons/Coulomb.

1.1 Capacitance

The voltage Vo between the plates is the work per unit charge needed to move a small amount of charge from one plate to the other; it is obtained by integrating the electric field across the distance between plates. We are assuming the field is uniform with r, so the integral reduces to

Vo = r Eo = r Q/A

(1.2)

where Vo is in units of erg/(esu of charge). Equation (1.2) is often written as

Q = Co Vo

(1.3)

where

Co = A/r

(1.4)

is called the capacitance; it has units of (esu of charge)2/erg = cm. Equation (1.3) asserts that the voltage across a parallel-plate capacitor is proportional to the charge. Capacitors are used as devices for storing charge. For example, if we make a capacitor with a large area A and small separation r, then its capacitance is large; such a device can store a large charge while developing a relatively small voltage.

1.2 Energy of a Capacitor

The energy Uo stored in a parallel-plate capacitor is the work required to charge the capacitor. At a particular value of the voltage, the work to increase the charge by a small amount dQ is

dUo = Vo dQ

(1.5)

Using Vo = Q/Co from (1.3), this becomes

dUo = Q dQ /Co

(1.6)

where the capacitance Co is a constant for a particular device. Integrating (1.6) yields

Uo = Q2 /2Co = CoVo2/2

(1.7)

Thus, for a particular device, for which Co is fixed, the energy increases with the square of the charge or the square of the voltage.

1.3 Force Between Plates

The force between the plates is the negative work per unit length required to increase the separation between the plates,

F = –dU /dr

(1.8)

For a fixed amount of charge, substituting (1.7) into (1.8) gives

Eq.(1.9)

(1.9)

And using (1.4) for C, (1.9) yields

Eq.(1.10)

(1.10)

The negative sign indicates that the force between the plates is attractive.

Consider a situation in which the area A of each plate is fixed and we remove any electrical connections between the charged plates and other objects, so the charge Q on each plate remains constant. Then (1.10) states that the force is independent of the separation r. This is because we have assumed the electric field Eo in (1.1) is uniform, independent of r.

2. The Dielectric Constant

In the capacitors described in Section 1, the plates were separated by a vacuum, and we reminded ourselves of this by adding a subscript "o" to properties of the capacitor. Now consider filling the space between the plates with a gas or liquid dielectric—a substance that is a poor conductor of electricity.

2.1 Effects of a Dielectric on the Capacitor

If we keep the charge Q fixed and measure the voltage across the plate, we find it is reduced by a factor e > 1, from Vo to V,

V = Vo/e = rQ/eA      (dielectrics)

(2.1)

By (1.3) if Q is constant and the voltage decreases, then the capacitance must increase by the same factor,

C = eCo = eA/r

(2.2)

Thus, large commercial capacitors are filled with an oil to enhance their ability to store charge. The factor e is called the dielectric constant,

e = C/Co

(2.3)

The dielectric constant is a dimensionless function of state; it depends on the temperature, pressure, composition, and phase of the dielectric material. On substituting (2.2) into (1.7), we see that, for constant Q, adding a dielectric between the plates decreases the energy of the capacitor by e

U = Uo/e

(2.4)

Similarly, (2.2) plus (1.10) shows that the force also decreases by e

F = Fo/e

(2.5)

2.2 Effects of the Capacitor on the Dielectric

But what happens in the dielectric fluid that reduces the voltage from Vo to V, energy from Uo to U, and the force from Fo to F?

When a fluid fills the space between the charged plates of a parallel-plate capacitor, the external field Eo tends to pull positive charges in the fluid toward the negative plate and negative charges toward the positive plate, as in Figure 2. The dielectric fluid is said to be polarized. The field Eo is that created by the charges on the plates and is the same regardless of whether or not a dielectric is present [1]. If we take the space between the plates to be our system, then the external field Eo constitutes an independent intensive variable whose value affects thermodynamic properties of the system.

Figure 2

Figure 2. Schematic of a parallel-plate capacitor with a dielectric fluid between the plates. The charges on the plates create an external field Eo that polarizes the fluid. The polarization is a second field that opposes Eo, so that the net field in the fluid is reduced from Eo to E.

Without the fluid, the voltage and field are related by (1.2), Vo = rEo; likewise, in the presence of the fluid, V = rE. Using these in (2.1) gives

E = Eo/e

(2.6)

For the same charge on the plates, the dielectric fluid reduces the field strength by a factor of the dielectric constant.

But when a dielectric fluid fills the space between the plates, the polarization of the dielectric creates a field that opposes the external field; it is conventional to call this field the polarization and represent it by 4πP. Then, at any point in the fluid between the plates, the net field is

E = Eo – 4πP

(2.7)

If we can determine the polarization P, then we use (2.6) and (2.7) to obtain the dielectric constant e. The contributions to the polarization are affected by the kinds of molecules in the fluid and the thermodynamic state of the fluid.

3. Basic Thermodynamics

Consider a parallel-plate capacitor as described in Section 1 with A the area of each plate and r the distance between plates. The capacitor is immersed in a fluid having dielectric constant e. The fluid between the plates is the system; it has volume Ar. This system is open to mass transfer with the surrounding fluid; the system is in equilibrium with that surrounding fluid and we assume no electric fields are present in the surrounding fluid.

3.1 First Law

The electrical energy stored in the system is the work required to charge the capacitor from 0 to Q. As the charge on the plates increases, the voltage V and field strength E increase, increasing the polarization P of the dielectric. The energy added to the system by charging is stored in the field and the polarization. Recall that E and P are both fields, but their product has dimensions of energy on a unit volume basis:

(force/charge) × (force/charge) = (charge/area) × (force/charge)
= energy/volume

In writing an expression for the work, some authors use the electric field E as the generalized force and use the polarization P as the generalized displacement; others reverse the roles of these two variables. Instead of either of these choices, we follow Davidson [2] and use the voltage V as the generalized force and the charge Q as the generalized displacement. The voltage is sometimes called the potential or the electromotive force (EMF).

Therefore, we write the work done by a battery in adding a small amount of charge dQ to the capacitor as

dW  =  V dQ

(3.1)

This is work done on the system, and we adopt a sign convention in which work done on a system is positive. Therefore, the first law for the change in total energy of the system is

dU  =   T dS + V dQ + μi dN

(3.2)

where μi is the chemical potential of the dielectric. During the charging process, the volume of the system is constant, so no volume term appears. If we prefer temperature and voltage as independent variables, we can use d(TS) = TdS + SdT and d(VQ) = VdQ + QdV to write

d(U – TS – QV)  =  –S dT – Q dV + μi dN

(3.3)

Using (2.1) and (2.2), we can replace the charge Q on the rhs in favor of the voltage V,

d(U – TS – QV) = –S dT – eCoV dV +μi dN

(3.4)

where Co = A/r is the capacitance when the fluid is removed.

3.2 Maxwell-like Relations

The result (3.4) yields two instructive Maxwell-like relations. One is

Eq.(3.5)

(3.5)

Experimentally we find that the dielectric constant e usually increases with increasing density, so

Eq.(3.6)

(3.6)

This means that, as the capacitor is charged isothermally, the chemical potential of the system decreases, leading to

μi, sys < μi, sur

(3.7)

This creates a driving force for diffusion so that, as the voltage increases, mass moves from the surrounding fluid into the system. Thus, isothermal charging increases the density of fluid between the plates.

The second instructive relation is

Eq.(3.8)

(3.8)

For nonpolar fluids, e is nearly independent of T, and then S is nearly constant. But for polar fluids, e decreases with increasing temperature, and then

Eq.(3.9)

(3.9)

Thus, isothermal charging causes a decrease in the entropy of a polar fluid. Such a decrease can only occur by heat transfer, so to keep T constant, heat must be transferred from the system to the surrounding fluid.

We note that the pressure P of the system is imposed by the surrounding fluid. Thus, (3.9) can be written as

Eq.(3.10)

(3.10)

We also know that most systems have

(∂S/∂P)T < 0;   hence,   (∂P/∂V)T > 0

That is, for polar dielectrics, the pressure increases on isothermal charging. The inverse relation must also apply,

Eq.(3.11)

(3.11)

so if we isothermally increase the pressure of a polar dielectric, the voltage between the plates increases. This is called the piezoelectric effect, first discovered by the Curie brothers in 1880 [3]. Typically, the voltage increases by a few hundredths of a volt when the pressure is increased by 1 bar. This effect provides a way for converting mechanical action into electrical signals; such piezoelectric transducers often employ a quartz crystal as the dielectric. They are widely used in pressure gages, strain gages, microphones, etc.

Literature Cited

[ 1 ] T. L. Hill, An Introduction to Statistical Thermodynamics, Addison-Wesley, Reading, MA, 1960.

[ 2 ] N. Davidson, Statistical Mechanics, McGraw-Hill, New York, 1962.

[ 3 ] P. W. Frosbergh, Jr., "Piezoelectricity, Electrostriction and Ferroelectricity," in Encyclopedia of Physics, vol. 17, S. Fluegge, ed., Springer-Verlag, Berlin, 1956.