MODELING MECHANICAL AND ELECTRICAL
2.1 FUNDAMENTAL RELATIONSHIPS IN MECHANICS AND ELECTROSTATICS
2.1.4 Electronic Constitutive Properties of Conducting and Insulating Materials
For a surface of finite area, the total current passing through the surface is equal to the integral over the surface, or
i =
Surf
J·ds, (2.57)
where the time dependence of the current is implicit in the expression. If we integrate over a closed surface, the continuity of charge specifies that
Surf
J·ds= −
Vol
∂ρv
∂t dVol. (2.58)
The negative sign can be understood by considering that the unit normal of the surface is defined as outward; therfore, a position current flow through the closed surface implies that the volume charge isdecreasing. The negative sign in equation (2.58) reflects this definition.
The divergence theorem is applied to write both sides of equation (2.58) as a volume integral,
Vol
∇ ·JdVol= −
Vol
∂ρv
∂t dVol. (2.59)
which yields the definition of continuity of charge at a point,
∇ ·J= −∂ρv
∂t . (2.60)
2.1.4 Electronic Constitutive Properties of Conducting
or an insulator. These expressions are derived by incorporating a constitutive law that defines the relationship between the electrical state variables of the system.
In a conductor the current density is related to the electric field through thecon- ductivityof the material,σ:
J=σE. (2.61)
Equation (2.61) is a statement of Ohm’s law expressed between current per area and electric field. The material parameterσ specifies the quality of the conducting material. As equation (2.61) demonstrates, higher conductivity indicates that a larger current flow will occur for a prescribed electric field. Conducting materials are used in all types of electronics applications. Examples of good conductors, those that have high conductivity, are copper and brass. These materials are often used as wiring and interconnects in electronics. In the field of smart materials, shape memory alloys are good conductors, and we will see that we can use a material’s conductivity as a stimulus for thermomechanical actuation.
Insulators do not contain a large proportion of mobile charge but are very useful in electronic applications and, as we shall see, smart material applications. Insulators are characterized by an atomic structure that contains a large proportion ofbound charge, charge that will not conduct through a material but will reorient in the presence of an applied electric field. Bound charge can be visualized as a pair of point charges of equal and opposite chargeQseparated by a distanced. At the center of the distance connecting the charge pair is a pivot, which is similar to a mechanical pivot but allows the charges to rotate about the point (Figure 2.8).
The pair of point charges separated by a distance is called anelectronic dipole, or simply adipole. Placing a dipole in an electric field will produce a motion of both charges butwill not result in charge conduction, due to the fact that the charges are bound. Thedipole momentassociated with this pair of point charges is defined as
p=Qd, (2.62)
where the vectordis defined as the vector from the negative charge to the positive charge. If there arendipole moments per unit volume, thepolarizationof the material
+ –
E
E
+
– +
– + –
+ –
+ –
+ –
+ –
+ –
+ – + –
+
– +
– + –
Figure 2.8 Representation of bound charge within a dielectric insulator as a group of dipoles.
is defined as
P=lim 1 Vol
n i=1
pi. (2.63)
The units of the dipole moment are C·m; therefore, the polarization (which is the number of dipole moments per unit volume) has units of charge per area, C/m2.
Once again let us define a surface with unit normaldsthat is pointing outward.
Imagine that the application of an electric field causesn dipole moments to rotate such that charge crosses the surface. The differential amount ofbound charge, Qb, that crosses the surface is equal to
d Qb =n Qd·ds=P·ds. (2.64) If we perform the integration over a closed surface, the total bound charge within the closed surface is equal to
Qb= −
Surf
P·ds. (2.65)
The question that we want to answer is: How does the existence of a polarization within the material change our expressions for the electrostatics of the material? To answer this question, consider a volume of material that contains both bound charge, Qb, and free (or mobile) charge,Q. The total charge in the volume is
Qt =Qb+Q. (2.66)
Applying Gauss’s law to the volume to solve for the total charge, we have Qt=Qb+Q=
Surf
ε0E·ds. (2.67)
Substituting equation (2.65) into equation (2.67) and rearranging the terms yields Q=
Surf
(ε0E+P)·ds. (2.68)
Equation (2.68) is an important expression because it tells us that polarization in the material can be thought of as an additional term of the electric displacement of the material. This can be expressed mathematically by rewriting the electric displacement in equation (2.47) as
D=ε0E+P, (2.69)
which we see makes it equivalent to equation (2.68).
In many materials the relationship between polarization and electric field is as- sumed to be a linear relationship. In this case we can write the polarization as the product of a constant and the electric field vector,
P=(εR−1)ε0E, (2.70)
whereεRis defined as therelative permittivity, which is a unitless parameter that is always greater than 1. The somewhat strange way of writing the constant of propor- tionality in equation (2.70) is due to the fact that when we substitute the expression into equation (2.69), we obtain
D=ε0E+(εR−1)ε0E=εRε0E. (2.71) Defining thepermittivity of the materialas the product of the relative permittivity and the permittivity of free space,
ε=εRε0, (2.72)
we can relationship between electric displacement and electric field as
D=εE. (2.73)
This analysis demonstrates that bound charge in a material can be treated mathemat- ically as an increase in the permittivity, which, in turn, produces an increase in the electric displacement for an applied electric field. The relative permittivity is a mate- rial constant that can range from between 2 and 10 for some polymer materials to the range 1000 to 5000 for some ceramics. Smart materials that exhibit dielectric proper- ties include piezoelectric materials such as the ones we study in upcoming chapters.
Consider once again a cube of material as shown in Figure 2.2. Defining a Cartesian coordinate system as we did for the analysis of stress and strain allows us to write a relationship between the electric displacement in three dimensions and the applied electric field. This expression is written in indicial notation,
Di =εi jEj, (2.74)
where the indices range from 1 to 3. Equation (2.74) represents the constitutive relationship between electric field and electric displacement for a linear material. For an anisotropic dielectric material, equation (2.74) is written
D1=ε11E1+ε12E2+ε13E3
D2=ε21E1+ε22E2+ε23E3 (2.75) D3=ε31E1+ε32E2+ε33E3.
In an anisotropic material an electric field applied in one coordinate direction can produce electric displacement in an orthogonal coordinate direction. In an isotropic
dielectric material
εi j =0 i = j, (2.76)
and the constitutive relationship is written as D1=ε11E1
D2=ε22E2 (2.77)
D3=ε33E3.
Sometimes the constitutive relationship for an isotropic linear dielectric material is written in a compact notation that uses only a single subscript:
Dk=εkEk k=1,2,3. (2.78)
Example 2.5 The dielectric properties of an insulator material have been measured and shown to have slight anisotropy. The relative dielectric properties have been measured to be
ε11=ε22=1200 ε23=ε32=150
ε33=1800.
Compute the electric displacement to the applied electric fields (a) E=1× 106xˆ1V/m and (b)E=1×106xˆ3V/m.
Solution (a) Applying equation (2.75) and substituting in the relative permittivity values and the electric field yields
D1 =(1200ε0F/m)(1×106V/m)+(0)(0)+(0)(0)
D2 =(0)(1×106V/m)+(1200ε0F/m)(0)+(150ε0F/m)(0) D3 =(0)(1×106V/m)+(150ε0F/m)(0)+(1800ε0F/m)(0).
Performing the computations yields the electric displacement vector for the material:
D1 =0.0106 C/m2 D2 =0
D3 =0.
(b) Substituting in the electric field vector in the 3 direction produces D1=(1200ε0F/m)(0)+(0)(0)+(0)(1×106V/m)
D2=(0)(0)+(1200ε0F/m)(0)+(150ε0F/m)(1×106V/m) D3=(0)(0)+(150ε0F/m)(0)+(1800ε0F/m)(1×106V/m),
which yields
D1=0
D2=0.0013 C/m2 D3=0.0159 C/m2.
This example illustrates that the electric field applied in the direction that is isotropic (in this case, the 1 direction) produces electric displacement in the same direction.
Applying the field in a direction that exhibits anisotropy (the 3 direction) produces an electric displacement vector that has components in orthogonal directions.