MODELING MECHANICAL AND ELECTRICAL
2.1 FUNDAMENTAL RELATIONSHIPS IN MECHANICS AND ELECTROSTATICS
2.1.3 Electrostatics
S5=0
S6= (2)(1+0.3)
62×109Pa2×106Pa, which equals
S1 =61.2×10−6m/m S2 =40.3×10−6m/m S3 = −43.5×10−6m/m S4 =0
S5 =0
S6 =83.9×10−6m/m.
As stated earlier, the unit 1×10−6m/m is called amicrostrain(µstrain).
Alternative forms of the governing equilibrium equations are obtained by com- bining equation (2.25) with the constitutive laws for a material. Assuming that the material is linear elastic, the relationship between stress and strain is expressed by equation (2.31). Substituting this expression into equation (2.25) yields
LucS+fV =0. (2.38)
The strain–displacement equation, equation (2.15), is now substituted in equa- tion (2.38) to produce
LucLuu+fV =0. (2.39)
The equilibrium expressions are now expressed in displacement form. Equation (2.39) represents a three-by-three set of equations that when solved yield a displacement vector that satisfies the equations of equilibrium. The boundary conditions expressed in equation (2.27) must also be satisfied for the solution to be admissible.
1 2
3
Q1 Q2
r1 r2
r2–r1
F2 R12 = |r2–r1|
Figure 2.4 System of two point charges showing the electrostatic force vector.
electric fields. In our discussion of electrostatics we use the same procedure as in our discussion of mechanics. First we define the basic quantities of electrostatics and then define the governing equations that enable the solution of specific problems.
Let us begin by defining charge as the fundamental electrical quantity. In its sim- plest form, consider a point charge defined at a location in space in a Cartesian coordinate system. The size of the point charge is quantified by the amount of charge, Q1, where charge is specified in the unit coulombs, which is given the symbol C. The vector from the origin of the coordinate system to the point charge is denotedr. Now consider the case where there are two point charges of size Q1 and Q2 located at pointsr1andr2, respectively, as shown in Figure 2.4. Coulomb’s law states that the magnitude of the force between any two objects in free space separated by a distance much larger than their dimension is
f = 1 4πε0
Q1Q2
R2 , (2.40)
whereR2is the square of the distance between objects. The variableε0, thepermittivity of free space, has the value 8.854×10−12 F/m. For the point charges shown in Figure 2.4, the force on point charge 2 due to the existence of point charge 1 is
f2= Q1Q2
4πε0R122
r2−r1
|r2−r1|. (2.41)
The force vector on point charge 2 is shown in Figure 2.4 for the case when the sign of the charges is the same. The force on point charge 1 is equal and opposite to the force on point charge 2; therefore,f1= −f2.
For the moment let us define point charge 2 as atest chargeand move the charge around the free space while keeping point charge 1 fixed. The force induced on point charge 2 is defined by equation (2.41). The magnitude of the force is proportional to the inverse of the distance between the charges squared and is always in the direction
1 2
3
Q1 ds E
Figure 2.5 Electric field flux through a differential surface element.
of a vector that points from the location of Q1 toQ2. We define the electric field intensity, or simply theelectric field, as the forcef2 normalized with respect to the size of the test charge, which in this case isQ2. Thus,
E= f2
Q2 = Q1 4πε0R122
r2−r1
|r2−r1|. (2.42)
The electric field has units of force per unit charge, N/C. Shortly we will define the unitvolts(V), which is equivalent to the product of force and distance per unit charge, or N·m/C. Using these definitions, we can also define the units of electric field as volts per unit length, or V/m.
The electric field produced by a unit of charge is a vector quantity whose magnitude is proportional to the size of the point charge and inversely proportional to the distance from the point charge. The electric field vector points directly away from the point charge location along a radial line with the center atr1. The electric field generated by a point charge can be visualized as a set of electric fieldlinesthat emanate from the location of the charge. If we specify a surface at some distance away from the point charge and count the number of electric field lines that cross the surface (Figure 2.5) and normalize the result with respect to the surface area, we can define theelectric flux intensity, orelectric displacement, as the vectorD. The direction of the vectorD is the direction of the electric field lines that cross the surface. Faraday demonstrated that the electric displacement is related to the electric field in a free space through the expression
D=ε0E= Q1 4πR212
r2−r1
|r2−r1|. (2.43)
Consider a system that consists of multiple point charges Qi located within a volume. The total charge is denotedQand is expressed as
Q=
NQ
i=1
Qi. (2.44)
Coulomb’s law is linear and therefore the electric field and electric displacement at any pointxwithin the space can be expressed as a summation of the electric fields and electric displacements due to each individual point charge:
E(x)=
NQ
i=1
Qi
4πε0R2i x rx−ri
|rx−ri| (2.45)
D(x)=
NQ
i=1
Qi
4πR2i x rx−ri
|rx−ri|. (2.46)
Example 2.3 A fixed charge of+Q coulombs has been placed in free space at x1=a/2 and a second fixed charge of−Qcoulombs has been placed atx1= −a/2.
Determine the electric field at the location (a/2,a/2).
Solution Since Coulomb’s law is linear, we can compute the electric field to each charge separately and add them to obtain the combined electric field at the point.
This is stated in equation (2.46). Defining the negative charge asQ1and the positive charge asQ2, the position vectors for the problem are
r1 = −a 2xˆ1
r2 = a 2xˆ1
rx = a 2xˆ1+a
2xˆ2.
Computing the electric field from the negative charge located at (−a/2,0) requires computation of
rx−r1 = a 2xˆ1+a
2xˆ2−
−a 2xˆ1
=axˆ1+a 2xˆ2. The magnitudeR1xis
|rx−r1| =R1x=
a2+a2 4 =a√
5 2 .
The direction of the electric field from charge 1 (the negative charge) to the test point is
rx−r1
|rx−r1| = axˆ1+(a/2) ˆx2
a√ 5/2
= 2
√5xˆ1+ 1
√5xˆ2. The electric field at the test point due to charge 1 is
E1x = −Q 4πε0(5a2/4)
2
√5xˆ1+ 1
√5xˆ2
= −Q a2πε0
2 5√
5xˆ1+ 1 5√
5xˆ2 .
The electric field due to charge 2 can be found in the same manner, or, by inspection, E2x = Q
a2πε0
ˆ x2.
The electric field is the sum of the electric field due to the individual charges,
E= Q
a2πε0
− 2 5√
5xˆ1+
1− 1 5√
5 xˆ2
= Q a2πε0
− 2 5√
5xˆ1+5√ 5−1 5√
5 xˆ2
.
If we compute the exact solution out to three decimal places, we have
E= Q
a2πε0
(−0.179 ˆx1+0.911 ˆx2).
Comparing the total electric field vector with the field vector due only to charge 2, we see that the field due to charge 1 tends to bend the electric field line back toward the negative charge. The electric field still has the largest component in the ˆx2direction, due to the fact that the positive charge is closer, but there is still a noticeable effect, due to the negative charge at (−a/2,0).
Now let us consider a system of point charges. If we draw a closed surface around this system of point charges, as shown in Figure 2.6, at any point on the surface we can define a differential unit of area whose unit normal is perpendicular to the surface at that point. We call this unit normal ds. The electric displacement that is in the direction of the unit normal at pointxis writtenD(x)·ds. One of the fundamental
2
1 3
ds D(x)
Figure 2.6 System of point charges illustrating the closed surface associated with applying Gauss’s law.
laws of electrostatics, Gauss’s law, states that the integral ofD(x)·dsoveranyclosed surface is equal to the total charge enclosed within that surface. Mathematically, Gauss’s law states that
Surf
D(x)·ds=Q. (2.47)
If we assume that the system of charges is not merely a set of discrete point charges but a continuous function of charge per unit volume,ρv(x), the total charge within a volume is the volume integral
Q=
Vol
ρv(x)dVol, (2.48)
and Gauss’s law can be restated as
Surf
D(x)·ds=
Vol
ρv(x)dVol. (2.49)
Applying the divergence theorem to the surface integral of equation (2.49),
Surf
D(x)·ds=
Vol
∇ ·D(x)dVol, (2.50)
allows us to write Gauss’s law as
Vol
∇ ·D(x)dVol=
Vol
ρv(x)dVol. (2.51)
Now that both sides of the equation are volume integrals, we can equate the integrands.
This leads to an expression of the point form of the first of Maxwell’s equations,
∇ ·D(x)=ρv(x). (2.52)
Combining equations (2.52) and (2.20) yields a less compact version of Maxwell’s first equation:
∂D1
∂x1 +∂D2
∂x2 +∂D3
∂x3 =ρv(x). (2.53)
The charge density can also be related to the electric field by recalling that elec- tric field and electric displacement are related through equation (2.43). Introducing equation (2.43) into equations (2.52) and (2.53) yields
∇ ·E(x)= ρv(x) ε0
(2.54)
∂E1
∂x1
+∂E2
∂x2
+∂E3
∂x3
= ρv(x) ε0
.
Example 2.4 The electric displacement field in free space over the range −a <
x1<aand−b<x2<bis defined as D(x)=cosh(10x1/a)
cosh(10) xˆ1+cosh (10x2/b) cosh(10) xˆ2.
Compute (a) the electric displacement vector at (a,0) and (a,b), and (b) the expression for the volume charge using the point form of Gauss’s law.
Solution (a) Evaluating the electric displacement vector at (a,0) yields D(a,0)= cosh[10(a)/a]
cosh(10) xˆ1+cosh[10(0)/b]
cosh(10) xˆ2
=xˆ1+ 1 cosh(10)xˆ2 and evaluating at (a,b) yields
D(a,b)= cosh[10(a)/a]
cosh(10) xˆ1+cosh[10(b)/b]
cosh(10) xˆ2
=xˆ1+xˆ2.
The term cosh(10) is on the order of 10,000; therefore, the electric displacement vector is approximately in the 1 unit direction at (a,0). At the corner of the region the electric displacement vector makes a 45◦angle to the horizontal axis.
E
E
J
Figure 2.7 System of volume point charges conducting through a material.
(b) The point form of Gauss’s law is shown in equation (2.52) and expanded in equation (2.53). Using the expanded form, we need to obtain the partial derivatives
∂D1
∂x1
= 10 a
sinh (10x1/a) cosh(10)
∂D2
∂x2
= 10 b
sinh (10x2/b) cosh(10) . The volume charge density is the summation of the two terms:
ρv(x)= 10 cosh(10)
sinh (10x1/a)
a +sinh (10x2/b) b
.
Thus far in our discussion we have focused on systems of fixed charges and related the charge distribution to electric field and electric displacement. Time-varying charge distributions produce electronic current through a material. Consider the material shown in Figure 2.7, which incorporates a volume charge that is depicted in the figure as a set of discrete point charges. If these charges are moving, the current is defined as the time rate of change of charge motion,
i(t)=d Q
dt . (2.55)
The units of current are C/s or amperes (A). Drawing a differential surface with unit normaldsallows us to determine the amount of current passing through this surface per unit area. We call this quantity thecurrent density, denote it asJ, and state it in units of current per area, A/m2 in SI units. The differential amount of current that passes through the surface is defined by the dot product of the current density and the differential surface element:
di=J·ds. (2.56)
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