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Electromags - Chapter 7
Electronics Engineering (CR 061)
Samar State University
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LESSON CONTENT
Current is symbolized by I, and therefore
(1)
Current is thus defined as the motion of positive charges, even though conduction in metals takes place through the motion of electrons, as we will see shortly. In field theory, we are usually interested in events occurring at a point rather than within a large region, and we find the concept of current density, measured in amperes per square meter (A/m 2 ), more useful. Current density is a vector represented by J.
The increment of current ∆I crossing an incremental surface ∆S normal to the current density is
and in the case where the current density is not perpendicular to the surface,
Total current is obtained by integrating,
(2)
Current density may be related to the velocity of volume charge density at a point. Consider the element of charge ∆Q = ρν∆ν = ρν∆S∆L, as shown in Figure 5. To simplify the explanation, assume that the charge element is oriented with its edges parallel to the coordinate axes and that it has only an x component of velocity. In the time interval ∆t, the element of charge has moved a distance ∆x, as indicated in Figure 5.1b have therefore moved a charge Q = ρν∆S∆x through a reference plane perpendicular to the direction of motion in a time increment ∆t, and the resulting current is
As we take the limit with respect to time, we have
where νx represents the x component of the velocity v. In terms of current density, we find
and in general
(3)
This last result shows clearly that charge in motion constitutes a current. We call this type of current a convection current, and J or ρνv is the convection current density. Note that the convection current density is related linearly to charge density as well as to velocity. The mass rate of flow of cars (cars per square foot per second) in the Holland Tunnel could be increased either by raising the density of cars per cubic foot, or by going to higher speeds, if the drivers were capable of doing so.
Example:
1) Given the current density J = − 104 [sin(2x)e− 2 yax + cos(2x)e− 2 yay] kA/m 2 :
a) Find the total current crossing the plane y = 1 in the ay direction in the region 0 < x < 1, 0 < z < 2: This is found through
b) Find the total current leaving the region 0 < x,x < 1, 2 < z < 3 by integrating J・dS over the
surface of the cube.
Note: current through the top and bottom surfaces will not exist, since J has no z component. Also note that there will be no current through the x = 0 plane, since Jx = 0 there. Current will pass through the three remaining surfaces, and will be found through.
- A certain current density is given in cylindrical coordinates as J= 100e-2z(ρaρ + az) A/m 2. Find the total current passing through each of these surfaces: (a) z=0, 0≤ρ≤1, in the az direction; (b) z=1, 0 ≤ρ≤1, in the az direction; (c) closed cylinder defined by 0≤z≤1, 0≤ ρ≤1, in an outward direction. a) z = 0, 0 ≤ ρ ≤ 1, in the az direction:
where aρ
・ az = 0. b)
z = 1, 0 ≤ ρ ≤ 1, in the az direction:
a) closed cylinder defined by 0 ≤ z ≤ 1, 0 ≤ ρ ≤ 1, in an outward direction:
- Continuity of Current (Hayt & Buck, 2001)
We next seek the volume charge density by integrating with respect to t. Because ρν is given by a partial derivative with respect to time, the “constant” of integration may be a function of r :
If we assume that ρν → 0 as t →∞, then K(r ) = 0, and
We may now use J = ρνv to find the velocity,
The velocity is greater at r = 6 than it is at r = 5, and we see that some (unspecified) force is accelerating the charge density in an outward direction.
In summary, we have a current density that is inversely proportional to r, a charge density that is inversely proportional to r 2 , and a velocity and total current that are proportional to r. All quantities vary as e−t.
Example: The current density in a certain region is approximated by J= (0/r) exp (-10 6 t)ar A/m 2 in spherical coordinates. (a) At t= 1μs, how much current is crossing the surface r=5? (b) Repeat for r=6. (c) Use the continuity equation to find ρv(r,t) assuming that ρv→0 as t→∞. (d) Find an expression for the velocity of the charge density.
a) At t= 1μs, how much current is crossing the surface r=5? Ia = JrS = 4π(5) 2 (0. 1 /5)e− 1 = 2πe− 1 = 2 A.
b) Repeat for r=6. Ib = JrS = 4π(6) 2 (0. 1 /6)e− 1 = 2πe− 1 = 2 A.
c) Use the continuity equation to find ρv(r, t), under the assumption that ρv → 0 as t→∞:
d) Find an expression for the velocity of the charge density.
- Metallic Conductors (Hayt & Buck, 2001)
Physicists describe the behavior of the electrons surrounding the positive atomic nucleus in terms of the total energy of the electron with respect to a zero reference level for an electron at an
infinite distance from the nucleus. The total energy is the sum of the kinetic and potential energies, and because energy must be given to an electron to pull it away from the nucleus, the energy of every electron in the atom is a negative quantity. Even though this picture has some limitations, it is convenient to associate these energy values with orbits surrounding the nucleus, the more negative energies corresponding to orbits of smaller radius. According to the quantum theory, only certain discrete energy levels, or energy states, are permissible in a given atom, and an electron must therefore absorb or emit discrete amounts of energy, or quanta, in passing from one level to another. A normal atom at absolute zero temperature has an electron occupying every one of the lower energy shells, starting outward from the nucleus and continuing until the supply of electrons is exhausted. In a crystalline solid, such as a metal or a diamond, atoms are packed closely together, many more electrons are present, and many more permissible energy levels are available because of the interaction forces between adjacent atoms. We find that the allowed energies of electrons are grouped into broad ranges, or “bands,” each band consisting of very numerous, closely spaced, discrete levels. At a temperature of absolute zero, the normal solid also has every level occupied, starting with the lowest and proceeding in order until all the electrons are located. The electrons with the highest (least negative) energy levels, the valence electrons, are located in the valence band. If there are permissible higher-energy levels in the valence band, or if the valence band merges smoothly into a conduction band, then additional kinetic energy may be given to the valence electrons by an external field, resulting in an electron flow. The solid is called a metallic conductor. The filled valence band and the unfilled conduction ban for a conductor at absolute zero temperature are suggested by the sketch in Figure 5.
If, however, the electron with the greatest energy occupies the top level in the valence band and a gap exists between the valence band and the conduction band, then the electron cannot accept additional energy in small amounts, and the material is an insulator. This band structure is indicated
in Figure 5. Note that if a relatively large amount of energy can be transferred to the electron, it may be sufficiently excited to jump the gap into the next band where conduction can occur easily. Here the insulator breaks down. An intermediate condition occurs when only a small “forbidden region” separates the two bands, as illustrated by Figure 5. Small amounts of energy in the form of heat, light, or an electric field may raise the energy of the electrons at the top of the filled band and provide the basis for conduction. These materials are insulators which display many of the properties of conductors and are called semiconductors. Let us first consider the conductor. Here the valence electrons, or conduction, or free, electrons, move under the influence of an electric field. With a field E, an electron having a charge Q = −e will experience a force
In free space, the electron would accelerate and continuously increase its velocity (and energy); in the crystalline material, the progress of the electron is impeded by continual collisions with the thermally excited crystalline lattice structure, and a constant average velocity is soon attained. This
The application of Ohm’s law in point form to a macroscopic (visible to the naked eye) region leads to a more familiar form. Initially, assume that J and E are uniform, as they are in the cylindrical region shown in Figure 5. Because they are uniform, (10)
and
(11)
or
Thus
o
r
The ratio of the potential difference between the two ends of the cylinder to the current entering the more positive end, however, is recognized from elementary circuit theory as the resistance of the cylinder, and therefore
(12) where
(13)
Equation (12) is, of course, known as Ohm’s law, and Eq. (13) enables us to compute the resistance R, measured in ohms (abbreviated as Ω), of conducting objects which possess uniform fields. If the fields are not uniform, the resistance may still be defined as the ratio of V to I , where V is the potential difference between two specified equipotential surfaces in the material and I is the total current crossing the more positive surface into the material. From the general integral relationships in Eqs.
(10) and (11), and from Ohm’s law (8), we may write this general expression for resistance when the fields are nonuniform,
(14) The line integral is taken between two equipotential surfaces in the conductor, and the surface integral is evaluated over the more positive of these two equipotentials. We cannot solve these nonuniform problems at this time.
Example:
- As an example of the determination of the resistance of a cylinder, we find the resistance of a 1-mile length of #16 copper wire, which has a diameter of 0 in.
Solution: The diameter of the wire is 0×0 = 1×10−3 m, the area of the cross section is π(1×10−3/2) 2 = 1×10−6 m 2 , and the length is 1609 m. Using a conductivity of 5 .80 × 10 7 S/m, the resistance of the wire is, therefore,
This wire can safely carry about 10 A dc, corresponding to a current density of 10/(1. 308 × 10 − 6 ) = 7. 65 × 106 A/m 2 , or 7 A/mm 2 .With this current, the potential difference between the two ends of the wire is 212 V, the electric field intensity is 0 V/m, the drift velocity is 0 m/s, or a little more than one furlong a week, and the freeelectron charge density is − 1. 81 × 1010 C/m 3 , or about one electron within a cube two angstroms on a side.
- a) Using data tabulated in Appendix C, calculate the required diameter for a 2-m long nichrome wire that will dissipate an average power of 450 W when 120 V rms at 60 Hz is applied to it:
Solution: The required resistance will be
Thus the diameter will be
b) Calculate the rms current density in the wire: The rms current will be I = 450/120 = 3 .75A. Thus
- Two perfectly-conducting cylindrical surfaces of length l are located at ρ = 3 and ρ = 5 cm. The total current passing radially outward through the medium between the cylinders is 3 A dc. (a) Find the voltage and resistance between the cylinders, and E in the region between the cylinders, if a conducting material having σ = 0 S/m is present for 3 < ρ < 5
cm. (b) Show that integrating the power dissipated per unit volume over the volume gives the total dissipated power.
Summarizing for electrostatics, no charge and no electric field may exist at any point within a conducting material. Charge may, however, appear on the surface as a surface charge density, and our next investigation concerns the fields external to the conductor.
We wish to relate these external fields to the charge on the surface of the conductor. The problem is a simple one, and we may first talk our way to the solution with a little mathematics.
If the external electric field intensity is decomposed into two components, one tangential and one normal to the conductor surface, the tangential component is seen to be zero. If it were not zero, a tangential force would be applied to the elements of the surface charge, resulting in their motion and nonstatic conditions. Because static conditions are assumed, the tangential electric field intensity and electric flux density are zero.
Gauss’s law answers our questions concerning the normal component. The electric flux leaving a small increment of surface must be equal to the charge residing on that incremental surface. The flux cannot penetrate into the conductor, for the total field there is zero. It must then leave the surface normally. Quantitatively, we may say that the electric flux density in coulombs per square meter leaving the surface normally is equal to the surface charge density in coulombs per square meter, or DN = ρS.
If we use some of our previously derived results in making a more careful analysis (and incidentally introducing a general method which we must use later), we should set up a boundary between a conductor and free space (Figure 5) showing tangential and normal components of D and E on the free-space side of the boundary. Both fields are zero in the conductor. The tangential field may be determined by applying Section 4, Eq. (21),
around the small closed path abcda. The integral must be broken up into four parts
Remembering that E = 0 within the conductor, we let the length from a to b or c to d be Δw and from b to c or d to a be Δh, and obtain
As we allow Δh to approach zero, keeping Δw small but finite, it makes no difference whether or not the normal fields are equal at a and b, for Δh causes these products to become negligibly small. Hence, EtΔw = 0 and, therefore, Et = 0.
The condition on the normal field is found most readily by considering DN rather than EN and choosing a small cylinder as the gaussian surface. Let the height be Δh and the area of the top and bottom faces be ΔS. Again, we let Δh approach zero. Using Gauss’s law,
we integrate over the three distinct surfaces
and find that the last two are zero (for different reasons). Then
or
These are the desired boundary conditions for the conductor-to-free-space boundary in electrostatics,
(15)
(16)
The electric flux leaves the conductor in a direction normal to the surface, and the value of the electric flux density is numerically equal to the surface charge density. Equations (15) and (16) can be more formally expressed using the vector fields
(17)
(18)
where n is the unit normal vector at the surface that points away from the conductor, as shown in Figure 5, and where both operations are evaluated at the conductor surface, s. Taking the cross product or the dot product of either field quantity with n gives the tangential or the normal component of the field, respectively.
An immediate and important consequence of a zero tangential electric field intensity is the fact that a conductor surface is an equipotential surface. The evaluation of the potential difference between any two points on the surface by the line integral leads to a zero result, because the path may be chosen on the surface itself where E · dL = 0.
To summarize the principles which apply to conductors in electrostatic fields, we may state that 1. The static electric field intensity inside a conductor is zero. 2. The static electric field intensity at the surface of a conductor is everywhere directed normal to that surface. 3. The conductor surface is an equipotential surface.
Using these three principles, there are a number of quantities that may be calculated at a conductor boundary, given a knowledge of the potential field.
Example:
- Given the potential, V = 100(x 2 − y 2 ) and a point P(2,− 1 , 3) that is stipulated to lie on a conductor-to-free-space boundary, find V, E, D, and ρS at P, and also the equation of the conductor surface. Solution: The potential at point P is
Note that if we had taken the region to the left of the equipotential surface as the conductor, the E field would terminate on the surface charge and we would let ρS = −3 nC/m 2.
- Let us determine the equation of the streamline passing through P.
Solution: We see that
Thus,
The line (or surface) through P is obtained when C 2 = (2)(−1) = −2. Thus, the streamline is the trace of another hyperbolic cylinder, xy = − 2
This is also shown on Figure 5.
Then
- Let V = 10(ρ + 1)z 2 cosφ V in free space. (a) Let the equipotential surface V = 20V define a conductor surface. Find the equation of the conductor surface. (b) Find ρ and E at that point on the conductor surface where φ= 0π and z= 1. (c) Find |ρs| at that point. Solution: a) Let the equipotential surface V = 20V define a conductor surface. Find the equation of the conductor surface: Set the given potential function equal to 20, to find:
(ρ + 1)z 2 cos φ = 2
b) Find ρ and E at that point on the conductor surface where φ = 0. 2 π and z = 1: At the given values of φ and z, we solve the equation of the surface found in part a for ρ, obtaining ρ =. 10.
c) Find |ρs| at that point: Since E is at the perfectly-conducting surface, it will be normal to the surface, so we may write:
- The Nature of Dielectric Materials (Hayt & Buck, 2001)
and
Therefore,
ln y + ln x = C 1
xy = C 2
A dielectric in an electric field can be viewed as a free-space arrangement of microscopic electric dipoles, each of which is composed of a positive and a negative charge whose centers do not quite coincide. These are not free charges, and they cannot contribute to the conduction process. Rather, they are bound in place by atomic and molecular forces and can only shift positions slightly in response to external fields. They are called bound charges, in contrast to the free charges that determine conductivity. The bound charges can be treated as any other sources of the electrostatic field. Therefore, we would not need to introduce the dielectric constant as a new parameter or to deal with permittivities different from the permittivity of free space; however, the alternative would be to consider every charge within a piece of dielectric material. This is too great a price to pay for using all our previous equations in an unmodified form, and we shall therefore spend some time theorizing about dielectrics in a qualitative way; introducing polarization P,
permittivity 𝐀, and relative permittivity 𝐀r; and developing some quantitative relationships involving
these new parameters.
The characteristic that all dielectric materials have in common, whether they are Solid, liquid, or gas, and whether or not they are crystalline in nature, is their ability to store electric energy. This storage takes place by means of a shift in the relative positions of the internal, bound positive and negative charges against the normal molecular and atomic forces.
This displacement against a restraining force is analogous to lifting a weight or stretching a spring and represents potential energy. The source of the energy is the external field, the motion of the shifting charges resulting perhaps in a transient current through a battery that is producing the field.
The actual mechanism of the charge displacement differs in the various dielectric materials. Some molecules, termed polar molecules, have a permanent displacement existing between the centers of “gravity” of the positive and negative charges, and each pair of charges acts as a dipole. Normally the dipoles are oriented in a random way throughout the interior of the material, and the action of the external field is to align these molecules, to some extent, in the same direction. A sufficiently strong field may even produce an additional displacement between the positive and negative charges.
A nonpolar molecule does not have this dipole arrangement until after a field is applied. The negative and positive charges shift in opposite directions against their mutual attraction and produce a dipole that is aligned with the electric field.
Either type of dipole may be described by its dipole moment p,
(20) where Q is the positive one of the two bound charges composing the dipole, and d is the vector from the negative to the positive charge. We note again that the units of p are coulomb-meters.
If there are n dipoles per unit volume and we deal with a volume ∆ν, then there are n∆ν dipoles, and the total dipole moment is obtained by the vector sum,
If the dipoles are aligned in the same general direction, p total may have a significant value. However, a random orientation may cause p total to be essentially zero. We now define the polarization P as the dipole moment per unit volume,
(21)
with units of coulombs per square meter. We will treat P as a typical continuous field, even though it is obvious that it is essentially undefined at points within an atom or molecule. Instead, we should think of its value at any point as an average value taken over a sample volume ∆ν—large enough to
where
and Q is the total free charge enclosed by the surface S. Note that the free charge appears without a subscript because it is the most important type of charge and will appear in Maxwell’s equations.
Combining these last three equations, we obtain an expression for the free charge enclosed,
(24)
D is now defined in more general terms than was done in IM no. 5,
(25)
There is thus an added term to D that appears when polarizable material is present. Thus,
(26)
where Q is the free charge enclosed.
Utilizing the several volume charge densities, we have
With the help of the divergence theorem, we may therefore transform Eqs. (22), (23), and (26) into the equivalent divergence relationships,
(27)
We will emphasize only Eq. (26) and (27), the two expressions involving the free charge, in the work that follows.
In order to make any real use of these new concepts, it is necessary to know the relationship between the electric field intensity E and the polarization P that results. This relationship will, of course, be a function of the type of material, and we will essentially limit our discussion to those isotropic materials for which E and P are linearly related. In an isotropic material, the vectors E and P are always parallel, regardless of the orientation of the field. Although most engineering dielectrics are linear for moderate-to-large field strengths and are also isotropic, single crystals may be anisotropic. The periodic nature of crystalline materials causes dipole moments to be formed most easily along the crystal axes, and not necessarily in the direction of the applied field.
In ferroelectric materials, the relationship between P and E not only is nonlinear, but also shows hysteresis effects; that is, the polarization produced by a given electric field intensity depends on the past history of the sample. Important examples of this type of dielectric are barium titanate, often used in ceramic capacitors, and Rochelle salt.
The linear relationship between P and E is
(28)
where χe (chi) is a dimensionless quantity called the electric susceptibility of the material. Using this
relationship in Eq. (25), we have
The expression within the parentheses is now defined as ( ) This is another dimensionless quantity, and it is known as the relative permittivity, or dielectric constant of the material. Thus,
(30)
where
(31)
and 𝐀 is the permittivity. The dielectric constants (εr or sometimes denoted as k) are given for some
representative materials in Appendix C.
Anisotropic dielectric materials cannot be described in terms of a simple susceptibility or permittivity parameter. Instead, we find that each component of D may be a function of every
component of E, and D = 𝐀E becomes a matrix equation where D and E are each 3 × 1 column
matrices and 𝐀 is a 3 × 3 square matrix. Expanding the matrix equation gives
Note that the elements of the matrix depend on the selection of the coordinate axes in the anisotropic material. Certain choices of axis directions lead to simpler matrices.
D and E (and P) are no longer parallel, and although D = 𝐀 0 E + P remains a valid equation for
anisotropic materials, we may continue to use D = 𝐀E only by interpreting it as a matrix equation.
We will concentrate our attention on linear isotropic materials and reserve the general case for a more advanced text.
In summary, then, we now have a relationship between D and E that depends on the dielectric material present,
(30) where (31)
This electric flux density is still related to the free charge by either the point or integral form of Gauss’s law:
Consider two conductors embedded in a homogeneous dielectric (Figure 6). Conductor M 2 carries a total positive charge Q, and M1 carries an equal negative charge. There are no other charges present, and the total charge of the system is zero. We now know that the charge is carried on the surface as a surface charge density and also that the electric field is normal to the conductor surface. Each conductor is, moreover, an equipotential surface. Because M 2 carries the positive charge, the electric flux is directed from M 2 to M 1 , and M 2 is at the more positive potential. In other words, work must be done to carry a positive charge from M 1 to M 2.
Let us designate the potential difference between M 2 and M 1 as V 0. We may now define the capacitance of this two-conductor system as the ratio of the magnitude of the total charge on either conductor to the magnitude of the potential difference between conductors,
(1)
In general terms, we determine Q by a surface integral over the positive conductors, and we find V 0 by carrying a unit positive charge from the negative to the positive surface,
(2)
The capacitance is independent of the potential and total charge, for their ratio is constant. If the charge density is increased by a factor of N, Gauss’s law indicates that the electric flux density or electric field intensity also increases by N, as does the potential difference. The capacitance is a function only of the physical dimensions of the system of conductors and of the permittivity of the homogeneous dielectric.
Capacitance is measured in farads (F), where a farad is defined as one coulomb per volt. Common values of capacitance are apt to be very small fractions of a farad, and consequently more practical units are the microfarad (μF), the nanofarad (nF), and the picofarad (pF).
We can apply the definition of capacitance to a simple two-conductor system in which the conductors are identical, infinite parallel planes with separation d (Figure 6). Choosing the lower conducting plane at z = 0 and the upper one at z = d, a uniform sheet of surface charge ±ρS on each conductor leads to the uniform field [Section 2, Eq. (18)]
where the permittivity of the homogeneous dielectric
is 𝐀, and
Note that this result could be obtained by applying the boundary condition at a conducting surface (Eq. (18), Chapter 5) at either one of the plate surfaces. Referring to the surfaces and their unit normal vectors in Fig. 6, where nl = az and nu = −az , we find on the lower plane:
On the upper plane, we get the same result
This is a key advantage of the conductor boundary condition, in that we need to apply it only to a single boundary to obtain the total field there (arising from all other sources). The potential difference between lower and upper planes is
Since the total charge on either plane is infinite, the capacitance is infinite. A more practical answer is obtained by considering planes, each of area S, whose linear dimensions are much greater than their separation d. The electric field and charge distribution are then almost uniform at all points not adjacent to the edges, and this latter region contributes only a small percentage of the total capacitance, allowing us to write the familiar result
(3)
More rigorously, we might consider Eq. (3) as the capacitance of a portion of the infinite-plane arrangement having a surface area S (or sometimes denoted as A). Methods of calculating the effect of the unknown and nonuniform distribution near the edges must wait until we are able to solve more complicated potential problems.
Example:
Electromags - Chapter 7
Course: Electronics Engineering (CR 061)
University: Samar State University
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