Textbook and Syllabus Textbook: Contents:

Textbook and Syllabus Textbook: Contents:
Engineering Electromagnetics Textbook and Syllabus Textbook: “Engineering Electromagnetics”, William H. Hayt, Jr. and John A. Buck, McGraw-Hill, 2006. Contents: Chapter 1: Vector Analysis Chapter 2: Coulomb’s Law and Electric Field Intensity Chapter 3: Electric Flux Density, Gauss’ Law, and Divergence Chapter 4: Energy and Potential Chapter 5: Conductors and Dielectrics Chapter 6: Capacitance Chapter 7: The Steady Magnetic Field Chapter 8: Magnetic Forces, Materials, and Inductance Chapter 9: Time-Varying Fields and Maxwell’s Equations Chapter 10: Transmission Lines Chapter 11: The Uniform Plane Wave

Chapter 1 Vector Analysis

What is Electromagnetics?
Engineering Electromagnetics What is Electromagnetics? Electric field Produced by the presence of electrically charged particles, and gives rise to the electric force. Magnetic field Produced by the motion of electric charges, or electric current, and gives rise to the magnetic force associated with magnets.

Electromagnetic Wave Spectrum
Engineering Electromagnetics Electromagnetic Wave Spectrum

Why do we learn Engineering Electromagnetics
Electric and magnetic field exist nearly everywhere.

Engineering Electromagnetics
Applications Electromagnetic principles find application in various disciplines such as microwaves, x-rays, antennas, electric machines, plasmas, etc.

Engineering Electromagnetics
Applications Electromagnetic fields are used in induction heaters for melting, forging, annealing, surface hardening, and soldering operation. Electromagnetic devices include transformers, radio, television, mobile phones, radars, lasers, etc.

Applications Transrapid Train
Engineering Electromagnetics Applications Transrapid Train A magnetic traveling field moves the vehicle without contact. The speed can be continuously regulated by varying the frequency of the alternating current.

Chapter 1 Vector Analysis Scalars and Vectors Scalar refers to a quantity whose value may be represented by a single (positive or negative) real number. Some examples include distance, temperature, mass, density, pressure, volume, and time. A vector quantity has both a magnitude and a direction in space. We especially concerned with two- and three-dimensional spaces only. Displacement, velocity, acceleration, and force are examples of vectors. Scalar notation: A or A (italic or plain) Vector notation: A or A (bold or plain with arrow) →

Chapter 1 Vector Analysis Vector Algebra

Rectangular Coordinate System
Chapter 1 Vector Analysis Rectangular Coordinate System Differential surface units: Differential volume unit :

Vector Components and Unit Vectors
Chapter 1 Vector Analysis Vector Components and Unit Vectors

Vector Components and Unit Vectors
Chapter 1 Vector Analysis Vector Components and Unit Vectors For any vector B, : Magnitude of B Unit vector in the direction of B Example Given points M(–1,2,1) and N(3,–3,0), find RMN and aMN.

Chapter 1 Vector Analysis The Dot Product Given two vectors A and B, the dot product, or scalar product, is defines as the product of the magnitude of A, the magnitude of B, and the cosine of the smaller angle between them: The dot product is a scalar, and it obeys the commutative law: For any vector and ,

Chapter 1 Vector Analysis The Dot Product One of the most important applications of the dot product is that of finding the component of a vector in a given direction. The scalar component of B in the direction of the unit vector a is Ba The vector component of B in the direction of the unit vector a is (Ba)a

The Dot Product Example
Chapter 1 Vector Analysis The Dot Product Example The three vertices of a triangle are located at A(6,–1,2), B(–2,3,–4), and C(–3,1,5). Find: (a) RAB; (b) RAC; (c) the angle θBAC at vertex A; (d) the vector projection of RAB on RAC.

Chapter 1 Vector Analysis The Cross Product Given two vectors A and B, the magnitude of the cross product, or vector product, written as AB, is defines as the product of the magnitude of A, the magnitude of B, and the sine of the smaller angle between them. The direction of AB is perpendicular to the plane containing A and B and is in the direction of advance of a right-handed screw as A is turned into B. The cross product is a vector, and it is not commutative:

The Cross Product Example
Chapter 1 Vector Analysis The Cross Product Example Given A = 2ax–3ay+az and B = –4ax–2ay+5az, find AB.

The Cylindrical Coordinate System
Chapter 1 Vector Analysis The Cylindrical Coordinate System

Chapter 1 Vector Analysis The Cylindrical Coordinate System Differential surface units: Relation between the rectangular and the cylindrical coordinate systems Differential volume unit :

Chapter 1 Vector Analysis The Cylindrical Coordinate System Dot products of unit vectors in cylindrical and rectangular coordinate systems

The Spherical Coordinate System
Chapter 1 Vector Analysis The Spherical Coordinate System

Chapter 1 Vector Analysis The Spherical Coordinate System Differential surface units: Differential volume unit :

Chapter 1 Vector Analysis The Spherical Coordinate System Relation between the rectangular and the spherical coordinate systems Dot products of unit vectors in spherical and rectangular coordinate systems

Chapter 1 Vector Analysis The Spherical Coordinate System Example Given the two points, C(–3,2,1) and D(r = 5, θ = 20°, Φ = –70°), find: (a) the spherical coordinates of C; (b) the rectangular coordinates of D.

Coulomb’s Law and Electric Field Intensity
Chapter 2 Coulomb’s Law and Electric Field Intensity

The Experimental Law of Coulomb
Chapter 2 Coulomb’s Law and Electric Field Intensity The Experimental Law of Coulomb In 1600, Dr. Gilbert, a physician from England, published the first major classification of electric and non-electric materials. He stated that glass, sulfur, amber, and some other materials “not only draw to themselves straw, and chaff, but all metals, wood, leaves, stone, earths, even water and oil.” In the following century, a French Army Engineer, Colonel Charles Coulomb, performed an elaborate series of experiments using devices invented by himself. Coulomb could determine quantitatively the force exerted between two objects, each having a static charge of electricity. He wrote seven important treatises on electric and magnetism, developed a theory of attraction and repulsion between bodies of the opposite and the same electrical charge.

Chapter 2 Coulomb’s Law and Electric Field Intensity The Experimental Law of Coulomb Coulomb stated that the force between two very small objects separated in vacuum or free space by a distance which is large compared to their size is proportional to the charge on each and inversely proportional to the square of the distance between them. In SI Units, the quantities of charge Q are measured in coulombs (C), the separation R in meters (m), and the force F should be newtons (N). This will be achieved if the constant of proportionality k is written as:

Chapter 2 Coulomb’s Law and Electric Field Intensity The Experimental Law of Coulomb The permittivity of free space ε is measured in farads per meter (F/m), and has the magnitude of: The Coulomb’s law is now: The force F acts along the line joining the two charges. It is repulsive if the charges are alike in sign and attractive if the are of opposite sign.

Chapter 2 Coulomb’s Law and Electric Field Intensity The Experimental Law of Coulomb In vector form, Coulomb’s law is written as: F2 is the force on Q2, for the case where Q1 and Q2 have the same sign, while a12 is the unit vector in the direction of R12, the line segment from Q1 to Q2.

Chapter 2 Coulomb’s Law and Electric Field Intensity The Experimental Law of Coulomb Example A charge Q1 = 310–4 C at M(1,2,3) and a charge of Q2 = –10–4 C at N(2,0,5) are located in a vacuum. Determine the force exerted on Q2 by Q1.

Electric Field Intensity
Chapter 2 Coulomb’s Law and Electric Field Intensity Electric Field Intensity Let us consider one charge, say Q1, fixed in position in space. Now, imagine that we can introduce a second charge, Qt, as a “unit test charge”, that we can move around. We know that there exists everywhere a force on this second charge ► This second charge is displaying the existence of a force field. The force on it is given by Coulomb’s law as: Writing this force as a “force per unit charge” gives: Vector Field, Electric Field Intensity

Chapter 2 Coulomb’s Law and Electric Field Intensity Electric Field Intensity We define the electric field intensity as the vector of force on a unit positive test charge. Electric field intensity, E, is measured by the unit newtons per coulomb (N/C) or volts per meter (V/m). The field of a single point charge can be written as: aR is a unit vector in the direction from the point at which the point charge Q is located, to the point at which E is desired/measured.

Chapter 2 Coulomb’s Law and Electric Field Intensity Electric Field Intensity For a charge which is not at the origin of the coordinate, the electric field intensity is:

Chapter 2 Coulomb’s Law and Electric Field Intensity Electric Field Intensity The electric field intensity due to two point charges, say Q1 at r1 and Q2 at r2, is the sum of the electric field intensity on Qt caused by Q1 and Q2 acting alone (Superposition Principle).

Chapter 2 Coulomb’s Law and Electric Field Intensity Electric Field Intensity Example A charge Q1 of 2 μC is located at at P1(0,0,0) and a second charge of 3 μC is at P2(–1,2,3). Find E at M(3,–4,2).

Field Due to a Continuous Volume Charge Distribution
Chapter 2 Coulomb’s Law and Electric Field Intensity Field Due to a Continuous Volume Charge Distribution We denote the volume charge density by ρv, having the units of coulombs per cubic meter (C/m3). The small amount of charge ΔQ in a small volume Δv is We may define ρv mathematically by using a limit on the above equation: The total charge within some finite volume is obtained by integrating throughout that volume:

Chapter 2 Coulomb’s Law and Electric Field Intensity Field Due to a Continuous Volume Charge Distribution Example Find the total charge inside the volume indicated by ρv = 4xyz2, 0 ≤ ρ ≤ 2, 0 ≤ Φ ≤ π/2, 0 ≤ z ≤ 3. All values are in SI units.

Chapter 2 Coulomb’s Law and Electric Field Intensity Field Due to a Continuous Volume Charge Distribution The contributions of all the volume charge in a given region, let the volume element Δv approaches zero, is an integral in the form of:

Chapter 2 Coulomb’s Law and Electric Field Intensity Field of a Line Charge Now we consider a filamentlike distribution of volume charge density. It is convenient to treat the charge as a line charge of density ρL C/m. Let us assume a straight-line charge extending along the z axis in a cylindrical coordinate system from –∞ to +∞. We desire the electric field intensity E at any point resulting from a uniform line charge density ρL.

Chapter 2 Coulomb’s Law and Electric Field Intensity Field of a Line Charge The incremental field dE only has the components in aρ and az direction, and no aΦ direction. Why? The component dEz is the result of symmetrical contributions of line segments above and below the observation point P. Since the length is infinity, they are canceling each other ► dEz = 0. The component dEρ exists, and from the Coulomb’s law we know that dEρ will be inversely proportional to the distance to the line charge, ρ.

Field of a Line Charge Take P(0,y,0), Chapter 2
Coulomb’s Law and Electric Field Intensity Field of a Line Charge Take P(0,y,0),

Field of a Line Charge Now let us analyze the answer itself:
Chapter 2 Coulomb’s Law and Electric Field Intensity Field of a Line Charge Now let us analyze the answer itself: The field falls off inversely with the distance to the charged line, as compared with the point charge, where the field decreased with the square of the distance.

Field of a Line Charge Example D2.5.
Chapter 2 Coulomb’s Law and Electric Field Intensity Field of a Line Charge Example D2.5. Infinite uniform line charges of 5 nC/m lie along the (positive and negative) x and y axes in free space. Find E at: (a) PA(0,0,4); (b) PB(0,3,4). PB PA ρ is the shortest distance between an observation point and the line charge

Field of a Sheet of Charge
Chapter 2 Coulomb’s Law and Electric Field Intensity Field of a Sheet of Charge Another basic charge configuration is the infinite sheet of charge having a uniform density of ρS C/m2. The charge-distribution family is now complete: point (Q), line (ρL), surface (ρS), and volume (ρv). Let us examine a sheet of charge above, which is placed in the yz plane. The plane can be seen to be assembled from an infinite number of line charge, extending along the z axis, from –∞ to +∞.

Chapter 2 Coulomb’s Law and Electric Field Intensity Field of a Sheet of Charge For a differential width strip dy’, the line charge density is given by ρL = ρSdy’. The component dEz at P is zero, because the differential segments above and below the y axis will cancel each other. The component dEy at P is also zero, because the differential segments to the right and to the left of z axis will cancel each other. Only dEx is present, and this component is a function of x alone.

Chapter 2 Coulomb’s Law and Electric Field Intensity Field of a Sheet of Charge The contribution of a strip to Ex at P is given by: Adding the effects of all the strips,

Chapter 2 Coulomb’s Law and Electric Field Intensity Field of a Sheet of Charge Fact: The electric field is always directed away from the positive charge, into the negative charge. We now introduce a unit vector aN, which is normal to the sheet and directed away from it. The field of a sheet of charge is constant in magnitude and direction. It is not a function of distance.

Electric Flux Density, Gauss’ Law, and Divergence
Chapter 3 Electric Flux Density, Gauss’ Law, and Divergence

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density About 1837, the Director of the Royal Society in London, Michael Faraday, was interested in static electric fields and the effect of various insulating materials on these fields. This is the lead to his famous invention, the electric motor. He found that if he moved a magnet through a loop of wire, an electric current flowed in the wire. The current also flowed if the loop was moved over a stationary magnet. Changing magnetic field produces an electric field.

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density In his experiments, Faraday had a pair of concentric metallic spheres constructed, the outer one consisting of two hemispheres that could be firmly clamed together. He also prepared shells of insulating material (or dielectric material), which would occupy the entire volume between the concentric spheres.

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density Faraday found out, that there was a sort of “charge displacement” from the inner sphere to the outer sphere, which was independent of the medium. We refer to this flow as displacement, displacement flux, or simply electric flux. Where ψ is the electric flux, measured in coulombs, and Q is the total charge on the inner sphere, also in coulombs.

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density At the surface of the inner sphere, ψ coulombs of electric flux are produced by the given charge Q coulombs, and distributed uniformly over a surface having an area of 4πa2 m2. The density of the flux at this surface is ψ/4πa2 or Q/4πa2 C/m2. The new quantity, electric flux density, is measured in C/m2 and denoted with D. The direction of D at a point is the direction of the flux lines at that point. The magnitude of D is given by the number of flux lines crossing a surface normal to the lines divided by the surface area.

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density Referring again to the concentric spheres, the electric flux density is in the radial direction : At a distance r, where a ≤ r ≤ b, If we make the inner sphere smaller and smaller, it becomes a point charge while still retaining a charge of Q. The electrix flux density at a point r meters away is still given by:

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density Comparing with the previous chapter, the radial electric field intensity of a point charge in free space is: Therefore, in free space, the following relation applies: For a general volume charge distribution in free space:

Electric Flux Density Example
Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density Example Find the electric flux density at a point having a distance 3 m from a uniform line charge of 8 nC/m lying along the z axis in free space. For the value ρ = 3 m,

Electric Flux Density Example
Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density Example Calculate D at point P(6,8,–10) produced by a uniform surface charge density with ρs = 57.2 μC/m2 on the plane x = 9. At P(6,8,–10),

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law The results of Faraday’s experiments with the concentric spheres could be summed up as an experimental law by stating that the electric flux passing through any imaginary spherical surface lying between the two conducting spheres is equal to the charge enclosed within that imaginary surface. Faraday’s experiment can be generalized to the following statement, which is known as Gauss’s Law: “The electric flux passing through any closed surface is equal to the total charge enclosed by that surface.”

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law Imagine a distribution of charge, shown as a cloud of point charges, surrounded by a closed surface of any shape. If the total charge is Q, the Q coulombs of electric flux will pass through the enclosing surface. At every point on the surface the electric-flux-density vector D will have some value DS (subscript S means that D must be evaluated at the surface).

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law ΔS defines an incremental element of area with magnitude of ΔS and the direction normal to the plane, or tangent to the surface at the point in question. At any point P, where DS makes an angle θ with ΔS, then the flux crossing ΔS is the product of the normal components of DS and ΔS.

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law The resultant integral is a closed surface integral, with dS always involves the differentials of two coordinates ► The integral is a double integral. We can formulate the Gauss’s law mathematically as: The charge enclosed meant by the formula above might be several point charges, a line charge, a surface charge, or a volume charge distribution.

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law We now take the last form, written in terms of the charge distribution, to represent the other forms: Illustration. Let a point charge Q be placed at the origin of a spherical coordinate system, and choose a closed surface as a sphere of radius a. The electric field intensity due to the point charge has been found to be:

Gauss’s Law At the surface, r = a, Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law At the surface, r = a,

Application of Gauss’s Law: Some Symmetrical Charge Distributions
Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions Let us now consider how to use the Gauss’s law to calculate the electric field intensity DS: The solution will be easy if we are able to choose a closed surface which satisfies two conditions: DS is everywhere either normal or tangential to the closed surface, so that DSdS becomes either DSdS or zero, respectively. On that portion of the closed surface for which DSdS is not zero, DS is constant. For point charge ► The surface of a sphere. For line charge ► The surface of a cylinder.

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions From the previous discussion of the uniform line charge, only the radial component of D is present: The choice of a surface that fulfill the requirement is simple: a cylindrical surface. Dρ is every normal to the surface of a cylinder. It may then be closed by two plane surfaces normal to the z axis.

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions We know that the charge enclosed is ρLL,

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions The problem of a coaxial cable is almost identical with that of the line charge. Suppose that we have two coaxial cylindrical conductors, the inner of radius a and the outer of radius b, both with infinite length. We shall assume a charge distribution of ρS on the outer surface of the inner conductor. Choosing a circular cylinder of length L and radius ρ, a < ρ < b, as the gaussian surface, we find: The total charge on a length L of the inner conductor is:

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions For one meter length, the inner conductor has 2πaρS coulombs, hence ρL = 2πaρS, Everly line of electrix flux starting from the inner cylinder must terminate on the inner surface of the outer cylinder: Due to simplicity, noise immunity and broad bandwidth, coaxial cable is still the most common means of data transmission over short distances. If we use a cylinder of radius ρ > b, then the total charge enclosed will be zero. ► There is no external field,

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions Example A 50-cm length of coaxial cable has an inner radius of 1 mm and an outer radius of 4 mm. The space between conductors is assumed to be filled with air. The total charge on the inner conductor is 30 nC. Find the charge density on each conductor and the expressions for E and D fields.

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions

Application of Gauss’s Law: Differential Volume Element
Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element We are now going to apply the methods of Gauss’s law to a slightly different type of problem: a surface without symmetry. We have to choose such a very small closed surface that D is almost constant over the surface, and the small change in D may be adequately represented by using the first two terms of the Taylor’s-series expansion for D. The result will become more nearly correct as the volume enclosed by the gaussian surface decreases.

Taylor’s Series Expansion
Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Taylor’s Series Expansion A point near x0 Only the linear terms are used for the linearization

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element Consider any point P, located by a rectangular coordinate system. The value of D at the point P may be expressed in rectangular components: We now choose as our closed surface, the small rectangular box, centered at P, having sides of lengths Δx, Δy, and Δz, and apply Gauss’s law:

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element We will now consider the front surface in detail. The surface element is very small, thus D is essentially constant over this surface (a portion of the entire closed surface): The front face is at a distance of Δx/2 from P, and therefore:

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element We have now, for front surface: In the same way, the integral over the back surface can be found as:

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element If we combine the two integrals over the front and back surface, we have: Repeating the same process to the remaining surfaces, we find: These results may be collected to yield:

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element The previous equation is an approximation, which becomes better as Δv becomes smaller. For the moment, we have applied Gauss’s law to the closed surface surrounding the volume element Δv, with the result:

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element Example Let D = y2z3 ax + 2xyz3 ay + 3xy2z2 az pC/m2 in free space. (a) Find the total electric flux passing through the surface x = 3, 0 ≤ y ≤ 2, 0 ≤ z ≤ 1 in a direction away from the origin. (b) Find |E| at P(3,2,1). (c) Find the total charge contained in an incremental sphere having a radius of 2 μm centered at P(3,2,1). (a)

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element (b)

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Application of Gauss’s Law: Differential Volume Element (c)

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Divergence We shall now obtain an exact relationship, by allowing the volume element Δv to shrink to zero. The last term is the volume charge density ρv, so that:

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Divergence Let us no consider one information that can be obtained from the last equation: This equation is valid not only for electric flux density D, but also to any vector field A to find the surface integral for a small closed surface.

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Divergence This operation received a descriptive name, divergence. The divergence of A is defined as: “The divergence of the vector flux density A is the outflow of flux from a small closed surface per unit volume as the volume shrinks to zero.” A positive divergence of a vector quantity indicates a source of that vector quantity at that point. Similarly, a negative divergence indicates a sink.

Divergence Rectangular Cylindrical Spherical Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence Divergence Rectangular Cylindrical Spherical

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Divergence Example If D = e–xsiny ax – e–x cosy ay + 2z az, find div D at the origin and P(1,2,3) Regardles of location the divergence of D equals 2 C/m3.

Maxwell’s First Equation (Electrostatics)
Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence Maxwell’s First Equation (Electrostatics) We may now rewrite the expressions developed until now: Maxwell’s First Equation Point Form of Gauss’s Law This first of Maxwell’s four equations applies to electrostatics and steady magnetic field. Physically it states that the electric flux per unit volume leaving a vanishingly small volume unit is exactly equal to the volume charge density there.

The Vector Operator Ñ and The Divergence Theorem
Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence The Vector Operator Ñ and The Divergence Theorem Divergence is an operation on a vector yielding a scalar, just like the dot product. We define the del operator Ñ as a vector operator: Then, treating the del operator as an ordinary vector, we can write:

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence The Vector Operator Ñ and The Divergence Theorem The Ñ operator does not have a specific form in other coordinate systems than rectangular coordinate system. Nevertheless, Cylindrical Spherical

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence The Vector Operator Ñ and The Divergence Theorem We shall now give name to a theorem that we actually have obtained, the Divergence Theorem: The first and last terms constitute the divergence theorem: “The integral of the normal component of any vector field over a closed surface is equal to the integral of the divergence of this vector field throughout the volume enclosed by the closed surface.”

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence The Vector Operator Ñ and The Divergence Theorem Example Evaluate both sides of the divergence theorem for the field D = 2xy ax + x2 ay C/m2 and the rectangular parallelepiped fomed by the planes x = 0 and 1, y = 0 and 2, and z = 0 and 3. Divergence Theorem But

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence The Vector Operator Ñ and The Divergence Theorem

Chapter 4 Energy and Potential

Energy Expended in Moving a Point Charge in an Electric Field
Chapter 4 Energy and Potential Energy Expended in Moving a Point Charge in an Electric Field The electric field intensity was defined as the force on a unit test charge at that point where we wish to find the value of the electric field intensity. To move the test charge against the electric field, we have to exert a force equal and opposite in magnitude to that exerted by the field. ► We must expend energy or do work. To move the charge in the direction of the electric field, our energy expenditure turns out to be negative. ► We do not do the work, the field does.

Chapter 4 Energy and Potential Energy Expended in Moving a Point Charge in an Electric Field To move a charge Q a distance dL in an electric field E, the force on Q arising from the electric field is: The component of this force in the direction dL is: The force that we apply must be equal and opposite to the force exerted by the field: Differential work done by external source to Q is equal to: If E and L are perpendicular, the differential work will be zero

Chapter 4 Energy and Potential Energy Expended in Moving a Point Charge in an Electric Field The work required to move the charge a finite distance is determined by integration: The path must be specified beforehand The charge is assumed to be at rest at both initial and final positions W > 0 means we expend energy or do work W < 0 means the field expends energy or do work

Chapter 4 Energy and Potential The Line Integral The integral expression of previous equation is an example of a line integral, taking the form of integral along a prescribed path. Without using vector notation, we should have to write: EL: component of E along dL The work involved in moving a charge Q from B to A is approximately:

The Line Integral If we assume that the electric field is uniform,
Chapter 4 Energy and Potential The Line Integral If we assume that the electric field is uniform, Therefore, Since the summation can be interpreted as a line integral, the exact result for the uniform field can be obtained as: For the case of uniform E, W does not depend on the particular path selected along which the charge is carried

The Line Integral Example
Chapter 4 Energy and Potential The Line Integral Example Given the nonuniform field E = yax + xay +2az, determine the work expended in carrying 2 C from B(1,0,1) to A(0.8,0.6,1) along the shorter arc of the circle x2 + y2 = 1, z = 1. Differential path, rectangular coordinate Circle equation:

The Line Integral Example
Chapter 4 Energy and Potential The Line Integral Example Redo the example, but use the straight-line path from B to A. Line equation:

Differential Length Rectangular Cylindrical Spherical Chapter 4
Energy and Potential Differential Length Rectangular Cylindrical Spherical

Work and Path Near an Infinite Line Charge
Chapter 4 Energy and Potential Work and Path Near an Infinite Line Charge

Definition of Potential Difference and Potential
Chapter 4 Energy and Potential Definition of Potential Difference and Potential We already find the expression for the work W done by an external source in moving a charge Q from one point to another in an electric field E: Potential difference V is defined as the work done by an external source in moving a unit positive charge from one point to another in an electric field: We shall now set an agreement on the direction of movement. VAB signifies the potential difference between points A and B and is the work done in moving the unit charge from B (last named) to A (first named).

Chapter 4 Energy and Potential Definition of Potential Difference and Potential Potential difference is measured in joules per coulomb (J/C). However, volt (V) is defined as a more common unit. The potential difference between points A and B is: VAB is positive if work is done in carrying the positive charge from B to A From the line-charge example, we found that the work done in taking a charge Q from ρ = a to ρ = b was: Or, from ρ = b to ρ = a, Thus, the potential difference between points at ρ = a to ρ = b is:

Chapter 4 Energy and Potential Definition of Potential Difference and Potential For a point charge, we can find the potential difference between points A and B at radial distance rA and rB, choosing an origin at Q: rB > rA  VAB > 0, WAB > 0, Work expended by the external source (us) rB < rA  VAB < 0, WAB < 0, Work done by the electric field

Chapter 4 Energy and Potential Definition of Potential Difference and Potential It is often convenient to speak of potential, or absolute potential, of a point rather than the potential difference between two points. For this purpose, we must first specify the reference point which we consider to have zero potential. The most universal zero reference point is “ground”, which means the potential of the surface region of the earth. Another widely used reference point is “infinity.” For cylindrical coordinate, in discussing a coaxial cable, the outer conductor is selected as the zero reference for potential. If the potential at point A is VA and that at B is VB, then:

The Potential Field of a Point Charge
Chapter 4 Energy and Potential The Potential Field of a Point Charge In previous section we found an expression for the potential difference between two points located at r = rA and r = rB in the field of a point charge Q placed at the origin: Any initial and final values of θ or Φ will not affect the answer. As long as the radial distance between rA and rB is constant, any complicated path between two points will not change the results. This is because although dL has r, θ, and Φ components, the electric field E only has the radial r component.

Chapter 4 Energy and Potential The Potential Field of a Point Charge The potential difference between two points in the field of a point charge depends only on the distance of each point from the charge. Thus, the simplest way to define a zero reference for potential in this case is to let V = 0 at infinity. As the point r = rB recedes to infinity, the potential at rA becomes:

Chapter 4 Energy and Potential The Potential Field of a Point Charge Generally, Physically, Q/4πε0r joules of work must be done in carrying 1 coulomb charge from infinity to any point in a distance of r meters from the charge Q. We can also choose any point as a zero reference: with C1 may be selected so that V = 0 at any desired value of r.

Equipotential Surface
Chapter 4 Energy and Potential Equipotential Surface Equipotential surface is a surface composed of all those points having the same value of potential. No work is involved in moving a charge around on an equipotential surface. The equipotential surfaces in the potential field of a point charge are spheres centered at the point charge. The equipotential surfaces in the potential field of a line charge are cylindrical surfaces axed at the line charge. The equipotential surfaces in the potential field of a sheet of charge are surfaces parallel with the sheet of charge.

The Potential Field of a System of Charges: Conservative Property
Chapter 4 Energy and Potential The Potential Field of a System of Charges: Conservative Property We will now prove, that for a system of charges, the potential is also independent of the path taken. Continuing the discussion, the potential field at the point r due to a single point charge Q1 located at r1 is given by: The field is linear with respect to charge so that superposition is applicable. Thus, the potential arising from n point charges is:

Chapter 4 Energy and Potential The Potential Field of a System of Charges: Conservative Property If each point charge is now represented as a small element of continuous volume charge distribution ρvΔv, then: As the number of elements approach infinity, we obtain the integral expression: If the charge distribution takes from of a line charge or a surface charge,

Chapter 4 Energy and Potential The Potential Field of a System of Charges: Conservative Property As illustration, let us find V on the z axis for a uniform line charge ρL in the form of a ring, ρ = a, in the z = 0 plane. The potential arising from point charges or continuous charge distribution can be seen as the summation of potential arising from each charge or each differential charge. It is independent of the path chosen.

Chapter 4 Energy and Potential The Potential Field of a System of Charges: Conservative Property With zero reference at ∞, the expression for potential can be taken generally as: Or, for potential difference: Both expressions above are not dependent on the path chosen for the line integral, regardless of the source of the E field. Potential conservation in a simple dc-circuit problem in the form of Kirchhoff’s voltage law For static fields, no work is done in carrying the unit charge around any closed path.

Chapter 4 Energy and Potential Potential Gradient We have discussed two methods of determining potential: directly from the electric field intensity by means of a line integral, or from the basic charge distribution itself by a volume integral. In practical problems, however, we rarely know E or ρv. Preliminary information is much more likely to consist a description of two equipotential surface, and the goal is to find the electric field intensity.

Chapter 4 Energy and Potential Potential Gradient The general line-integral relationship between V and E is: For a very short element of length ΔL, E is essentially constant: Assuming a conservative field, for a given reference and starting point, the result of the integration is a function of the end point (x,y,z). We may pass to the limit and obtain:

Chapter 4 Energy and Potential Potential Gradient From the last equation, the maximum positive increment of potential, Δvmax, will occur when cosθ = –1, or ΔL points in the direction opposite to E. We can now conclude two characteristics of the relationship between E and V at any point: The magnitude of E is given by the maximum value of the rate of change of V with distance L. This maximum value of V is obtained when the direction of the distance increment is opposite to E.

Chapter 4 Energy and Potential Potential Gradient For the equipotential surfaces below, find the direction of E at P.

Chapter 4 Energy and Potential Potential Gradient Since the potential field information is more likely to be determined first, let us describe the direction of ΔL (which leads to a maximum increase in potential) in term of potential field. Let aN be a unit vector normal to the equipotential surface and directed toward the higher potential. The electric field intensity is then expressed in terms of the potential as: The maximum magnitude occurs when ΔL is in the aN direction. Thus we define dN as incremental length in aN direction,

Chapter 4 Energy and Potential Potential Gradient The mathematical operation to find the rate of change in a certain direction is called gradient. Now, the gradient of a scalar field T is defined as: Using the new term,

Chapter 4 Energy and Potential Potential Gradient Since V is a function of x, y, and z, the total differential is: But also, Both expression are true for any dx, dy, and dz. Thus: Note: Gradient of a scalar is a vector.

Potential Gradient Introducing the vector operator for gradient:
Chapter 4 Energy and Potential Potential Gradient Introducing the vector operator for gradient: We now can relate E and V as: Rectangular Cylindrical Spherical

Potential Gradient Example
Chapter 4 Energy and Potential Potential Gradient Example Given the potential field, V = 2x2y–5z, and a point P(–4,3,6), find V, E, direction of E, D, and ρv.

Chapter 4 Energy and Potential The Dipole The dipole fields form the basis for the behavior of dielectric materials in electric field. The dipole will be discussed now and will serve as an illustration about the importance of the potential concept presented previously. An electric dipole, or simply a dipole, is the name given to two point charges of equal magnitude and opposite sign, separated by a distance which is small compared to the distance to the point P at which we want to know the electric and potential fields.

Chapter 4 Energy and Potential The Dipole The distant point P is described by the spherical coordinates r, θ, Φ = 90°. The positive and negative point charges have separation d and described in rectangular coordinates (0,0, 0.5d) and (0,0,–0.5d).

The Dipole The total potential at P can be written as:
Chapter 4 Energy and Potential The Dipole The total potential at P can be written as: The plane z = 0 is the locus of points for which R1 = R2 ► The potential there is zero (as also all points at ∞).

The Dipole For a distant point, R1 ≈ R2 ≈ r, R2–R1 ≈ dcosθ
Chapter 4 Energy and Potential The Dipole For a distant point, R1 ≈ R2 ≈ r, R2–R1 ≈ dcosθ Using the gradient in spherical coordinates,

Plane at zero potential
Chapter 4 Energy and Potential The Dipole To obtain a plot of the potential field, we choose Qd/(4πε0) = 1 and thus cosθ = Vr2. The colored lines in the figure below indicate equipotentials for V = 0, +0.2, +0.4, +0.6, +0.8, and +1. r = 2.236 r = 1.880 Plane at zero potential 45°

Chapter 4 Energy and Potential The Dipole The potential field of the dipole may be simplified by making use of the dipole moment. If the vector length directed from –Q to +Q is identified as d, then the dipole moment is defined as Qd and is assigned the symbol p. Since dar = d cosθ , we then have: Dipole charges: Point charge:

Conductors and Dielectrics
Chapter 5 Conductors and Dielectrics

Current and Current Density
Chapter 5 Current and Conductors Current and Current Density Electric charges in motion constitute a current. The unit of current is the ampere (A), defined as a rate of movement of charge passing a given reference point (or crossing a given reference plane). Current is defined as the motion of positive charges, although conduction in metals takes place through the motion of electrons. Current density J is defined, measured in amperes per square meter (A/m2).

Chapter 5 Current and Conductors Current and Current Density The increment of current ΔI crossing an incremental surface ΔS normal to the current density is: If the current density is not perpendicular to the surface, Through integration, the total current is obtained:

Chapter 5 Current and Conductors Current and Current Density Current density may be related to the velocity of volume charge density at a point. An element of charge ΔQ = ρvΔSΔL moves along the x axis In the time interval Δt, the element of charge has moved a distance Δx The charge moving through a reference plane perpendicular to the direction of motion is ΔQ = ρvΔSΔx

Chapter 5 Current and Conductors Current and Current Density The limit of the moving charge with respect to time is: In terms of current density, we find: This last result shows clearly that charge in motion constitutes a current. We name it here convection current. J = ρvv is then called convection current density.

Chapter 5 Current and Conductors Continuity of Current The principle of conservation of charge: “Charges can be neither created nor destroyed.” But, equal amounts of positive and negative charge (pair of charges) may be simultaneously created, obtained by separation, destroyed, or lost by recombination. The Continuity Equation in Closed Surface Any outward flow of positive charge must be balanced by a decrease of positive charge (or perhaps an increase of negative charge) within the closed surface. If the charge inside the closed surface is denoted by Qi, then the rate of decrease is –dQi/dt and the principle of conservation of charge requires: The Integral Form of the Continuity Equation

Chapter 5 Current and Conductors Continuity of Current The differential form (or point form) of the continuity equation is obtained by using the divergence theorem: We next represent Qi by the volume integral of ρv: If we keep the surface constant, the derivative becomes a partial derivative. Writing it within the integral, The Differential Form (Point Form) of the Continuity Equation

Continuity of Current Example The current density is given by .
Chapter 5 Current and Conductors Continuity of Current Example The current density is given by Total outward current at time instant t = 1 s and r = 5 m. Total outward current at time instant t = 1 s and r = 6 m. Finding volume charge density:

Chapter 5 Current and Conductors Metallic Conductors The energy-band structure of three types of materials at 0 K is shown as follows: Energy in the form of heat, light, or an electric field may raise the energy of the electrons of the valence band, and in sufficient amount they will be excited and jump the energy gap into the conduction band.

Metallic Conductors First let us consider the conductor.
Chapter 5 Current and Conductors Metallic Conductors First let us consider the conductor. Here, the valence electrons (or free conductive electrons) move under the influence of an electric field E. An electron having a charge Q = –e will experiences a force: In the crystalline material, the progress of the electron is impeded by collisions with the lattice structure, and a constant average velocity is soon attained. This velocity vd is termed the drift velocity. It is linearly related to the electric field intensity by the mobility of the electron μe: The Point Form of Ohm’s Law

Chapter 5 Current and Conductors Metallic Conductors The application of Ohm’s law in point form to a macroscopic region leads to a more familiar form. Assuming J and E to be uniform, in a cylindrical region shown below, we can write:

Conductor Properties and Boundary Conditions
Chapter 5 Current and Conductors Conductor Properties and Boundary Conditions Property 1: The charge density within a conductor is zero (ρv = 0) and the surface charge density resides on the exterior surface. Property 2: In static conditions, no current may flow, thus the electric field intensity within the conductor is zero (E = 0). Now our next concern is the fields external to the conductor. The external electric field intensity and electric flux density are decomposed into the tangential components and the normal components.

Chapter 5 Current and Conductors Conductor Properties and Boundary Conditions The tangential component of the electric field intensity is seen to be zero Et = 0  Dt = 0.  If not, then a force will be applied to the surface charges, resulting in their motion and no static conditions. The normal component of the electric flux density leaving the surface is equal to the surface charge density in coulombs per square meter (DN = ρS).  According to Gauss’s law, the electric flux leaving an incremental surface is equal to the charge residing on that incremental surface.  The flux cannot penetrate into the conductor since the total field there is zero.  It must leave the surface normally.

Chapter 5 Current and Conductors Conductor Properties and Boundary Conditions

Chapter 5 Current and Conductors Conductor Properties and Boundary Conditions Example Given the potential V = 100(x2–y2) and a point P(2,–1,3) that is predefined 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. Carefully examine the surface direction

Chapter 5 Current and Conductors The Method of Images One important characteristic of the dipole field developed in Chapter 4 is the infinite plane at zero potential that exists midway between the two charges. Such a plane may be represented by a thin infinite conducting plane. The conductor is an equipotential surface at a potential V = 0. The electric field intensity, as for a plane, is normal to the surface.

Chapter 5 Current and Conductors The Method of Images Thus, we can replace the dipole configuration (left) with the single charge and conducting plane (right), without affecting the fields in the upper half of the figure. Now, we begin with a single charge above a conducting plane. ► The same fields above the plane can be maintained by removing the plane and locating a negative charge at a symmetrical location below the plane. This charge is called the image of the original charge, and it is the negative of that value.

The Method of Images The same procedure can be done again and again.
Chapter 5 Current and Conductors The Method of Images The same procedure can be done again and again. Any charge configuration above an infinite ground plane may be replaced by an arrangement composed of the given charge configuration, its image, and no conducting plane.

The Method of Images Example
Chapter 5 Current and Conductors The Method of Images Example Find the surface charge density at P(2,5,0) on the conducting plane z = 0 if there is a line charge of 30 nC/m located at x = 0, z = 3, as shown below. We remove the plane and install an image line charge The field at P may now be obtained by superposition of the known fields of the line charges

The Method of Images Normal to the plane x = 0, z = 3 P(2,5,0)
Chapter 5 Current and Conductors The Method of Images x = 0, z = 3 P(2,5,0) x = 0, z = –3 Normal to the plane

Chapter 5 Current and Conductors Semiconductors In an intrinsic semiconductor material, such as pure germanium or silicon, two types of current carriers are present: electrons and holes. The electrons are those from the top of the filled valence band which have received sufficient energy to cross the small forbidden band into conduction band. The forbidden-band energy gap in typical semiconductors is of the order of 1 eV. The vacancies left by the electrons represent unfilled energy states in the valence band. They may also move from atom to atom in the crystal. The vacancy is called a hole, and the properties of semiconductor are described by treating the hole as a positive charge of e, a mobility μh, and an effective mass comparable to that of the electron.

Semiconductors The conductivity of a semiconductor is described as:
Chapter 5 Current and Conductors Semiconductors The conductivity of a semiconductor is described as: As temperature increases, the mobilities decrease, but the charge densities increase very rapidly. As a result, the conductivity of silicon increases by a factor of 100 as the temperature increases from about 275 K to 330 K.

Chapter 5 Current and Conductors Semiconductors The conductivity of the intrinsic semiconductor increases with temperature, while that of a metallic conductor decreases with temperature. The intrinsic semiconductors also satisfy the point form of Ohm's law: the conductivity is reasonably constant with current density and with the direction of the current density.

Chapter 6 capacitance

The Nature of Dielectric Materials
Chapter 6 Dielectrics and Capacitance The Nature of Dielectric Materials A dielectric material in an electric field can be viewed as a free- space arrangement of microscopic electric dipoles, a pair of positive and negative charges whose centers do not quite coincide. These charges are not free charges, not contributing to the conduction process. They are called bound charges, can only shift positions slightly in response to external fields. All dielectric materials have the ability to store electric energy. This storage takes place by means of a shift (displacement) in the relative positions of the bound charges against the normal molecular and atomic forces.

Chapter 6 Dielectrics and Capacitance The Nature of Dielectric Materials The mechanism of this charge displacement differs in various dielectric materials. Polar molecules have a permanent displacement existing between the centers of “gravity” of the positive and negative charges, each pair of charges acts as a dipole. Dipoles are normally oriented randomly, and the action of the external field is to align these molecules in the same direction. Nonpolar molecules does not have 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 which is aligned with the electric field.

Chapter 6 Dielectrics and Capacitance The Nature of Dielectric Materials Either type of dipole may be described by its dipole moment p: If there are n dipoles per unit volume, then there are nΔv dipoles in a volume Δv. The total dipole moment is: We now define the polarization P as the dipole moment per unit volume: The immediate goal is to show that the bound-volume charge density acts like the free-volume charge density in producing an external field ► We shall obtain a result similar to Gauss’s law.

Chapter 6 Dielectrics and Capacitance The Nature of Dielectric Materials Take a dielectric containing nonpolar molecules. No molecules has p, and P = 0 throughout the material. Somewhere in the interior of the dielectric we select an incremental surface element ΔS, and apply an electric field E. The electric field produces a moment p = Qd in each molecule, such that p and d make an angle θ with ΔS. Due to E, any positive charges initially lying below the surface ΔS and within ½dcosθ must have crossed ΔS going upward. Any negative charges initially lying above the surface ΔS and within ½dcosθ must have crossed ΔS going downward.

Chapter 6 Dielectrics and Capacitance The Nature of Dielectric Materials For n molecules/m3, the net total charge (positive and negative) which crosses the elemental surface in upward direction is: The notation Qb means the bound charge. In terms of the polarization, we have: If we interpret ΔS as an element of a closed surface, then the direction of ΔS is outward. The net increase in the bound charge within the closed surface is:

Chapter 6 Dielectrics and Capacitance The Nature of Dielectric Materials Seeing some similarity to Gauss’s law, we may now generalize the definition of electric flux density so that it applies to media other than free space. We write Gauss’s law in terms of ε0E and QT, the total enclosed charge (bound charge plus free charge): Combining the last three equations: We may now define D in more general terms: There is an added term to D when a material is polarized

Chapter 6 Dielectrics and Capacitance The Nature of Dielectric Materials For equations with volume charge densities, we now have: With the help of the divergence theorem, we may transform the equations into equivalent divergence relationships:

Chapter 6 Dielectrics and Capacitance The Nature of Dielectric Materials To utilize the new concepts, it is necessary to know the relationship between E and P. This relationship will be a function of the type of material. We will limit the discussion to 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. The linear relationship between P and E can be described as: We now define: χe : electric susceptibility, a measure of how easily a dielectric polarizes in response to an electric field εr : relative permittivity

Chapter 6 Dielectrics and Capacitance The Nature of Dielectric Materials In summary, we now have a relationship between D and E which depends on the dielectric material present:

Chapter 6 Dielectrics and Capacitance The Nature of Dielectric Materials Example We locate a slab of Teflon in the region 0 ≤ x ≤ a, and assume free space where x < 0 and x > a. Outside the Teflon there is a uniform field Eout = E0ax V/m. Find the values for D, E, and P everywhere. No dielectric materials outside 0 ≤ x ≤ a No relations yet established over the boundary This will be discussed in the next section

Boundary Conditions for Perfect Dielectric Materials
Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials Consider the interface between two dielectrics having permittivities ε1 and ε2, as shown below. We first examine the tangential components around the small closed path on the left, with Δw<< :

Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials The tangential electric flux density is discontinuous, The boundary conditions on the normal components are found by applying Gauss’s law to the small cylinder shown at the right of the previous figure (net tangential flux is zero). ρS cannot be a bound surface charge density because the polarization already counted in by using dielectric constant different from unity ρS cannot be a free surface charge density, for no free charge available in the perfect dielectrics we are considering ρS exists only in special cases where it is deliberately placed there

Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials Except for this special case, we may assume ρS is zero on the interface: The normal component of electric flux density is continuous. It follows that:

Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials Combining the normal and the tangential components of D, After one division,

Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials The direction of E on each side of the boundary is identical with the direction of D, because D = εE.

Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials The relationship between D1 and D2 may be found from: The relationship between E1 and E2 may be found from:

Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials Example Complete the previous example by finding the fields within the Teflon. E only has normal component

Boundary Conditions Between a Conductor and a Dielectric
Chapter 6 Dielectrics and Capacitance Boundary Conditions Between a Conductor and a Dielectric The boundary conditions existing at the interface between a conductor and a dielectric are much simpler than those previously discussed. First, we know that D and E are both zero inside the conductor. Second, the tangential E and D components must both be zero to satisfy: Finally, the application of Gauss’s law shows once more that both D and E are normal to the conductor surface and that DN = ρS and EN = ρS/ε. The boundary conditions for conductor–free space are valid also for conductor–dielectric boundary, with ε0 replaced by ε.

Chapter 6 Dielectrics and Capacitance Boundary Conditions Between a Conductor and a Dielectric We will now spend a moment to examine one phenomena: “Any charge that is introduced internally within a conducting material will arrive at the surface as a surface charge.” Given Ohm’s law and the continuity equation (free charges only): We have:

Chapter 6 Dielectrics and Capacitance Boundary Conditions Between a Conductor and a Dielectric If we assume that the medium is homogenous, so that σ and ε are not functions of position, we will have: Using Maxwell’s first equation, we obtain; Making the rough assumption that σ is not a function of ρv, it leads to an easy solution that at least permits us to compare different conductors. The solution of the above equation is: ρ0 is the charge density at t = 0 Exponential decay with time constant of ε/σ

Chapter 6 Dielectrics and Capacitance Boundary Conditions Between a Conductor and a Dielectric Good conductors have low time constant. This means that the charge density within a good conductors will decay rapidly. We may then safely consider the charge density to be zero within a good conductor. In reality, no dielectric material is without some few free electrons (the charge density is thus not completely zero), but the charge introduced internally in any of them will eventually reach the surface. t ρv ρ0 ρ0/e ε/σ

Chapter 6 Dielectrics and Capacitance Capacitance Now let us consider two conductors embedded in a homogenous dielectric. Conductor M2 carries a total positive charge Q, and M2 carries an equal negative charge –Q. No other charges present  the total charge of the system is zero. The charge is carried on the surface as a surface charge density. The electric field is normal to the conductor surface. Each conductor is an equipotential surface

Chapter 6 Dielectrics and Capacitance Capacitance The electric flux is directed from M2 to M1, thus M2 is at the more positive potential. Works must be done to carry a positive charge from M1 to M2. Let us assign V0 as the potential difference between M2 and M1. 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 the conductors.

Chapter 6 Dielectrics and Capacitance Capacitance The capacitance is independent of the potential and total charge for their ratio is constant. If the charge density is increased by a factor, Gauss's law indicates that the electric flux density or electric field intensity also increases by the same factor, as does the potential difference. Capacitance is a function only of the physical dimensions of the system of conductors and of the permittivity of the homogenous dielectric. Capacitance is measured in farads (F), 1 F = 1 C/V.

Chapter 6 Dielectrics and Capacitance Capacitance We will now apply the definition of capacitance to a simple two- conductor system, where the conductors are identical, infinite parallel planes, and separated a distance d to each other. The charge on the lower plane is positive, since D is upward. The charge on the upper plane is negative,

Chapter 6 Dielectrics and Capacitance Capacitance The potential difference between lower and upper planes is: The total charge for an area S of either plane, both with linear dimensions much greater than their separation d, is: The capacitance of a portion of the infinite-plane arrangement, far from the edges, is:

Chapter 6 Dielectrics and Capacitance Capacitance Example Calculate the capacitance of a parallel-plate capacitor having a mica dielectric, εr = 6, a plate area of 10 in2, and a separation of 0.01 in.

Capacitance The total energy stored in the capacitor is: Chapter 6
Dielectrics and Capacitance Capacitance The total energy stored in the capacitor is:

Several Capacitance Examples
Chapter 6 Dielectrics and Capacitance Several Capacitance Examples As first example, consider a coaxial cable or coaxial capacitor of inner radius a, outer radius b, and length L. The capacitance is given by: Next, consider a spherical capacitor formed of two concentric spherical conducting shells of radius a and b, b>a.

Chapter 6 Dielectrics and Capacitance Several Capacitance Examples If we allow the outer sphere to become infinitely large, we obtain the capacitance of an isolated spherical conductor: A sphere about the size of a marble, with a diameter of 1 cm, will have: Coating this sphere with a different dielectric layer, for which ε = ε1, extending from r = a to r = r1,

Chapter 6 Dielectrics and Capacitance Several Capacitance Examples While the potential difference is: Therefore,

Chapter 6 Dielectrics and Capacitance Several Capacitance Examples A capacitor can be made up of several dielectrics. Consider a parallel-plate capacitor of area S and spacing d, d << linear dimension of S. The capacitance is ε1S/d, using a dielectric of permittivity ε1. Now, let us replace a part of this dielectric by another of permittivity ε2, placing the boundary between the two dielectrics parallel to the plates. Assuming a charge Q on one plate, ρS = Q/S, while DN1 = DN2, since D is only normal to the boundary. E1 = D1/ε1 = Q/(ε1S), E2 = D2/ε2 = Q/(ε2S). V1 = E1d1, V2 = E2d2.

Chapter 6 Dielectrics and Capacitance Several Capacitance Examples Another configuration is when the dielectric boundary were placed normal to the two conducting plates and the dielectrics occupied areas of S1 and S2. Assuming a charge Q on one plate, Q = ρS1S1 + ρS2S2. ρS1 = D1 = ε1E1, ρS2 = D2 = ε2E2. V0 = E1d = E2d.

Capacitance of a Two-Wire Line
Chapter 6 Dielectrics and Capacitance Capacitance of a Two-Wire Line The configuration of the two-wire line consists of two parallel conducting cylinders, each of circular cross section. We shall be able to find complete information about the electric field intensity, the potential field, the surface charge density distribution, and the capacitance. This arrangement is an important type of transmission line.

Chapter 6 Dielectrics and Capacitance Capacitance of a Two-Wire Line The potential field of two infinite line charges, with a positive line charge in the xz plane at x = a and a negative line at x = –a is shown below. The potential of a single line charge with zero reference at a radius of R0 is: The combined potential field can be written as:

Chapter 6 Dielectrics and Capacitance Capacitance of a Two-Wire Line We choose R10 = R20, thus placing the zero reference at equal distances from each line. Expressing R1 and R2 in terms of x and y, To recognize the equipotential surfaces, some algebraic manipulations are necessary. Choosing an equipotential surface V = V1, we define a dimensionless parameter K1 as:

Chapter 6 Dielectrics and Capacitance Capacitance of a Two-Wire Line After some multiplications and algebra, we obtain: The last equation shows that the V = V1 equipotential surface is independent of z and intersects the xz plane in a circle of radius b, The center of the circle is x = h, y = 0, where:

Chapter 6 Dielectrics and Capacitance Capacitance of a Two-Wire Line Let us no consider a zero-potential conducting plane located at x = 0, and a conducting cylinder of radius b and potential V0 with its axis located a distance h from the plane. Solving the last two equations for a and K1 in terms of b and h, The potential of the cylinder is V0, so that: Therefore,

Chapter 6 Dielectrics and Capacitance Capacitance of a Two-Wire Line Given h, b, and V0, we may determine a, K1, and ρL. The capacitance between the cylinder and the plane is now available. For a length L in the z direction,

Chapter 6 Dielectrics and Capacitance Capacitance of a Two-Wire Line Example The black circle shows the cross section of a cylinder of 5 m radius at a potential of 100 V in free space. Its axis is 13 m away from a plane at zero potential.

Chapter 6 Dielectrics and Capacitance Capacitance of a Two-Wire Line We may also identify the cylinder representing the 50 V equipotential surface by finding new values for K1, b, and h.

Chapter 6 Dielectrics and Capacitance Capacitance of a Two-Wire Line

Chapter 6 Dielectrics and Capacitance Capacitance of a Two-Wire Line + -

Chapter 6 Dielectrics and Capacitance Capacitance of a Two-Wire Line For the case of a conductor with b << h, then:

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