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**EEE 498/598 Overview of Electrical Engineering**

Lecture 4: Electrostatics: Electrostatic Shielding; Poisson’s and Laplace’s Equations; Capacitance; Dielectric Materials and Permittivity 1

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Lecture 4 Objectives To continue our study of electrostatics with electrostatic shielding; Poisson’s and Laplace’s equations; capacitance; and dielectric materials and permittivity. 2

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**Ungrounded Spherical Metallic Shell**

Consider a point charge at the center of a spherical metallic shell: Q a b Electrically neutral 3

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**Ungrounded Spherical Metallic Shell (Cont’d)**

The applied electric field is given by 4

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**Ungrounded Spherical Metallic Shell (Cont’d)**

The total electric field can be obtained using Gauss’s law together with our knowledge of how fields behave in a conductor. 5

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**Ungrounded Spherical Metallic Shell (Cont’d)**

(1) Assume from symmetry the form of the field (2) Construct a family of Gaussian surfaces spheres of radius r where 6

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**Ungrounded Spherical Metallic Shell (Cont’d)**

Here, we shall need to treat separately 3 sub-families of Gaussian surfaces: a b 1) 2) 3) 7

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**Ungrounded Spherical Metallic Shell (Cont’d)**

(3) Evaluate the total charge within the volume enclosed by each Gaussian surface 8

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**Ungrounded Spherical Metallic Shell (Cont’d)**

Gaussian surfaces for which Gaussian surfaces for which Gaussian surfaces for which 9

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**Ungrounded Spherical Metallic Shell (Cont’d)**

For Shell is electrically neutral: The net charge carried by shell is zero. 10

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**Ungrounded Spherical Metallic Shell (Cont’d)**

For since the electric field is zero inside conductor. A surface charge must exist on the inner surface and be given by 11

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**Ungrounded Spherical Metallic Shell (Cont’d)**

Since the conducting shell is initially neutral, a surface charge must also exist on the outer surface and be given by 12

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**Ungrounded Spherical Metallic Shell (Cont’d)**

(4) For each Gaussian surface, evaluate the integral surface area of Gaussian surface. magnitude of D on Gaussian surface. 13

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**Ungrounded Spherical Metallic Shell (Cont’d)**

(5) Solve for D on each Gaussian surface (6) Evaluate E as 14

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**Ungrounded Spherical Metallic Shell (Cont’d)**

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**Ungrounded Spherical Metallic Shell (Cont’d)**

The induced field is given by 16

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**Ungrounded Spherical Metallic Shell (Cont’d)**

total electric field Eapp r a b Eind 17

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**Ungrounded Spherical Metallic Shell (Cont’d)**

The electrostatic potential is obtained by taking the line integral of E. To do this correctly, we must start at infinity (the reference point or ground) and “move in” back toward the point charge. For r > b 18

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**Ungrounded Spherical Metallic Shell (Cont’d)**

Since the conductor is an equipotential body (and potential is a continuous function), we have for 19

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**Ungrounded Spherical Metallic Shell (Cont’d)**

For 20

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**Ungrounded Spherical Metallic Shell (Cont’d)**

V No metallic shell r a b 21

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**Grounded Spherical Metallic Shell**

When the conducting sphere is grounded, we can consider it and ground to be one huge conducting body at ground (zero) potential. Electrons migrate from the ground, so that the conducting sphere now has an excess charge exactly equal to -Q. This charge appears in the form of a surface charge density on the inner surface of the sphere. 22

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**Grounded Spherical Metallic Shell**

There is no longer a surface charge on the outer surface of the sphere. The total field outside the sphere is zero. The electrostatic potential of the sphere is zero. Q a b - 23

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**Grounded Spherical Metallic Shell (Cont’d)**

total electric field Eapp R a b Eind 24

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**Grounded Spherical Metallic Shell (Cont’d)**

V Grounded metallic shell acts as a shield. R a b 25

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**The Need for Poisson’s and Laplace’s Equations**

So far, we have studied two approaches for finding the electric field and electrostatic potential due to a given charge distribution. 26

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**The Need for Poisson’s and Laplace’s Equations (Cont’d)**

Method 1: given the position of all the charges, find the electric field and electrostatic potential using (A) 27

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**The Need for Poisson’s and Laplace’s Equations (Cont’d)**

(B) Method 1 is valid only for charges in free space. 28

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**The Need for Poisson’s and Laplace’s Equations (Cont’d)**

Method 2: Find the electric field and electrostatic potential using Gauss’s Law Method 2 works only for symmetric charge distributions, but we can have materials other than free space present. 29

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**The Need for Poisson’s and Laplace’s Equations (Cont’d)**

Consider the following problem: What are E and V in the region? Conducting bodies Neither Method 1 nor Method 2 can be used! 30

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**The Need for Poisson’s and Laplace’s Equations (Cont’d)**

Poisson’s equation is a differential equation for the electrostatic potential V. Poisson’s equation and the boundary conditions applicable to the particular geometry form a boundary-value problem that can be solved either analytically for some geometries or numerically for any geometry. After the electrostatic potential is evaluated, the electric field is obtained using 31

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**Derivation of Poisson’s Equation**

For now, we shall assume the only materials present are free space and conductors on which the electrostatic potential is specified. However, Poisson’s equation can be generalized for other materials (dielectric and magnetic as well). 32

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**Derivation of Poisson’s Equation (Cont’d)**

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**Derivation of Poisson’s Equation (Cont’d)**

2 is the Laplacian operator. The Laplacian of a scalar function is a scalar function equal to the divergence of the gradient of the original scalar function. 34

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**Laplacian Operator in Cartesian, Cylindrical, and Spherical Coordinates**

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Laplace’s Equation Laplace’s equation is the homogeneous form of Poisson’s equation. We use Laplace’s equation to solve problems where potentials are specified on conducting bodies, but no charge exists in the free space region. Laplace’s equation 36

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Uniqueness Theorem A solution to Poisson’s or Laplace’s equation that satisfies the given boundary conditions is the unique (i.e., the one and only correct) solution to the problem. 37

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**Potential Between Coaxial Cylinders Using Laplace’s Equation**

Two conducting coaxial cylinders exist such that a b x y + V0 38

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**Potential Between Coaxial Cylinders Using Laplace’s Equation (Cont’d)**

Assume from symmetry that 39

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**Potential Between Coaxial Cylinders Using Laplace’s Equation (Cont’d)**

Two successive integrations yield The two constants are obtained from the two BCs: 40

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**Potential Between Coaxial Cylinders Using Laplace’s Equation (Cont’d)**

Solving for C1 and C2, we obtain: The potential is 41

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**Potential Between Coaxial Cylinders Using Laplace’s Equation (Cont’d)**

The electric field between the plates is given by: The surface charge densities on the inner and outer conductors are given by 42

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**Capacitance of a Two Conductor System**

The capacitance of a two conductor system is the ratio of the total charge on one of the conductors to the potential difference between that conductor and the other conductor. + V12 = V2-V1 V2 V1 - 43

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**Capacitance of a Two Conductor System**

Capacitance is a positive quantity measured in units of Farads. Capacitance is a measure of the ability of a conductor configuration to store charge. 44

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**Capacitance of a Two Conductor System**

The capacitance of an isolated conductor can be considered to be equal to the capacitance of a two conductor system where the second conductor is an infinite distance away from the first and at ground potential. 45

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Capacitors A capacitor is an electrical device consisting of two conductors separated by free space or another conducting medium. To evaluate the capacitance of a two conductor system, we must find either the charge on each conductor in terms of an assumed potential difference between the conductors, or the potential difference between the conductors for an assumed charge on the conductors. 46

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Capacitors (Cont’d) The former method is the more general but requires solution of Laplace’s equation. The latter method is useful in cases where the symmetry of the problem allows us to use Gauss’s law to find the electric field from a given charge distribution. 47

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**Parallel-Plate Capacitor**

Determine an approximate expression for the capacitance of a parallel-plate capacitor by neglecting fringing. Conductor 2 d A Conductor 1 48

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**Parallel-Plate Capacitor (Cont’d)**

“Neglecting fringing” means to assume that the field that exists in the real problem is the same as for the infinite problem. z z = d V = V12 z = 0 V = 0 49

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**Parallel-Plate Capacitor (Cont’d)**

Determine the potential between the plates by solving Laplace’s equation. 50

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**Parallel-Plate Capacitor (Cont’d)**

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**Parallel-Plate Capacitor (Cont’d)**

Evaluate the electric field between the plates 52

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**Parallel-Plate Capacitor (Cont’d)**

Evaluate the surface charge on conductor 2 Evaluate the total charge on conductor 2 53

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**Parallel-Plate Capacitor (Cont’d)**

Evaluate the capacitance 54

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Dielectric Materials A dielectric (insulator) is a medium which possess no (or very few) free electrons to provide currents due to an impressed electric field. Although there is no macroscopic migration of charge when a dielectric is placed in an electric field, microscopic displacements (on the order of the size of atoms or molecules) of charge occur resulting in the appearance of induced electric dipoles. 55

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**Dielectric Materials (Cont’d)**

A dielectric is said to be polarized when induced electric dipoles are present. Although all substances are polarizable to some extent, the effects of polarization become important only for insulating materials. The presence of induced electric dipoles within the dielectric causes the electric field both inside and outside the material to be modified. 56

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Polarizability Polarizability is a measure of the ability of a material to become polarized in the presence of an applied electric field. Polarization occurs in both polar and nonpolar materials. 57

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**Electronic Polarizability**

cloud nucleus In the absence of an applied electric field, the positively charged nucleus is surrounded by a spherical electron cloud with equal and opposite charge. Outside the atom, the electric field is zero. 58

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**Electronic Polarizability (Cont’d)**

Eapp In the presence of an applied electric field, the electron cloud is distorted such that it is displaced in a direction (w.r.t. the nucleus) opposite to that of the applied electric field. 59

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**Electronic Polarizability (Cont’d)**

The net effect is that each atom becomes a small charge dipole which affects the total electric field both inside and outside the material. dipole moment (C-m) polarizability (F-m2) 60

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Ionic Polarizability negative ion positive ion In the absence of an applied electric field, the ionic molecules are randomly oriented such that the net dipole moment within any small volume is zero. 61

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**Ionic Polarizability (Cont’d)**

Eapp In the presence of an applied electric field, the dipoles tend to align themselves with the applied electric field. 62

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**Ionic Polarizability (Cont’d)**

The net effect is that each ionic molecule is a small charge dipole which aligns with the applied electric field and influences the total electric field both inside and outside the material. dipole moment (C-m) polarizability (F-m2) 63

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**Orientational Polarizability**

In the absence of an applied electric field, the polar molecules are randomly oriented such that the net dipole moment within any small volume is zero. 64

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**Orientational Polarizability (Cont’d)**

Eapp In the presence of an applied electric field, the dipoles tend to align themselves with the applied electric field. 65

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**Orientational Polarizability (Cont’d)**

The net effect is that each polar molecule is a small charge dipole which aligns with the applied electric field and influences the total electric field both inside and outside the material. dipole moment (C-m) polarizability (F-m2) 66

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**Polarization Per Unit Volume**

The total polarization of a given material may arise from a combination of electronic, ionic, and orientational polarizability. The polarization per unit volume is given by 67

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**Polarization Per Unit Volume (Cont’d)**

P is the polarization per unit volume. (C/m2) N is the number of dipoles per unit volume. (m-3) p is the average dipole moment of the dipoles in the medium. (C-m) aT is the average polarizability of the dipoles in the medium. (F-m2) 68

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**Polarization Per Unit Volume (Cont’d)**

Eloc is the total electric field that actually exists at each dipole location. For gases Eloc = E where E is the total macroscopic field. For solids 69

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**Polarization Per Unit Volume (Cont’d)**

From the macroscopic point of view, it suffices to use electron susceptibility (dimensionless) 70

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Dielectric Materials The effect of an applied electric field on a dielectric material is to create a net dipole moment per unit volume P. The dipole moment distribution sets up induced secondary fields: 71

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**Volume and Surface Bound Charge Densities**

A volume distribution of dipoles may be represented as an equivalent volume (qevb) and surface (qesb) distribution of bound charge. These charge distributions are related to the dipole moment distribution: 72

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**Gauss’s Law in Dielectrics**

Gauss’s law in differential form in free space: Gauss’s law in differential form in dielectric: 73

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**Displacement Flux Density**

Hence, the displacement flux density vector is given by 74

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**General Forms of Gauss’s Law**

Gauss’s law in differential form: Gauss’s law in integral form: 75

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**Permittivity Concept Assuming that we have**

The parameter e is the electric permittivity or the dielectric constant of the material. 76

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**Permittivity Concept (Cont’d)**

The concepts of polarizability and dipole moment distribution are introduced to relate microscopic phenomena to the macroscopic fields. The introduction of permittivity eliminates the need for us to explicitly consider microscopic effects. Knowing the permittivity of a dielectric tells us all we need to know from the point of view of macroscopic electromagnetics. 77

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**Permittivity Concept (Cont’d)**

For the most part in macroscopic electromagnetics, we specify the permittivity of the material and if necessary calculate the dipole moment distribution within the medium by using 78

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**Relative Permittivity**

The relative permittivity of a dielectric is the ratio of the permittivity of the dielectric to the permittivity of free space 79

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