UNIVERSITI MALAYSIA PERLIS EKT 2414 ELECTROMAGNETIC THEORY CHAPTER

UNIVERSITI MALAYSIA PERLIS EKT 241/4: ELECTROMAGNETIC THEORY CHAPTER 3 – ELECTROSTATICS 1

Chapter Outline § § § Maxwell’s Equations Charge and Current Distributions Coulomb’s Law Gauss’s Law Electric Scalar Potential Electrical Properties of Materials Conductors & Dielectrics Electric Boundary Conditions Capacitance Electrostatic Potential Energy Image Method 2

Maxwell’s equations: Where; E = electric field intensity D = electric flux density ρv = electric charge density per unit volume H = magnetic field intensity B = magnetic flux density 3

Maxwell’s equations: Relationship: D=εE B=µH ε = µ = electrical permittivity of the material magnetic permeability of the material 4

Maxwell’s equations • For static case, ∂/∂t = 0. • Maxwell’s equations is reduced to: Electrostatics Magnetostatics 5

Charge and current distributions § Charge may be distributed over a volume, a surface or a line. § Electric field due to continuous charge distributions: 6

Charge and current distributions §Volume charge density, ρv is defined as: §Total charge Q contained in a volume V is: 7

Charge and current distributions • Surface charge density • Total charge Q on a surface: 8

Charge and current distributions • Line charge density • Total charge Q along a line 9

Example 1 Calculate the total charge Q contained in a cylindrical tube of charge oriented along the zaxis. The line charge density is , where z is the distance in meters from the bottom end of the tube. The tube length is 10 cm. 10

Solution to Example 1 The total charge Q is: 11

Example 2 Find the total charge over the volume with volume charge density: 12

Solution to Example 2 The total charge Q: 13

Current densities • Current density, J is defined as: • Where: u = mean velocity of moving charges • For surface S, total current flowing through is 14

Current densities • There are 2 types of current: 1) Convection current § generated by actual movement of electrically charged matter; does NOT obey Ohm’s law § E. g movement of charged particles in cathode ray tube 2) Conduction current § atoms of conducting material do NOT move; obeys Ohm’s law § E. g movement of electrons in a metal wire 15

Coulomb’s Law Coulomb’s law for a point charge: Where; R = distance between P and q = unit vector from q to P ε = electrical permittivity of the medium containing the observation point P 16

Force acting on a charge • In the presence of an electric field E at a given point in space, • F is the force acting on a test charge q’ when that charge is placed at that given point in space • The electric field E is maybe due to a single charge or a distribution of many charges Units: F in Newtons (N), q’ in Coulombs (C) 17

For acting on a charge § For a material with electrical permittivity, ε: D=εE where: ε = εR ε 0 = 8. 85 × 10− 12 ≈ (1/36π) × 10− 9 (F/m) § For most material and under most condition, ε is constant, independent of the magnitude and direction of E 18

E-field due to multipoint charges • At point P, the electric field E 1 due to q 1 alone: • At point P, the electric field E 1 due to q 2 alone: 19

E-field due to multipoint charges • Total electric field E at point P due to two charges: 20

E-field due to multipoint charges • In general for case of N point of charges, 21

Example 3 Two point charges with and are located in free space at (1, 3, − 1) and (− 3, 1, − 2), respectively in a Cartesian coordinate system. Find: (a) the electric field E at (3, 1, − 2) (b) the force on a 8 × 10− 5 C charge located at that point. All distances are in meters. 22

Solution to Example 3 • The electric field E with ε = ε 0 (free space) is given by: • The vectors are: 23

Solution to Example 3 • a) Hence, • b) We have 24

E-field due to charge distribution continuous distribution • Total electric field due to 3 types of continuous charge distribution: 25

E-field of a ring of charge • Electric field due to a ring of charge is: 26

Example 4 Find the electric field at a point P(0, 0, h) in free space at a height h on the z-axis due to a circular disk of charge in the x–y plane with uniform charge density ρs as shown. Then evaluate E for the infinite-sheet case by letting a→∞. 27

Solution to Example 4 • A ring of radius r and width dr has an area ds = 2πrdr • The charge is: • The field due to the ring is: 28

Solution to Example 4 stop • The total electric field at P is With plus sign corresponds to h>0, minus sign corresponds to h<0. • For an infinite sheet of charge with a =∞, 29

Gauss’s law • Electric flux density D through an enclosing surface is proportional to enclosed charge Q. • Differential and integral form of Gauss’s law: 30

Example 5 • Use Gauss’s law to obtain an expression for E in free space due to an infinitely long line of charge with uniform charge density ρl along the z-axis. 31

Solution to Example 5 Construct a cylindrical Gaussian surface. The integral is: Equating both equations, and re-arrange, we get: 32

Solution to Example 5 Then, use , we get: Note: unit vector is inserted for E due to the fact that E is a vector in direction. For application example refer to tutorial 2 33

Electric scalar potential • Electric potential energy is required to move a unit charge between 2 points • The presence of an electric field between two points give rise to voltage difference 34

Electric Potential as a function of electric field • Integrating along any path between point P 1 and P 2, we get: Potential difference between P 1 and P 2 , regardless of path 1, 2 or 3 : 35

Electric Potential as a function of electric field • Kirchhoff’s voltage law states that the net voltage drop around a closed loop is zero. • Line integral E around closed contour C is: • The electric potential V at any point is given by: • Integration path between point P 1 and P 2 is arbitrary 36

Electric potential due to point charges • For a point charge located at the origin of a spherical coordinate systems, the electric field at a distance R: • The electric potential between two end points: 37

Electric potential due to point charges • For charge Q located other than origin, specified by a source position vector R 1, then V at the observation vector R becomes: • For N discrete point charges, electric potential is: 38

Electric potential due to continuous distributions • For a continuous charge distribution: 39

Electric field as a function of electric potential • To find E for any charge distribution easily, where: = gradient of V 40

Poisson’s & Laplace’s equations • Differential form of Gauss’s law: • This may be written as: • Then, using , we get: 41

Poisson’s & Laplace’s equations • Hence: • Poisson’s and Laplace’s equations are used to find V where boundaries are known: • Example: the region between the plates of a capacitor with a specified voltage difference across it. (we will see in capacitance topic) 42

Conductivity • Conductivity – characterizes the ease with which charges can move freely in a material. • Perfect dielectric, σ = 0. Charges do not move inside the material • Perfect conductor, σ = ∞. Charges move freely throughout the material 43 •

Conductivity • Drift velocity of electrons, in a conducting material is in the opposite direction to the externally applied electric field E: • Hole drift velocity, is in the same direction as the applied electric field E: where: µe = electron mobility (m 2/V. s) µh = hole mobility (m 2/V. s) 44

Conductivity • Conductivity of a material, σ, is defined as: where ρve = volume charge density of free electrons ρvh = volume charge density of free holes Ne = number of free electrons per unit volume Nh = number of free holes per unit volume e = absolute charge = 1. 6 × 10− 19 (C) 45

Conductivity • Conductivities of different materials: 46

Ohm’s Law stop • Point form of Ohm’s law states that: Where: J = current density σ = conductivity E = electric field intensity • Properties for perfect dielectric and conductor: 47

Example 6 A 2 -mm-diameter copper wire with conductivity of 5. 8 × 107 S/m and electron mobility of 0. 0032 (m 2/V·s) is subjected to an electric field of 20 (m. V/m). Find : The free electron volume charge density and the current density, total current flowing in the wire and the electron drift velocity, and the volume density of the free electrons. 48

Solution to Example 6 a) b) c) d) e) 49

Resistance • The resistance R of a conductor of length l and uniform cross section A (linear resistor) as shown in the figure below: 50

Resistance • For a resistor of arbitrary shape, the resistance R is: • The conductance G of a linear resistor: 51

Joule’s Law • Joule’s law states that for a volume v, the total dissipated power P is: • For a linear resistor, the dissipated power P: 52

Dielectrics • Conductor has free electrons. • Dielectric electrons are strongly bounded to the atom. • In a dielectric, an externally applied electric field, Eext cannot cause mass migration of charges since none are able to move freely. • But, Eext can polarize the atoms or molecules in the material. • The polarization is represented by an electric dipole. 53

Dielectrics • Fig (a) - Eext is absent: the center of the electron cloud is co-located with the center of the nucleus • Fig (b) - Eext is present: the two centers are separated by a distance d 54 • Fig (c) – an electric dipole caused by Eext

Electric Boundary Conditions • Electric field maybe continuous in each of two dissimilar media • But, the E-field maybe discontinuous at the boundary between them • Boundary conditions specify how the tangential and normal components of the field in one medium are related to the components in other medium across the boundary • Two dissimilar media could be: two different dielectrics, or a conductor and a dielectric, or two conductors 55

• Note the field will apply a force on both the positively charged nucleus and the negatively charged electron. However, these forces will move these particles in opposite directions • Note, an electric dipole has been created !

D, flux density is proportionally increase as polarization increase through induction of permittivity, ε of the material relating the E and D permittivity, ε = proportional to the permittivity of free space, ε 0 ε = εr ε 0

• However, the electron may be break free from the atom, creating a positive ion and a free electron. • We call these free charges, and the electric field will cause them to move in opposite directions : • Moving charge is electric current J(r ).

Electric Boundary Conditions • Electric field maybe continuous in each of two dissimilar media Dielectric and dielectric • But, the E-field maybe discontinuous at the boundary between them • Boundary conditions specify how the tangential Dielectric and conductor and normal components of the field in one condition medium are related to the components in other medium across the boundary Conductor • Two dissimilar media could be: and two conductor different dielectrics, or a conductor and a dielectric, or two conductors

Dielectric- dielectric boundary • Interface between two dielectric media 60

Dielectric- dielectric boundary • Based on the figure in previous slide: • First boundary condition related to the tangential components of the electric field E is: • Second boundary condition related to the normal components of the electric field E is: • OR 61

z ε 1 E 2 Exy 1 ε 2 Ez 1 Exy 2 Ez 2 xy -plane

xy • Solution: STOP HERE ε 1=2ε 0 ε 2=5ε 0 1) Exy 1=Exy 2 thus, Exy 2 = 3 ax+4 ay 2) Ez 1 = 5 az, but, Ez 2 = ? ? E 2 Ө 1 z E 1 Example 1: 1) Find E 2 in the dielectric, when E 1 = 3 ax+4 ay+5 az, 2) And find Ө 1 and Ө 2. 2ε 0(5 az) = 5ε 0(Ez 2) Ez 2 = 2 az thus, E 2 =3 az+4 ay+2 az

z ε 1=2ε 0 Ө 2 ε 2=8ε 0 E 1 E 2 xy Ө 1 Find E 1 if E 2 = 2 x -3 y +3 z with s = 3. 54 x 10 -11(C/m 2) And find Ө 1 and Ө 2

Perfect conductor • When a conducting slab is placed in an external electric field, • Charges that accumulate on the conductor surfaces induces an internal electric field • Hence, total field inside conductor is zero. 65

Dielectric-conductor boundary • Based on the figure in previous slide: • In a perfect conductor, • Hence, • This requires the tangential and normal components of E 2 and D 2 to be zero. 66

Dielectric-conductor boundary • Assume medium 1 is a dielectric • Medium 2 is a perfect conductor 67

Dielectric-conductor boundary • The fields in the dielectric medium, at the boundary with the conductor is. • Since , it follows that. • Using the equation, , • we get: • Hence, boundary condition at conductor surface: where = normal vector pointing outward 68

Conductor- conductor boundary • Boundary between two conducting media: • Using the 1 st and 2 nd boundary conditions: and 69

xy J 2 Jz 1 Jxy 1 Jz 2 J 1 Jxy 2 z

Conductor- conductor boundary • In conducting media, electric fields give rise to current densities. • From , we have: and • The normal component of J has be continuous across the boundary between two different media under electrostatic conditions. 71

Conductor- conductor boundary • Hence, upon setting , we found the boundary condition for conductor- conductor boundary: 72

Capacitance • Capacitor – two conducting bodies separated by a dielectric medium 73

Capacitance • Capacitor – two conducting bodies separated by a dielectric medium ρs = Q / A ε 1 E ρs = surface charge density Q = charge (+ve / -ve) A = surface Area E E E = ρs/ε ε 2 E=0/V=0 on the surface

Capacitance • Capacitance is defined as: where: V = potential difference (V) Q = charge (C) C = capacitance (F) 75

Capacitance • Capacitance is defined as: where: V = potential difference (V) Q = charge (C) C = capacitance (F)

Example 7 Obtain an expression for the capacitance C of a parallel-plate capacitor comprised of two parallel plates each of surface area A and separated by a distance d. The capacitor is filled with a dielectric material with permittivity ε.

Solution to Example 7 • The charge density on the upper plate is ρs = Q/A. Hence, • magnitude of E at the dielectric-conductor boundary: • The voltage difference is • Hence, the capacitance is: 78

Electrostatic potential energy • Assume a capacitor with plates of good conductors – zero resistance, • Dielectric between two conductors has negligible conductivity, σ ≈ 0 – no current can flow through dielectric • No ohmic losses occur anywhere in capacitor • When a source is connected to a capacitor, energy is stored in capacitor • Charging-up energy is stored in the form of electrostatic potential energy in the dielectric medium 79

Electrostatic potential energy • Electrostatic potential energy, • The capacitance: • Hence, We for a parallel plate capacitor: 80

Image Method • Image theory states that a charge Q above a grounded perfectly conducting plane is equal to Q and its image –Q with ground plane removed. 81

Example 8 Use image theory to determine E at an arbitrary point P (x, y, z) in the region z > 0 due to a charge Q in free space at a distance d above a grounded conducting plane. 82

Solution to Example 8 • Charge Q is at (0, 0, d) and its image −Q is at (0, 0, −d) in Cartesian coordinates. Using Coulomb’s law, E at point P(x, y, z) due to two point charges: 83