EKT 241/4: ELECTROMAGNETIC THEORY

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

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

Maxwell’s equations 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

Maxwell’s equations Maxwell’s equations: Relationship: D = ε E B = µ H ε = electrical permittivity of the material µ = magnetic permeability of the material

Maxwell’s equations • For staticcase, ∂/∂t = 0. • Maxwell’s equations is reduced to: ElectrostaticsMagnetostatics

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

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

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

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

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

Solution to Example 1 The total charge Q is:

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

Solution to Example 2 The total charge Q:

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

Current densities • There are 2 types of current: • 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 • Conduction current • atoms of conducting material do NOT move; obeys Ohm’s law • E.g movement of electrons in a metal wire

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

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)

For acting on a charge • For a material with electrical permittivity, ε: D = εE where: ε = εR ε0 ε0 = 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

E-field due to multipoint charges • At point P, the electric field E1 due to q1 alone: • At point P, the electric field E1 due to q2 alone:

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

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

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.

Solution to Example 3 • The electric field E with ε = ε0 (free space) is given by: • The vectors are:

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

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

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

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→∞.

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:

Solution to Example 4stop • 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 =∞,

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:

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 ρlalong the z-axis.

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

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

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

Electric Potential as a function of electric field • Integrating along any path between point P1 and P2, we get: Potential difference between P1 and P2 , regardless of path 1, 2 or 3 :

Electric Potential as a function of electric field • Kirchhoff’s voltage lawstates that the net voltage drop around a closed loop is zero. • Line integral E around closed contourC is: • The electric potential Vat any point is given by: • Integration path between point P1 and P2 is arbitrary

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:

Electric potential due to point charges • For charge Q located other than origin, specified by a source position vector R1, then V at the observation vector R becomes: • For N discrete point charges, electric potential is:

Electric potential due tocontinuous distributions • For a continuous charge distribution:

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

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

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)

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

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 (m2/V.s) µh= hole mobility (m2/V.s)

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)

Conductivity • Conductivities of different materials:

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:

Example 6 A 2-mm-diameter copper wire with conductivity of 5.8 × 107 S/m and electron mobility of 0.0032 (m2/V·s) is subjected to an electric field of 20 (mV/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.

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

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

EKT 241/4: ELECTROMAGNETIC THEORY