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Explore Maxwell’s equations and their applications in electrostatics and magnetostatics, including charge distributions, electric fields, and conductivity concepts, with detailed examples and solutions.
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3. Electrostatics Ruzelita Ngadiran
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 God said: And there was light!
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
Charge Distributions Volume charge density: Total Charge in a Volume Surface and Line Charge Densities
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 Density For a surface with any orientation: J is called the current density
Coulomb’s Law Electric field at point P due to single charge Electric force on a test charge placed at P Electric flux density D
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:
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
Electric Field Due to Charge Distributions Field due to:
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 4 • 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 Application of the divergence theorem gives:
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.
Applying Gauss’s Law Construct an imaginary Gaussian cylinder of radius r and height h:
Electric Scalar Potential Minimum force needed to move charge against E field:
Electric Potential Due to Charges For a point charge, V at range R is: In electric circuits, we usually select a convenient node that we call ground and assign it zero reference voltage. In free space and material media, we choose infinity as reference with V = 0. Hence, at a point P For continuous charge distributions:
Poisson’s & Laplace’s Equations In the absence of charges:
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:
Conductivity ve = volume charge density of electrons he = volume charge density of holes e = electron mobility h = hole mobility Ne = number of electrons per unit volume Nh = number of holes per unit volume
Conduction Current Conduction current density: Note how wide the range is, over 24 orders of magnitude
Resistance Longitudinal Resistor For any conductor: