3. Electrostatics Ruzelita Ngadiran

Chapter 4 Overview

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

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 z is 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

Convection vs. Conduction

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 = ε E 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:

Electric Field Due to 2 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

Electric Field Due to Charge Distributions

Cont.

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 ρl along 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 Scalar Potential

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:

Relating E to V

Cont.

(cont.)

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:

G’=0 if the insulating material is air or a perfect dielectric with zero conductivity.

Joule’s Law The power dissipated in a volume containing electric field E and current density J is: For a coaxial cable: For a resistor, Joule’s law reduces to:

Piezoresistivity The Greek word piezein means to press R0 = resistance when F = 0 F = applied force A0 = cross-section when F = 0  = piezoresistive coefficient of material

Piezoresistors

Wheatstone Bridge Wheatstone bridge is a high sensitivity circuit for measuring small changes in resistance

Dielectric Materials

Polarization Field P = electric flux density induced by E

Electric Breakdown Electric Breakdown

Boundary Conditions

Summary of Boundary Conditions Remember E = 0 in a good conductor

Conductors Net electric field inside a conductor is zero

Field Lines at Conductor Boundary At conductor boundary, E field direction is always perpendicular to conductor surface

Capacitance

Capacitance For any two-conductor configuration: For any resistor:

Application of Gauss’s law gives: Q is total charge on inside of outer cylinder, and –Q is on outside surface of inner cylinder

Tech Brief 8: Supercapacitors For a traditional parallel-plate capacitor, what is the maximum attainable energy density? Mica has one of the highest dielectric strengths ~2 x 10**8 V/m. If we select a voltage rating of 1 V and a breakdown voltage of 2 V (50% safety), this will require that d be no smaller than 10 nm. For mica,  = 60 and  = 3 x 10**3 kg/m3 . Hence: W = 90 J/kg = 2.5 x10**‒2 Wh/kg. By comparison, a lithium-ion battery has W = 1.5 x 10**2 Wh/kg, almost 4 orders of magnitude greater Energy density is given by:  = permittivity of insulation material V = applied voltage  = density of insulation material d = separation between plates

A supercapacitor is a “hybrid” battery/capacitor

Users of Supercapacitors

Energy Comparison

Electrostatic Potential Energy Electrostatic potential energy density (Joules/volume) Energy stored in a capacitor Total electrostatic energy stored in a volume

Image Method Image method simplifies calculation for E and V due to charges near conducting planes. For each charge Q, add an image charge –Q Remove conducting plane Calculate field due to all charges