4. ELECTROSTATICSApplied EM by Ulaby, Michielssen and Ravaioli

Chapter 4: ElectrostaticsMaxwells EquationsCharge and Current distributionsCoulombs and Gausss LawsElectric scalar potentialConductors, dielectrics, and capacitanceElectric boundary conditionsPotential energy

Chapter 4: Electrostatics

Maxwells EquationsGod said: And there was light!

Charge DistributionsVolume charge density:Total Charge in a VolumeSurface and Line Charge DensitiesTotal Charge in a surface and line

Examples1. Given the line charge density: 2. Given the surface charge density: Cylindrical Coordinate System : Cartesian Coordinate System:

ExerciseA square plate residing in the x-y plane is situated in the space defined by 0 x 3m and 0 y 3m. Find the total charge on the plate if the surface charge density is:33Cartesian Coordinate System:

ExerciseFind the total charge contained in a volume shown the figure below. The volume charge density is:xyz103Cylindrical Coordinate System : dr = 0 3

Current DensityFor a surface with any orientation:J is called the current density

Convection vs. Conduction

Coulombs LawElectric field at point P due to single chargeElectric force on a test charge placed at PCoulombs LawREE

Electric flux density: DElectric flux density DNote: The density of the field lines is the same regardless of the material that you are in.

Use Coulombs Law to find E fieldAssume that we try to find the E field at point P.Define an origin ( coordinate system).Write vector Rs from origin to Charge pointWrite vector Rp from origin to Field PointFind vector from Charge point to Field Point RspApply Coulombs Law to find Field due to the Charge pointSum or Integrate to find total E at the point PPq1q2xyzRpRs1Rs2R1R2

Electric Field Due to 2 Charges

Example 4-3: Electric Field due to two point charge

Electric Field Due to Charge DistributionsField due to:

E field due to straight line chargesLet us assume

E field due to Surface ChargesLet us assume

E field due to Volume ChargesLet us assume

Cylindrical System: Surface ChargesLet us assume E(rp,p,zp) = ?xy1114zUsing Table 3-2 convert Cylindrical to Rectangular, and then back to Cylindrical

Cont.

Cont.

Example 4-5 cont.

ExampleAn infinite sheet with uniform surface charge density is located at z=0 (x-y plane), and another infinite sheet with density is located at z = 2m, both in free space. Determine E everywhere.

Gausss LawGausss law The total outward electric flux through any closed surface is equal to the total charge enclosed by the surface.Gausss Law is written in equation form as:

Example: Gausss law and point charge at origin Given a point charge at the origin, show that Gausss law is valid on a spherical surface of radius R.Solution:From Gausss law :

We know:

and

Gausss Law to find E fieldGausss law is only for the case which the charge distribution has symmetry properties while Coulombs Law is for general case.Use Gausss Law to find D first , and then find E by D=E.Major Points:Point (differential) form of Gausss Law:Integrate over a Volume:Use Divergence theory:Integrate form of Gausss Law

Applying Gausss LawAssume that we try to find the E field at point P.Be sure that the charge is symmetric (D must be equal in magnitude everywhere on the Gaussian Surface. If not, use Coulombs Law)Choose a Gaussian surface that has the field point P. The volume containing charges is inside the surface.Do the integral on the left side (This is always the same)Do the integral on the right side to find QSolve for D, and then find E.

Example: Using Gausss law to determine E

Example: Using Gausss law to determine E (conts)Solution:

Example: Using Gausss law to determine E (conts)Therefore:(2) Determine E field inside the charged sphere

Example 4-6Construct an imaginary Gaussian cylinder of radius r and height h:

4.5 Electric Scalar PotentialMinimum force needed to move charge against E field:Potential difference or Voltage V between two points in a circuit (electric field) is defined as the amount of work required to move a unit positive charge from one to the other.

Electric Scalar Potential

Example: potential difference in E field of a point chargeSolution:We choose an inward radial path from point A to point BThe absolute potential at point B is found by taking the limit as RA goes to infinity.

Electric Potential Due to ChargesIn 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 PFor a point charge, V at range R is:For continuous charge distributions:Note: R is the distance between the integration point and observation point.

Relating E to V

Cont.

(cont.)

Poissons & Laplaces EquationsIn the absence of charges:

E fields in Material SpaceWhen an electric field is applied to a gas, solid, or liquid material, the charge associated with the material atoms will be affected.Charge that is not free to move under the influence of the field will be displaced (polarized) bound chargeCharge not bound by other forces will be set in motion (electric current) free charge

ConductorsConduction current density:Note how wide the range is, over 24 orders of magnitude

Conductivityve = volume charge density of electronshe = volume charge density of holese = electron mobilityh = hole mobilityNe = number of electrons per unit volumeNh = number of holes per unit volume

The drift velocity of the electron in a conductor is the product of the E field and the conductor mobility:

ResistanceFor any conductor:Longitudinal ResistorThe conductance is

Joules LawThe power dissipated in a volume containing electric field E and current density J is:For a resistor, Joules law reduces to:For a coaxial cable:

Tech Brief 7: Resistive SensorsAn electrical sensor is a device capable of responding to an applied stimulus by generating an electrical signal whose voltage, current, or some other attribute is related to the intensity of the stimulus.

Typical stimuli : temperature, pressure, position, distance, motion, velocity, acceleration, concentration (of a gas or liquid), blood flow, etc.

Sensing process relies on measuring resistance, capacitance, inductance, induced electromotive force (emf), oscillation frequency or time delay, etc.

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

Dielectric MaterialsNonconducting materials are commonly designated as insulators or dielectrics.When an electric field is applied to a dielectric atom, an effect known as polarization results.A polarized dielectric atom may be modeled as equivalent electric dipole.

Dielectric MaterialsEach small equivalent electric dipole sets up a small electric field, pointing from + to -, which is opposite in direction to E.This induced electric field called a polarization field.The total (net) electric field in the dielectric material is smaller than E.This is only for nonpolar molecules.

Polarization FieldP = electric flux density induced by E:When a dielectric material is polarized, the field flux is

P is called the electric polarization field.: relative permittivity

Electric Breakdown Electric Breakdown

4.8 Boundary ConditionsTangential componentNormal componentConsider a closed rectangular loop abcda in Figure 4-18.Apply the conservative property of E field:Letting

Boundary Conditions: normal componentsTangential componentNormal componentNote: The normal component of D changes abruptly at a charged boundary,Consider a closed small cylinder in Figure 4-18.Apply Gausss lawLetting

Summary of Boundary ConditionsRemember E = 0 in a good conductor

Dielectric-Conductor boundaryNet electric field inside a conductor is zerosConsider a boundary: medium 1 is dielectric and medium 2 is a perfect conductor.perfect conductordielectric

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

Perfect conductor under static conditions

ExampleDetermine E and D everywhere for the charge-free boundary shown below given

Solution:Using boundary condition to determine E2t and E2nThen E2 = E2t+E2nThen D2 = 2E2

4.9 CapacitanceA capacitor is an energy storage device that stores energy in an electric field.A capacitor consists of two conductors separated by an insulating medium.Application of a voltage between two conductors causes a charge separation. This charge separation produces an E field within insulating medium such that energy is stored in the capacitorThe capacitance depends only on the geometry of the conductors and permittivity of insulating material.

CapacitanceFor any two-conductor configuration:For any resistor:

Obtain an expression for the capacitance C of a parallel-plate capacitor composed of two parallel plates each of surface area A and separated by a distance d. The permittivity of the insolating material is .Determine the breakdown voltage if d = 1cm and the dielectric material is quartzSolution:Applying a voltage V to two conductors, charges will accumulate on the top and bottom plates.Charges will induce a E filed between two conductors

Application of Gausss law gives:Q is total charge on inside of outer cylinder, and Q is on outside surface of inner cylinder

Electrostatic Potential EnergyThe energy stored in dielectric medium in the form of electrostatic potential energyThe total stored energy is related to Q, C, and VTotal electrostatic energy stored in a volumeEnergy stored in a capacitorElectrostatic potential energy density (Joules/volume)

Methods to find ESo far, we have learned three ways to find E field

Coulombs Law: need to know the charge distributionGausss Law: need to choose a proper Gauss surfacePoissons or Laplaces Law: math skills

How to find E, V for the following case?

Image MethodImage method simplifies calculation for E and V due to charges near conducting planes.

For each charge Q, add an image charge QRemove conducting planeCalculate field due to all charges

Example 4-13Question: how to find the surface charge density of the surface of the conducting plane?

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