Coulomb 2017 mk

Nguyễn Công Phương Engineering Electromagnetics Coulomb’s Law & Electric Field Intensity Contents I II III IV V VI VII VIII IX X XI XII XIII XIV XV Introduction Vector Analysis Coulomb’s Law & Electric Field Intensity Electric Flux Density, Gauss’ Law & Divergence Energy & Potential Current & Conductors Dielectrics & Capacitance Poisson’s & Laplace’s Equations The Steady Magnetic Field Magnetic Forces & Inductance Time – Varying Fields & Maxwell’s Equations Transmission Lines The Uniform Plane Wave Plane Wave Reflection & Dispersion Guided Waves & Radiation Engineering Electromagnetics - sites.google.com/site/ncpdhbkhn Coulomb’s Law & Electric Field Intensity Coulomb’s Law Electric Field Intensity Field Due to a Continuous Volume Charge Distribution Field of a Line Charge Field of a Sheet Charge Sketches of Fields Applications Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn Coulomb's Law (1) Q1Q2 F =k R – In free space – between very small objects (compared to the separation R) – Q1 & Q2 are the positive/negative quantities of charge − k= 4πε – ε0: permittivity of free space, − 12 ε = 8.854 × 10 = 10−9 F/m 36π https://w ww.teylersmuseum.nl/nl/collectie/instrument en/fk-0556-electrometer-coulomb-balance Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn Coulomb's Law (2) Q1Q2 F =k R k= 4πε Q1Q2 →F = 4πε R Q1 a12 R12 Q2 F2 r2 r1 Q1 & Q2: alike in sign Origin Q1Q2 F2 = a12 4πε R12 R12 = r2 − r1 R12 R12 r2 − r1 a12 = = = R12 R12 r2 − r1 Q1 a12 F2 R12 r1 Origin Q2 r2 Q1 & Q2: opposite sign Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn Ex Coulomb's Law (3) Given Q1 = 4.10-4 C at A(3, 2, 1) & Q2 = – 3.10-4 C at B(1, 0, 2) in vacuum Find the force exerted on Q2 by Q1 Q1Q2 F2 = a12 4πε R12 R12 = r2 − r1 = (1 − 3)a x + (0 − 2)a y + (2 − 1)a z = −2a x − 2a y + a z R12 = (−2) + ( −2) + 12 = R12 −2a x − 2a y + a z a12 = = R12 4.10−4 ( −3.10−4 ) −2a x − 2a y + a z → F2 = × = 80a x + 80a y − 40az N 4π 10−932 36π Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn Coulomb’s Law & Electric Field Intensity Coulomb’s Law Electric Field Intensity Field Due to a Continuous Volume Charge Distribution Field of a Line Charge Field of a Sheet Charge Sketches of Fields Applications Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn Electric Field Intensity (1) • Consider a fixed Q1 & a test Qt Q1Qt Ft Q1 Ft = a → = a 1t 1t 4πε R1t Qt 4πε R1t • Electric Field Intensity: the vector force on 1C • Unit: V/m • EFI due to a single point charge Q in a vacuum: E= Q 4πε R aR – R: from Q to the point of E – aR: unit vector of R Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn Electric Field Intensity (2) E= Q 4πε R aR • If Q is at the center of a spherical coordinate system: E= Q 4πε r ar • If Q is at the center of a rectangular coordinate system:   Q x y z  E= ax + ay + az  2 2 2 2  4πε ( x + y + z )  x2 + y + z x + y + z x + y + z   Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn E= Q 4πε R Electric Field Intensity (3) aR 3.5 2.5 1.5 0.5 4 -1 -1 -2 -2 -3 -3 -4 -4 Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 10 Sheet Charge (5) x >a – ρS ρS E+ = ax 2ε ρS E- = − ax 2ε a x → E = E+ + E− = ρS E= aN 2ε 0< x