Electric field generated by an electric dipole

Similar documents
Review. Electrostatic. Dr. Ray Kwok SJSU

F = net force on the system (newton) F,F and F. = different forces working. E = Electric field strength (volt / meter)

Chapter 22: Electric Fields. 22-1: What is physics? General physics II (22102) Dr. Iyad SAADEDDIN. 22-2: The Electric Field (E)

Review for Midterm-1

Today s Plan. Electric Dipoles. More on Gauss Law. Comment on PDF copies of Lectures. Final iclicker roll-call

2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum

Flux. Area Vector. Flux of Electric Field. Gauss s Law

Module 05: Gauss s s Law a

Welcome to Physics 272

Gauss Law. Physics 231 Lecture 2-1

Electrostatics (Electric Charges and Field) #2 2010

CHAPTER 25 ELECTRIC POTENTIAL

Objectives: After finishing this unit you should be able to:

MAGNETIC FIELD INTRODUCTION

Electrostatics. 1. Show does the force between two point charges change if the dielectric constant of the medium in which they are kept increase?

Hopefully Helpful Hints for Gauss s Law

Continuous Charge Distributions: Electric Field and Electric Flux

Force and Work: Reminder

Electromagnetism Physics 15b

ELECTROSTATICS::BHSEC MCQ 1. A. B. C. D.

Chapter 21: Gauss s Law

Physics 235 Chapter 5. Chapter 5 Gravitation

Physics 2212 GH Quiz #2 Solutions Spring 2016

University Physics (PHY 2326)

(r) = 1. Example: Electric Potential Energy. Summary. Potential due to a Group of Point Charges 9/10/12 1 R V(r) + + V(r) kq. Chapter 23.

[Griffiths Ch.1-3] 2008/11/18, 10:10am 12:00am, 1. (6%, 7%, 7%) Suppose the potential at the surface of a hollow hemisphere is specified, as shown

A moving charged particle creates a magnetic field vector at every point in space except at its position.

( ) Make-up Tests. From Last Time. Electric Field Flux. o The Electric Field Flux through a bit of area is

Chapter 22 The Electric Field II: Continuous Charge Distributions

II. Electric Field. II. Electric Field. A. Faraday Lines of Force. B. Electric Field. C. Gauss Law. 1. Sir Isaac Newton ( ) A.

Physics 122, Fall October 2012

Physics 11 Chapter 20: Electric Fields and Forces

Φ E = E A E A = p212c22: 1

Chapter 4. Newton s Laws of Motion

Module 5: Gauss s Law 1

Review: Electrostatics and Magnetostatics

(Sample 3) Exam 1 - Physics Patel SPRING 1998 FORM CODE - A (solution key at end of exam)

Ch 30 - Sources of Magnetic Field! The Biot-Savart Law! = k m. r 2. Example 1! Example 2!

Gauss s Law Simulation Activities

PHYS 2135 Exam I February 13, 2018

2 E. on each of these two surfaces. r r r r. Q E E ε. 2 2 Qencl encl right left 0

EELE 3331 Electromagnetic I Chapter 4. Electrostatic fields. Islamic University of Gaza Electrical Engineering Department Dr.

anubhavclasses.wordpress.com CBSE Solved Test Papers PHYSICS Class XII Chapter : Electrostatics

THE MAGNETIC FIELD. This handout covers: The magnetic force between two moving charges. The magnetic field, B, and magnetic field lines

Magnetic Fields Due to Currents

Chapter 23: GAUSS LAW 343

17.1 Electric Potential Energy. Equipotential Lines. PE = energy associated with an arrangement of objects that exert forces on each other

Your Comments. Do we still get the 80% back on homework? It doesn't seem to be showing that. Also, this is really starting to make sense to me!

PHY 114 A General Physics II 11 AM-12:15 PM TR Olin 101

Algebra-based Physics II


13. The electric field can be calculated by Eq. 21-4a, and that can be solved for the magnitude of the charge N C m 8.

Potential Energy. The change U in the potential energy. is defined to equal to the negative of the work. done by a conservative force

Charges, Coulomb s Law, and Electric Fields

PHYS 1444 Lecture #5

Static Electric Fields. Coulomb s Law Ε = 4πε. Gauss s Law. Electric Potential. Electrical Properties of Materials. Dielectrics. Capacitance E.

Qualifying Examination Electricity and Magnetism Solutions January 12, 2006

Introduction: Vectors and Integrals

Chapter 4 Gauss s Law

Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 1 -

From last times. MTE1 results. Quiz 1. GAUSS LAW for any closed surface. What is the Electric Flux? How to calculate Electric Flux?

16.1 Permanent magnets

18.1 Origin of Electricity 18.2 Charged Objects and Electric Force

Chapter 31 Faraday s Law

EM-2. 1 Coulomb s law, electric field, potential field, superposition q. Electric field of a point charge (1)

Section 26 The Laws of Rotational Motion

! E da = 4πkQ enc, has E under the integral sign, so it is not ordinarily an

Current, Resistance and

UNIT 3:Electrostatics

Look over Chapter 22 sections 1-8 Examples 2, 4, 5, Look over Chapter 16 sections 7-9 examples 6, 7, 8, 9. Things To Know 1/22/2008 PHYS 2212

Your Comments. Conductors and Insulators with Gauss's law please...so basically everything!

Sources of the Magnetic Field. Moving charges currents Ampere s Law Gauss Law in magnetism Magnetic materials

Unit 7: Sources of magnetic field

Module 18: Outline. Magnetic Dipoles Magnetic Torques

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS

3.8.1 Electric Potential Due to a System of Two Charges. Figure Electric dipole

Electrostatic Potential

The Law of Biot-Savart & RHR P θ

Chapter 26: Magnetism: Force and Field

7.2.1 Basic relations for Torsion of Circular Members

Kinetic energy, work, and potential energy. Work, the transfer of energy: force acting through distance: or or

Chapter 13 Gravitation

Phys102 Second Major-182 Zero Version Monday, March 25, 2019 Page: 1

ECE 3318 Applied Electricity and Magnetism. Spring Prof. David R. Jackson ECE Dept. Notes 13

Force between two parallel current wires and Newton s. third law

Magnetic Field. Conference 6. Physics 102 General Physics II

Phys-272 Lecture 17. Motional Electromotive Force (emf) Induced Electric Fields Displacement Currents Maxwell s Equations

PHYS 1444 Section 501 Lecture #7

X ELECTRIC FIELDS AND MATTER

Gauss s Law: Circuits

FI 2201 Electromagnetism

Section 1: Main results of Electrostatics and Magnetostatics. Electrostatics

Electric Field. y s +q. Point charge: Uniformly charged sphere: Dipole: for r>>s :! ! E = 1. q 1 r 2 ˆr. E sphere. at <0,r,0> at <0,0,r>

Chapter 7-8 Rotational Motion

To Feel a Force Chapter 7 Static equilibrium - torque and friction

Moment. F r F r d. Magnitude of moment depends on magnitude of F and the length d

( )( )( ) ( ) + ( ) ( ) ( )

ECE 3318 Applied Electricity and Magnetism. Spring Prof. David R. Jackson Dept. Of ECE. Notes 20 Dielectrics

$ i. !((( dv vol. Physics 8.02 Quiz One Equations Fall q 1 q 2 r 2 C = 2 C! V 2 = Q 2 2C F = 4!" or. r ˆ = points from source q to observer

Electrostatics. 3) positive object: lack of electrons negative object: excess of electrons

Transcription:

Electic field geneated by an electic dipole ( x) 2 (22-7) We will detemine the electic field E geneated by the electic dipole shown in the figue using the pinciple of supeposition. The positive chage geneates at P an electic field whose magnitude 1+ ; 1 2x ( ) 2 4πε o 1 q ( + ) 2 4πε o + The negative chage ceates an electic field with magnitude E = E = 1 q

( x) 2 1+ ; 1 2x (22-7) The net electic field at P is: E 1 q = 4πε o q 2 2 + 1 q q = 4 πε o /2 /2 ( z d ) ( z+ d ) 2 2 q d d E = 1 1 2 + 4πε oz 2z 2z d We assume: = 1 2z 2 2 q d d E = 1 1 2 4πε oz + z z qd 1 p = = 3 3 2πε z 2πε z o o E = E E ( + ) ( )

Electic dipole : Two chages that ae equal in magnitude but of opposite signal. ( ) ( ) kq ( ) kq k q k q E = ˆ + ˆ= + 2 + 2 3 + 3 + + d d + = 2 = + 2 3 3 2 3 3 d d d ( ) 2 + = + = + + = 2 + 3 + +q d/2 ρ + ρ ρ E ρ ( ) =? >>d 2 4 3 3 o 2 2 2 2 d/2 ρ 2 d 2 d d = d + = 1 + 2 2 4 4 q 2 3 d d d 1>>Δx n 1, 2 >> ( 1+ Δx 2 2 ) 1+ nδx 2 4 3 3 3 3 3 1 ( ) 2 d kq d kq v 1 3 d 3( d ˆˆ = = = ) d ( ) 2 = 3 2 = 3 3 2 3 2 d d 3 3 d d E = kq 1+ kq 1 + 2 2 3 2 2 2

v Dipole moment : p = qd v kq ( ) kp 1. d E = d = 3 3 v v 2. / / d, d > 0 kq kq v kpˆ E = 3 [ 3dˆ dˆ] = 2d 2 3 3 = v v 3. / / d, d < 0 kq kq v kpˆ E = 3 [ 3( d ) ˆ dˆ] = 2d = 2, ( ˆ= dˆ ) 3 3

Effect of dielectic mateial E 0 = k 0 q 2 fee q bound q fee ε 0 ( k 0 : vacuum) q fee E ρ induced E kq 1 q kq E = E E = = = = 0 0 fee fee fee 0 induced 2 2 2 ε ε 4πε0ε

Foces and toques exeted on electic dipoles by a unifom electic field Conside the electic dipole shown in the figue in the pesence of a unifom (constant magnitude and diection) electic field E along the x-axis. The electic field exets a foce F + = qe on the positive chage and a foce F = qe on the negatice chage. The net foce on the dipole F = qe qe = 0 net (22-14)

The net toque geneated by F+ and F about the dipole cente is: d d τ = τ+ + τ = F+ sinθ F sinθ = qedsinθ = pesinθ 2 2 In vecto fom: τ = p E The electic dipole in a unifom electic field does not move but can otate about its cente. F 0 τ = p E net = (22-14)

Potential enegy of an electic dipole in a unifom electic field U U = pecosθ U = p E B 180 θ θ U = τdθ = pesinθdθ θ 90 90 θ U = pe sinθdθ = pecosθ = p E 90 p E At point A ( θ = 0) U has a minimum value U min = pe It is a position of stable equilibium At point B ( θ = 180 ) U has a maximum A (22-15) value U max =+ pe It is a position of p unstab E le equilibium

Chapte 24 Gauss s Law

Electic Flux Electic flux is the poduct of the magnitude of the electic field and the suface aea, A, pependicula to the field Φ E = EA

Electic Flux, Geneal Aea The electic flux is popotional to the numbe of electic field lines penetating some suface The field lines may make some angle θ with the pependicula to the suface Then Φ E = EA cos θ

Electic Flux, Intepeting the Equation The flux is a maximum when the suface is pependicula to the field The flux is zeo when the suface is paallel to the field If the field vaies ove the suface, Φ = EA cos θ is valid fo only a small element of the aea

Electic Flux, Geneal In the moe geneal case, look at a small aea element ΔΦ = EΔ A cosθ = E ΔA E i i i i i In geneal, this becomes Φ = lim E Δ A = d E i i ΔA 0 i E A suface

Electic Flux, final The suface integal means the integal must be evaluated ove the suface in question In geneal, the value of the flux will depend both on the field patten and on the suface The units of electic flux will be N. m 2 /C 2

Electic Flux, Closed Suface Assume a closed suface The vectos ΔA i point in diffeent diections At each point, they ae pependicula to the suface By convention, they point outwad

Active Figue 24.4 (SLIDESHOW MODE ONLY)

Flux Though Closed Suface, cont. At (1), the field lines ae cossing the suface fom the inside to the outside; θ < 90 o, Φ is positive At (2), the field lines gaze suface; θ = 90 o, Φ = 0 At (3), the field lines ae cossing the suface fom the outside to the inside;180 o > θ > 90 o, Φ is negative

Flux Though Closed Suface, final The net flux though the suface is popotional to the net numbe of lines leaving the suface This net numbe of lines is the numbe of lines leaving the suface minus the numbe enteing the suface If E n is the component of E pependicula to the suface, then Φ E = E da = EndA 旄

Gauss s Law, Intoduction Gauss s law is an expession of the geneal elationship between the net electic flux though a closed suface and the chage enclosed by the suface The closed suface is often called a gaussian suface Gauss s law is of fundamental impotance in the study of electic fields

PI Electostatics CT9 A cylindical piece of insulating mateial is placed in an extenal electic field, as shown. The net electic flux passing though the suface of the cylinde is 1 positive 2 negative 3 zeo