Physics 2113 Lecture 14: WED 18 FEB

Similar documents
Phys102 General Physics II

Physics for Scientists & Engineers 2

Physics 114 Exam 2 Fall 2014 Solutions. Name:

Electric Potential Energy & Potential. Electric Potential Energy. Potential Energy. Potential Energy. Example: Charge launcher

Physics 114 Exam 2 Spring Name:

Study Guide For Exam Two

ˆ (0.10 m) E ( N m /C ) 36 ˆj ( j C m)

CONDUCTORS AND INSULATORS

PHY2049 Exam 2 solutions Fall 2016 Solution:

PES 1120 Spring 2014, Spendier Lecture 6/Page 1

PHYSICS 231 Review problems for midterm 2

Physics 114 Exam 3 Spring Name:

Chapter 3 and Chapter 4

24-2: Electric Potential Energy. 24-1: What is physics

CHAPTER 10 ROTATIONAL MOTION

Electricity and Magnetism - Physics 121 Lecture 10 - Sources of Magnetic Fields (Currents) Y&F Chapter 28, Sec. 1-7

Fields, Charges, and Field Lines

Week3, Chapter 4. Position and Displacement. Motion in Two Dimensions. Instantaneous Velocity. Average Velocity

MAGNETISM MAGNETIC DIPOLES

Section 8.3 Polar Form of Complex Numbers

Please review the following statement: I certify that I have not given unauthorized aid nor have I received aid in the completion of this exam.

Chapter 2: Electric Energy and Capacitance

τ rf = Iα I point = mr 2 L35 F 11/14/14 a*er lecture 1

Energy and Energy Transfer

Chapter 3. r r. Position, Velocity, and Acceleration Revisited

Angular Momentum and Fixed Axis Rotation. 8.01t Nov 10, 2004

PHYSICS - CLUTCH CH 28: INDUCTION AND INDUCTANCE.

Dr. Fritz Wilhelm, Physics 230 E:\Excel files\230 lecture\ch26 capacitance.docx 1 of 13 Last saved: 12/27/2008; 8:40 PM. Homework: See website.

Physics 181. Particle Systems

PHYS Week 5. Reading Journals today from tables. WebAssign due Wed nite

Week 6, Chapter 7 Sect 1-5

Change. Flamenco Chuck Keyser. Updates 11/26/2017, 11/28/2017, 11/29/2017, 12/05/2017. Most Recent Update 12/22/2017

Physics 53. Rotational Motion 3. Sir, I have found you an argument, but I am not obliged to find you an understanding.

PHYS 1101 Practice problem set 12, Chapter 32: 21, 22, 24, 57, 61, 83 Chapter 33: 7, 12, 32, 38, 44, 49, 76

Boundaries, Near-field Optics

PY2101 Classical Mechanics Dr. Síle Nic Chormaic, Room 215 D Kane Bldg

Physics 207: Lecture 20. Today s Agenda Homework for Monday

PHYSICS - CLUTCH 1E CH 28: INDUCTION AND INDUCTANCE.

Mechanics Physics 151

kq r 2 2kQ 2kQ (A) (B) (C) (D)

Chapter 8. Potential Energy and Conservation of Energy

Lecture 16. Chapter 11. Energy Dissipation Linear Momentum. Physics I. Department of Physics and Applied Physics

Physics 111: Mechanics Lecture 11

VEKTORANALYS GAUSS THEOREM STOKES THEOREM. and. Kursvecka 3. Kapitel 6 7 Sidor 51 82

Rotational Dynamics. Physics 1425 Lecture 19. Michael Fowler, UVa

Week 9 Chapter 10 Section 1-5

EMU Physics Department

8.022 (E&M) Lecture 4

A Tale of Friction Basic Rollercoaster Physics. Fahrenheit Rollercoaster, Hershey, PA max height = 121 ft max speed = 58 mph

VEKTORANALYS. GAUSS s THEOREM and STOKES s THEOREM. Kursvecka 3. Kapitel 6-7 Sidor 51-82

Chapter 21 Chapter 24. Electric Potential. Copyright 2014 John Wiley & Sons, Inc. All rights reserved.

ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.

Physics 106 Lecture 6 Conservation of Angular Momentum SJ 7 th Ed.: Chap 11.4

Part C Dynamics and Statics of Rigid Body. Chapter 5 Rotation of a Rigid Body About a Fixed Axis

Confirmation of Gauss s law

So far: simple (planar) geometries

10/9/2003 PHY Lecture 11 1

Introduction to circuit analysis. Classification of Materials

Spring 2002 Lecture #13

Field and Wave Electromagnetic. Chapter.4

Numerical Transient Heat Conduction Experiment

Linear Momentum. Center of Mass.

= 1.23 m/s 2 [W] Required: t. Solution:!t = = 17 m/s [W]! m/s [W] (two extra digits carried) = 2.1 m/s [W]

Supplemental Instruction sessions next week

Dynamics of Rotational Motion

Gravitational Acceleration: A case of constant acceleration (approx. 2 hr.) (6/7/11)

Physics 141. Lecture 14. Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 14, Page 1

Physics 207: Lecture 27. Announcements

Mechanics Physics 151

Physics 207 Lecture 13. Lecture 13

K = 100 J. [kg (m/s) ] K = mv = (0.15)(36.5) !!! Lethal energies. m [kg ] J s (Joule) Kinetic Energy (energy of motion) E or KE.

Chapter 20 Electric Potential and Electric potential Energy

Lecture #4 Capacitors and Inductors Energy Stored in C and L Equivalent Circuits Thevenin Norton

Please review the following statement: I certify that I have not given unauthorized aid nor have I received aid in the completion of this exam.

CHAPTER 8 Potential Energy and Conservation of Energy

CHAPTER 7 ENERGY BALANCES SYSTEM SYSTEM. * What is energy? * Forms of Energy. - Kinetic energy (KE) - Potential energy (PE) PE = mgz

Math1110 (Spring 2009) Prelim 3 - Solutions

Physics Lecture: 15 FRI 20 FEB

measurement and the charge the electric field will be calculated using E =. The direction of the field will be the . But, so

v c motion is neither created nor destroyed, but transferred via interactions. Fri. Wed (.18,.19) Introducing Potential Energy RE 6.

PHYS 1443 Section 004 Lecture #12 Thursday, Oct. 2, 2014

MTE1 results. Mean 75% = 90/120

Please review the following statement: I certify that I have not given unauthorized aid nor have I received aid in the completion of this exam.

First Law: A body at rest remains at rest, a body in motion continues to move at constant velocity, unless acted upon by an external force.

Chapter 7. Potential Energy and Conservation of Energy

Physics 2102 Spring 2007 Lecture 10 Current and Resistance

Electricity and Magnetism Review Faraday s Law

Please review the following statement: I certify that I have not given unauthorized aid nor have I received aid in the completion of this exam.

Physics 1202: Lecture 11 Today s Agenda

How Differential Equations Arise. Newton s Second Law of Motion

Chapter 07: Kinetic Energy and Work

MTH 263 Practice Test #1 Spring 1999

PART I: MULTIPLE CHOICE (32 questions, each multiple choice question has a 2-point value, 64 points total).

Mechanics Cycle 3 Chapter 9++ Chapter 9++

Problem Points Score Total 100

DC Circuits. Crossing the emf in this direction +ΔV

Physics 4B. Question and 3 tie (clockwise), then 2 and 5 tie (zero), then 4 and 6 tie (counterclockwise) B i. ( T / s) = 1.74 V.

(b) This is about one-sixth the magnitude of the Earth s field. It will affect the compass reading.

COMPLEX NUMBERS AND QUADRATIC EQUATIONS

Chapter 11: Angular Momentum

Transcription:

Physcs 2113 Jonathan Dowlng Physcs 2113 Lecture 14: WED 18 FEB Electrc Potental II Danger!

Electrc Potental Energy, Unts : Electrc Potental Potental Energy = U = [J] = Joules Electrc Potental = V = U/q = [J/C] = [Nm/C] = [V] = Volts Electrc Feld = E = [N/C] = [V/m] = Volts per meter F = qe (Force s charge tmes Feld) U = qv (Potental Energy s charge tmes Potental) Electron Volt = 1eV = Work Needed to Move an Electron Through a Potental Dfference of 1V: W = qδv = e x 1V = 1.60 10 19 C x 1J/C = 1.60 10 19 J

Electrc Potental Energy = Joules Electrc potental energy dfference ΔU between two ponts = work needed to move a charge between the two ponts: ΔU = U f U = W dw =! F d! s F = q 0! E f f! W = dw = F d s! f!! = q 0 E d s ΔU = U f U = W = q 0 f! E d! s

Electrc Potental Voltage = Volts = Joules/ Coulomb! Electrc potental voltage! dfference ΔV between two ponts = work per unt charge needed to move a charge between the two ponts: ΔV = V f V = W/q = ΔU/q dw =! F d! s! F = q 0! E f W = dw = f q 0! E d! s ΔV = V f V = W q 0 = f! E d! s

Equal-Potental = Equpotental Surfaces W Δ V = V f V = = E!! ds q The Electrc Feld s Tangent to the Feld Lnes Equpotental Surfaces are Perpendcular to Feld Lnes 0 f E!" E!" E!" E!" Work Is Needed to Move a Charge Along a Feld Lne. No Work Is Needed to Move a Charge Along an Equpotental Surface (Or Back to the Surface Where t Started). Electrc Feld Lnes Always Pont Towards Equpotental Surfaces Wth Lower Potental.

Electrc Feld Lnes and Equpotental Surfaces Why am I smlng? I m About to Be Struck by Lghtnng! http://www.cco.caltech.edu/~phys1/java/phys1/efeld/efeld.html

Conservatve Forces The potental dfference between two ponts s ndependent of the path taken to calculate t: electrc forces are conservatve. W ΔU Δ V = V f V = = = E!! ds q q 0 0 f

Electrc Potental of a Pont Charge f P!! V = E ds = E ds = R kq kq kq = dr = + = + 2 r r R R Brng magnary + test charge q 0 n from nfnty! Note: f q were a negatve charge, V would be negatve If q 0 and are both + charges as shown, s the Work needed to brng q 0 from to P + or?

Electrc Potental of Many Pont Charges Electrc potental s a SCALAR not a vector. q 4 Just calculate the potental due to each ndvdual pont charge, and add together! (Make sure you get the SIGNS correct!) q 5 r 4 r 5 r 3 r 1 P r 2 q 3 q 2 V = k q r q 1

3 D = 2d For speed set d = e = 1! (a) V P = +e d + +e 2d = 3 2 (b) V P = +e d + +e 2d = 3 2 (c) V P = +e d + +e 2d = 3 2 V P (a) = V P (b) = V P (c) e d = 3 / 2 e d = 3 / 2 e d = 3 / 2 No vectors! Just add wth sgn. One over dstance. Snce all charges same and all dstances same all potentals same.

ICPP: Postve and negatve charges of equal magntude Q are held n a crcle of radus r. V = k q r Q +Q 1. What s the electrc potental voltage at the center of each crcle? V A = ( ) /r = +kq/r k +3Q 2Q A V B = ( ) /r = 2kQ/r k +2Q 4Q V C = ( ) / r = 0 k +2Q 2Q 2. Draw an arrow representng the approxmate drecton of B the electrc feld at the center of each crcle. 3. Whch system has the hghest potental energy? C U A =+q 0 V A has largest postve un-canceled charge

Potental Energy of A System of Charges 4 pont charges (each +Q and equal mass) are connected by strngs, formng a square of sde L If all four strngs suddenly snap, what s the knetc energy of each charge when they are very far apart? +Q +Q Use conservaton of energy: Fnal knetc energy of all four charges = ntal potental energy stored = energy requred to assemble the system of charges +Q +Q If each charge has mass m, fnd the velocty of each charge long after the strng snaps. Let s do ths from scratch!

Potental Energy of A System of No energy needed to brng n frst charge: U 1 =0 Energy needed to brng n 2nd charge: U Charges: Soluton = QV = 2 1 Energy needed to brng n 3rd charge = kq U3 = QV = Q( V1 + V2 ) = + L kq L kq 2 2 2 Energy needed to brng n 4th charge = 2kQ U 4 = QV = Q( V1 + V2 + V3 ) = + L 2L kq 2 2 2L L +Q +Q 2L +Q +Q Total potental energy s sum of all the ndvdual terms shown on left hand sde = kq 2 ( ) L 4 + 2 So, fnal knetc energy of each charge =K=mv 2 /2 = kq 2 ( ) 4L 4 + 2

Potental Energy of a Dpole Δ U = Wapp = qδv +Q a Q What s the potental energy of a dpole? +Q Q a Frst: Brng charge +Q: no work nvolved, no potental energy. The charge +Q has created an electrc potental everywhere, V(r) = kq/r Second: The work needed to brng the charge Q to a dstance a from the charge +Q s W app = U = (-Q)V = ( Q)(+kQ/a) = -kq 2 /a The dpole has a negatve potental energy equal to -kq 2 /a: we had to do negatve work to buld the dpole (electrc feld dd postve work).

Electrc Potental of a Dpole (on axs) What s V at a pont at an axal dstance r away from the mdpont of a dpole (on sde of postve charge)? Q Q V = k k a a ( r ) ( r + ) 2 2 a a ( r + ) ( r ) kq 2 2 = a a ( r )( r + ) 2 2 Qa = 2 2 a 4πε 0( r ) 4 p a Q +Q r Far away, when r >> a: V = p 4πε r 0 2

IPPC: Electrc Potental on Perpendcular Bsector of Dpole You brng a charge of Q o = 3C from nfnty to a pont P on the perpendcular bsector of a dpole as shown. Is the work that you do: a) Postve? b) Negatve? c) Zero? a -Q +Q P d U = Q o V = Q o ( Q/d+Q/d) = 0 3C

4 P a V a = + V b = 0 V c = P b V a > V c > V b P c

Summary: Electrc potental: work needed to brng +1C from nfnty; unts V = Volt Electrc potental unquely defned for every pont n space -- ndependent of path! Electrc potental s a scalar add contrbutons from ndvdual pont charges We calculated the electrc potental produced by a sngle charge: V=kq/r, and by contnuous charge dstrbutons: dv=kdq/r Electrc potental energy: work used to buld the system, charge by charge. Use W=qV for each charge.

Mdterm Exam #1 AVG = 67; STDV=17 A: 90-100% B: 75-89% C: 60-74% D: 50-59% F: 49-0%