I would prefer to not start with and integrate. Why?

Slides:



Advertisements
Similar presentations
Lecture 6 Problems.
Advertisements

Continuous Charge Distributions
Chapter 24 Gauss’s Law.
I-2 Gauss’ Law Main Topics The Electric Flux. The Gauss’ Law. The Charge Density. Use the G. L. to calculate the field of a.
Chapter 24 Gauss’s Law.
Calculate Electric Potential: Example A conducting sphere of radius R 1 =0.5m is placed at the center of a conducting spherical shell of inner and outer.
1 Fall 2004 Physics 3 Tu-Th Section Claudio Campagnari Lecture 11: 2 Nov Web page:
E The net electric flux through a closed cylindrical surface is zero.
Electric Potential with Integration Potential Difference in a Variable E-field If E varies, we integrate the potential difference for a small displacement.
From Chapter 23 – Coulomb’s Law
a b c Gauss’ Law … made easy To solve the above equation for E, you have to be able to CHOOSE A CLOSED SURFACE such that the integral is TRIVIAL. (1)
A b c Gauss' Law.
Physics.
Electricity and Magnetism Review 1: Units 1-6
Faculty of Engineering Sciences Department of Basic Science 5/26/20161W3.
Chapter 22 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s.
1 W04D1 Electric Potential and Gauss’ Law Equipotential Lines Today’s Reading Assignment Course Notes: Sections ,
Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures.
Previous Lectures: Introduced to Coulomb’s law Learnt the superposition principle Showed how to calculate the electric field resulting from a series of.
Physics 2112 Unit 4: Gauss’ Law
Electric Potential Chapter 25. ELECTRIC POTENTIAL DIFFERENCE The fundamental definition of the electric potential V is given in terms of the electric.
(nz183.jpg)
Day 4: Electric Field Calculations for Continuous Charge Distributions A Uniform Distribution of Surface charge A Ring of Continuous Charge A Long Line.
Chapter 23 Electric Potential.
Wednesday, Sep. 14, PHYS Dr. Andrew Brandt PHYS 1444 – Section 04 Lecture #5 Chapter 21: E-field examples Chapter 22: Gauss’ Law Examples.
-Electric Potential due to Continuous Charge Distributions AP Physics C Mrs. Coyle.
Lecture 17: THU 18 MAR 10 Ampere’s law Physics 2102 Jonathan Dowling André Marie Ampère (1775 – 1836)
Lecture 5 Dr. Lobna Mohamed Abou El-Magd The Electric Potential.
A b c. Choose either or And E constant over surface is just the area of the Gaussian surface over which we are integrating. Gauss’ Law This equation can.
Electric potential §8-5 Electric potential Electrostatic field does work for moving charge --E-field possesses energy 1.Work done by electrostatic force.
Electric Potential Chapter 25 The Electric Potential
Tuesday, Sept. 20, 2011PHYS , Fall 2011 Dr. Jaehoon Yu 1 PHYS 1444 – Section 003 Lecture #9 Tuesday, Sept. 20, 2011 Dr. Jaehoon Yu Chapter 23 Electric.
Wednesday, Feb. 8, 2012PHYS , Spring 2012 Dr. Jaehoon Yu 1 PHYS 1444 – Section 004 Lecture #7 Wednesday, Feb. 8, 2012 Dr. Alden Stradeling Chapter.
Lecture 23: WED 11 MAR Ampere’s law Physics 2102 Jonathan Dowling André Marie Ampère (1775 – 1836)
Copyright © 2009 Pearson Education, Inc. Applications of Gauss’s Law.
Chapter 23 Electric Potential & Electric Potential Energy.
To be worked at the blackboard in lecture.
My office change was not reflected on the syllabus. It is now ESCN 2
Line integral of Electric field: Electric Potential
Exam 1: Tuesday, Feb 14, 5:00-6:00 PM
Line integral of Electric field: Electric Potential
Potential and E-field = external work done on charge to move it.
Electric potential of a charge distribution. Equipotentials.
Chapter 23 Electric Potential
Question 300 V/m 0 V/m 300 V/m A B 0.02m 0.03m 0.04m What is VB-VA?
Conductors and Gauss’s Law
Exam 1: Tuesday, Feb 14, 5:00-6:00 PM
Chapter 25 Electric Potential.
Capacitors: parallel plate, cylindrical, spherical.
Applied Electricity and Magnetism
E The net electric flux through a closed cylindrical surface is zero.
Lecture 27: MON 26 OCT Magnetic Fields Due to Currents II
That reminds me… must download the test prep HW.
Electric potential energy. Electric potential.
PHYS 1444 – Section 003 Lecture #9
Physics 2102 Lecture 16 Ampere’s law Physics 2102 Jonathan Dowling
TOPIC 4 Electrostatic Potential
That reminds me… must download the test prep HW.
Exam 1: Tuesday, Sept 19, 5:00-6:00 PM
Example: use Gauss’ Law to calculate the electric field due to a long line of charge, with linear charge density . This is easy using Gauss’ Law (remember.
Chapter 23 Electric Potential
A ring with radius R has a uniform charge density 
Gauss’s Law Electric Flux
PHYS 1444 – Section 003 Lecture #8
PHYS 1444 – Section 002 Lecture #9
Exercises on sheet similar to this
Norah Ali Al-moneef King Saud university
PHY 102: Lecture Symmetry 3.2 Concept of Flux
Chapter 23 Electric Potential.
PHYS 1444 – Section 02 Lecture #7
Presentation transcript:

I would prefer to not start with and integrate. Why? Example 5: calculate the potential at a point outside a very long insulating cylinder of radius R and positive uniform linear charge density . I would prefer to not start with and integrate. Why? Reason #1: a mathematical pain. What do I pick for my charge element dq? Little cubes? Ugh. Circular rings that I have to integrate from 0 to R and from - to  along the axis? Long cylindrical shells that I have to integrate from 0 to R? There must be an easier way. Reason #2: I suspect that we would eventually find (read your text) that V= at any finite distance from the cylinder. Not very useful. Reason #3: we have previously derived a simple expression for outside a cylinder of charge. This can easily be integrated to find V. To be worked at the blackboard in lecture…

r r=a  >0 E dr r=R R Start with and to calculate

r r=a  >0 E dr r=R R

If we let a be an arbitrary distance r, then If we take V=0 at r=R, then

Things to note: V is zero at the surface of the cylinder and decreases as you go further out. This makes sense! V decreases as you move away from positive charges. If we tried to use V=0 at r= then (V is infinite at any finite r). That’s why we can’t start with

Things to note: For >0 and r>R, Vr – VR <0. Our text’s convention is Vab = Va – Vb. This is explained on page 759. Thus VrR = Vr – VR is the potential difference between points r and R and for r>R, VrR < 0. In Physics 1135, Vba = Va – Vb. I like the Physics 1135 notation because it clearly shows where you start and end. But Vab has mathematical advantages which we will see in Chapter 24.

Vif = Vf – Vi so Vif = -Vif See your text for other examples of potentials calculated from charge distributions, as well as an alternate discussion of the electric field between charged parallel plates. Remember: worked examples in the text are “testable.” Make sure you know what Vab means, and how it relates to V. Vif = Vf – Vi so Vif = -Vif

Special Dispensation For tomorrow’s homework only: you may use the equation for the electric field of a long straight wire without first proving it: Of course, this is relevant only if a homework problem requires you to know the electric field of a long straight wire. You can also use this equation for the electric field outside a long cylinder that carries charge.

Homework Hint! Problems like 23.32 and 23.33: you must derive an expression for the potential outside a long conducting cylinder. See example 23.10. V is not zero at infinity in this case. Use If 23.32 and 23.33 are not assigned, don’t be disappointed. We can still get you on this in problems in chapter 24!

Homework Hints! In energy problems involving potentials, you may know the potential but not details of the charge distribution that produced it (or the charge distribution may be complex). In that case, you don’t want to attempt to calculate potential energy using . Instead, use If the electric field is zero everywhere in some region, what can you say about the potential in that region? Why?