Physics 212 Lecture 7, Slide 1Physics 212Lecture 7Today's Concept:Conductors and CapacitancePhysics 212 Lecture 7MusicWho is the Artist?A) Ricky SkaggsB) Bill MonroeC) Mark O’ConnorD) Chris ThileE) Mike MarshallWhy?Final Ellnora Tribute: Mandolin virtuoso who played with Michael Daveson guitar Friday afternoon.. Great show !!My Favorite? Hard to say.. There were many, but if I had to pick one, I’ll go with Richard Thompson.Physics 212 Lecture 7, Slide 3LOGISTICS1) EXAM 1: WED Sep. 21 at 7pmSign Up in Gradebook for Conflict Exam at 5:15pm if desired2) EXAM 1 PREPARATION?Old exams are on-line (“Practice Exams”), also “Worked Examples”BY THURS. SEP. 15 at 10:00 p.m.MATERIAL: Lectures 1 - 8Physics 212 Lecture 7, Slide 4Your Comments05We will do all of these in this lecture, including a calculation of a cylindrical capacitor“A little more about the neutral plate in the middle of capacitors.”“It would be nice to discuss examples of capacitors other than the parallel plates in a bit more depth.” “Can we see some capacitor demo's? Possibly very large capacitors being discharged?”“Can we please talk about homework problems a little bit at the end of the lecture? So that we can have an idea about how to start on the homework...”“I found most of this lecture to be confusing because it's something completely new. Go over it all please!”“I think this stuff is tough, but I can get a handle on it.”Physics 212 Lecture 7, Slide 5Conductors• Charges free to move• E = 0 in a conductor• Surface = Equipotential• E at surface perpendicular to surface5You did well on the questions on charge distributions on conductorsThe Main PointsPhysics 212 Lecture 7, Slide 6Checkpoint 1a“The surface of conductors are equipotentials so they are equal in potential despite the increased radius.““V=kQ/R! Since the Ra is greater than Rb, the Va<Vb”6Two spherical conductors are separated by a large distance. They each carry the same positive charge Q. Conductor A has a larger radius than conductor B.Compare the potential at the surface of conductor A with the potential at the surface of conductor B.A. VA> VBB. VA= VBC. VA< VBPhysics 212 Lecture 7, Slide 7Checkpoint 1b“No matter what the initial conditions are, when both spheres are making contact, their potential has to be equal since they are connected by a wire that makes them behave like a single conductor.”7The two conductors are now connected by a wire. How do the potentials at the conductor surfaces compare now?A. VA> VBB. VA= VBC. VA< VBPhysics 212 Lecture 7, Slide 8Checkpoint 1c“Charge will always move to a place with lower potential, and the larger sphere has a lower potential than the smaller sphere. ”“the charge would decrease in order to compensate for the lower charge on the particle B”8What happens to the charge on conductor A after it is connected to conductor B by the wire?A. QAincreases B. QAdecreases C. QAdoesn’t change“When you connect two conductors by a wire and charge moves between them as to make difference in potential of the system zero what is the charge of the wire? Or does it not matter?”Physics 212 Lecture 7, Slide 9Parallel Plate CapacitorTHE CAPACITOR QUESTIONS WERE TOUGH!THE PLAN:We’ll work through the example in the Prelecture and then do the Checkpoint questions.Two parallel plates of equal area carry equal and opposite charge Q0. The potential difference between the two plates is measured to be V0. An uncharged conducting plate (the green thing in the picture below) is slipped into the space between the plates without touching either one. The charge on the plates is adjusted to a new value Q1such that the potential difference between the plates remains the same.Physics 212 Lecture 7, Slide 10CapacitanceCapacitance is defined for any pair of spatially separated conductors9QCV≡How do we understand this definition ???+Q-Qd• Consider two conductors, one with excess charge = +Q and the other with excess charge = -Q• These charges create an electric field in the space between themE• This potential difference should be proportional to Q !!• The ratio of Q to the potential difference is the capacitance and only depends on the geometry of the conductorsV• We can integrate the electric field between them to find the potential difference between the conductorPhysics 212 Lecture 7, Slide 11First determine E field produced by charged conductors: Example (done in Prelecture 7)12+Q-QdESecond, integrate E to find the potential difference VxyAs promised, V is proportional to Q !!AQdQVQCoε/=≡dAC0ε=C determined by geometry !!oEεσ=What is σ ??A = area of plateAQ=σ∫⋅−=dydEV0dAQdyEEdyVod dε=∫ ∫=−−=0 0)(Physics 212 Lecture 7, Slide 12+Q0-Q0dInitial charge on capacitor = Q0Plates not connected to anything14Question Related to Checkpoint 2td+Q1-Q1Insert uncharged conductorCharge on capacitor now = Q1A. Q1< Q0B. Q1= Q0C. Q1> Q0How is Q1related to Q0??CHARGE CANNOT CHANGE !!Physics 212 Lecture 7, Slide 13A. +Q0B. -Q0C. 0D. Positive but the magnitude unknownE. Negative but the magnitude unknown17td+Q0-Q0What is the total charge induced on the bottom surface of the conductor?Where to Start??Physics 212 Lecture 7, Slide 14E+Q0-Q0EWHAT DO WE KNOW ???+Q0-Q019E = 0E must be = 0 in conductor !!Charges inside conductor move to cancel E field from top & bottom platesWHY ??Physics 212 Lecture 7, Slide 15Now calculate V as a function of distance from the bottom conductor. +Q0-Q0E = 0yVy21dtCalculate V∫⋅−=yydEyV0)(What is ∆V = V(d)?A) ∆V = E0dB) ∆V = E0(d – t)C) ∆V = E0(d + t)dtEy-E0The integral = area under the curvePhysics 212 Lecture 7, Slide 16Back to Checkpoint 2a“C has gone down so Q must go down as well to maintain same V.”“The charge remains the same since it is supplied through the battery or whatever”“It order to make up for the conductor interrupting its potential Q1 must be bigger.”A) Q1< QoB) Q1= QoC) Q1> QoTwo parallel plates of equal area carry equal and opposite charge Q0. The potential difference between the two plates is measured to be V0. An uncharged conducting plate (the green thing in the picture below) is slipped into the space between the plates without touching either one. The charge on the plates is adjusted to a new value Q1such that the potential difference between the plates remains the same.How do you get the same V0in ‘less space’?Physics 212 Lecture 7,