Physics 212 Lecture 7 Todays Concept Conductors and

Physics 212 Lecture 7 Today's Concept: Conductors and Capacitance Physics 212 Lecture 7, Slide 1

Music Who is the Artist? A) B) C) D) E) Eric Clapton Bill Frisell Jimmy Page Jeff Beck Buddy Guy Why? Starting on some circuits – electric guitar Physics 212 Lecture 3

LOGISTICS 1) EXAM 1: WED Feb. 15 at 7 pm Sign Up in Gradebook for Conflict Exam at 5: 15 pm if desired BY Mon. Feb. 13 at 10: 00 p. m. MATERIAL: Lectures 1 - 8 2) EXAM 1 PREPARATION? Old exams are on-line (“Practice Exams”), also “Worked Examples” and “Exam Prep Exercises” Physics 212 Lecture 7, Slide 3

Your Comments “I'm really lost. Can you explain the examples more in lecture please? ” “I enjoyed this lecture, seems pretty straight forward. ” “Soooooooooooooooooooo confused and the homework was sooooooooooooooooooooooooo hard” “How exactly does an electric field store energy in capacitors? ” “go over the conductor between the plates!” “PLEASE!!! EXPLAIN SOME OF THE HOMEWORK IN CLASS!!!” “I feel like all the examples were just derived from the prelecture, and I would like to go over some different examples to make sure I truly understand this stuff, and am not just repeating exactly what was previously told to me. ” True, but … We’ll go through the conductor between the plates for two separate cases We’ll do a problem much like a homework problem at the end of class 05 “If there is a train that could hold then entire capacity of students enrolled in Physics 212, that conductor must be pretty attracted to his job. ” “Ignorance is bliss” Physics 212 Lecture 7, Slide 4

Conductors You did well on the questions on charge distributions on conductors The Main Points • • 5 Charges free to move E = 0 in a conductor (even in a cavity) Surface = Equipotential E at surface perpendicular to surface Physics 212 Lecture 7, Slide 5

Checkpoint 1 a Two 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 > VB B. VA = VB C. VA < VB “larger area more charge” “Conductors with the same charge are equipotential“ “The radius of A is 4 B, and since V=k. Q/r you get 4 Va=Vb” 6 Physics 212 Lecture 7, Slide 6

Checkpoint 1 b The two conductors are now connected by a wire. How do the potentials at the conductor surfaces compare now? A. VA > VB B. VA = VB C. VA < VB “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. ” 7 Physics 212 Lecture 7, Slide 7

Checkpoint 1 c What happens to the charge on conductor A after it is connected to conductor B by the wire? A. QA increases B. QA decreases C. QA doesn’t change “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” “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? ” 8 Physics 212 Lecture 7, Slide 8

Parallel Plate Capacitor Two parallel plates of equal area carry equal and opposite charge Q 0. The potential difference between the two plates is measured to be V 0. 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 Q 1 such that the potential difference between the plates remains the same. THE CAPACITOR QUESTIONS WERE TOUGH! THE PLAN: We’ll work through the example in the Prelecture and then do the Checkpoint questions. Physics 212 Lecture 7, Slide 9

Capacitance is defined for any pair of spatially separated conductors How do we understand this definition ? ? ? • Consider two conductors, one with excess charge = +Q and the other with excess charge = -Q +Q d E V -Q • These charges create an electric field in the space between them • We can integrate the electric field between them to find the potential difference between the conductors • 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 conductors 9 Physics 212 Lecture 7, Slide 10

Example (done in Prelecture 7) First determine E field produced by charged conductors: What is s ? ? +Q y d x E -Q A = area of plate Second, integrate E to find the potential difference V As promised, V is proportional to Q !! st o m Al g n i th y er v e u yo ed e n HW r fo 1! C determined by geometry !! 12 Physics 212 Lecture 7, Slide 11

Question Related to Checkpoint 2 +Q 0 Initial charge on capacitor = Q 0 d -Q 0 Insert uncharged conductor Charge on capacitor now = Q 1 +Q 1 d t -Q 1 How is Q 1 related to Q 0 ? ? A. Q 1 < Q 0 B. Q 1 = Q 0 C. Q 1 > Q 0 14 Plates not connected to anything CHARGE CANNOT CHANGE !! Physics 212 Lecture 7, Slide 12

Where to Start? ? +Q 0 d t -Q 0 What is the total charge induced on the bottom surface of the conductor? A. B. C. D. E. 17 +Q 0 -Q 0 0 Positive but the magnitude unknown Negative but the magnitude unknown Physics 212 Lecture 7, Slide 13

WHY ? ? +Q 0 -Q 0 +Q 0 E E=0 E -Q 0 WHAT DO WE KNOW ? ? ? E must be = 0 in conductor !! Charges inside conductor move to cancel E field from top & bottom plates 19 Physics 212 Lecture 7, Slide 14

Calculate V Now calculate V as a function of distance from the bottom conductor +Q 0 y E d t y -E 0 d E=0 t V -Q 0 21 What is DV = V(d)? A) DV = E 0 d B) DV = E 0(d – t) C) DV = E 0(d + t) y The integral = area under the curve Physics 212 Lecture 7, Slide 15

Back to Checkpoint 2 a Two parallel plates of equal area carry equal and opposite charge Q 0. The potential difference between the two plates is measured to be V 0. 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 Q 1 such that the potential difference between the plates remains the same. A) Q 1 < Qo B) Q 1 = Qo C) Q 1 > Qo “The air space in between Q 0 is greater than Q 1 so Q 0 must be greater to achieve the same potential difference. ” “The potential difference is just the difference in charge between the plates. Adding a conductor in the center doesn't change that” “Q 1 needs a greater charge to have the same potential since part of its electric field is zero. ” How do you get the same V in ‘less space’? Physics 212 Lecture 7, Slide 16

Checkpoint 2 b Two parallel plates of equal area carry equal and opposite charge Q 0. The potential difference between the two plates is measured to be V 0. 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 Q 1 such that the potential difference between the plates remains the same. What happens to C 1 relative to C 0? A) C 1 > Co B) C 1 = Co C) C 1 < Co “Capacitance is directly proportional to the charge, so if in case 1, the charge is greater than in case 0, that means the Capacitance is greater. ” “Capacitance is equal to charge over voltage, both of which are the same. ” “C=(1/2)*((Q^2)/U). Thus, if Q decreases, then C will decrease. ” Physics 212 Lecture 7, Slide 17

Checkpoint 2 b Two parallel plates of equal area carry equal and opposite charge Q 0. The potential difference between the two plates is measured to be V 0. 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 Q 1 such that the potential difference between the plates remains the same. What happens to C 1 relative to C 0? A) C 1 > Co B) C 1 = Co C) C 1 < Co We can determine C from either case same V (Checkpoint) same Q (Prelecture) C depends only on geometry !! Same V: V 0 = E 0 d V 0 = E 1(d – t) E = Q/e 0 A C 0 = Q 0/E 0 d C 1 = Q 1/(E 1(d – t)) C 0 = e 0 A/d C 1 = e 0 A/(d – t) Physics 212 Lecture 7, Slide 18

Energy in Capacitors BANG 31 Physics 212 Lecture 7, Slide 19

Homework for Lec. 7&8 Physics 212 Lecture 7, Slide 20

Calculation cross-section a 4 a 3 A capacitor is constructed from two conducting cylindrical shells of radii a 1, a 2, a 3, and a 4 and length L (L >> ai). a 2 a 1 What is the capacitance C of this device ? metal • Conceptual Analysis: But what is Q and what is V? They are not given? ? • Important Point: C is a property of the object!! (concentric cylinders here) • Assume some Q (i. e. , +Q on one conductor and –Q on the other) • These charges create E field in region between conductors • This E field determines a potential difference V between the conductors • V should be proportional to Q; the ratio Q/V is the capacitance. 33 Physics 212 Lecture 7, Slide 21

cross-section a 4 a 3 Calculation A capacitor is constructed from two conducting cylindrical shells of radii a 1, a 2, a 3, and a 4 and length L (L >> ai). a 2 a 1 metal What is the capacitance C of this capacitor ? metal • Strategic Analysis: – – Put +Q on outer shell and –Q on inner shell Cylindrical symmetry: Use Gauss’ Law to calculate E everywhere Integrate E to get V Take ratio Q/V: should get expression only using geometric parameters (ai, L) Physics 212 Lecture 7, Slide 22

Calculation cross-section +Q + + + -Q + + + a 2 a 1 + a 4 a 3 + metal + + A capacitor is constructed from two conducting cylindrical shells of radii a 1, a 2, a 3, and a 4 and length L (L >> ai). What is the capacitance C of this capacitor ? + + + metal Where is +Q on outer conductor located? (A) at r=a 4 (B) at r=a 3 (C) both surfaces (D) throughout shell Why? Gauss’ law: We know that E = 0 in conductor (between a 3 and a 4) +Q must be on inside surface (a 3), so that Qenclosed = + Q – Q = 0 Physics 212 Lecture 7, Slide 23

Calculation cross-section +Q + + - - - - + + - -Q - + + - - metal - + + + + a 2 a 1 + a 4 a 3 A capacitor is constructed from two conducting cylindrical shells of radii a 1, a 2, a 3, and a 4 and length L (L >> ai). What is the capacitance C of this capacitor ? metal Where is -Q on inner conductor located? (A) at r=a 2 (B) at r=a 1 (C) both surfaces (D) throughout shell Why? Gauss’ law: We know that E = 0 in conductor (between a 1 and a 2) +Q must be on outer surface (a 2), so that Qenclosed = 0 Physics 212 Lecture 7, Slide 24

Calculation cross-section +Q A capacitor is constructed from two conducting cylindrical shells of radii a 1, a 2, a 3, and a 4 and length L (L >> ai). What is the capacitance C of this capacitor ? + + - - - - + + - -Q - + + - - metal - + + + + a 2 a 1 + a 4 a 3 metal a 2 < r < a 3: What is E(r)? (A) 0 (B) (C) (D) (E) Why? Gauss’ law: Direction: Radially In Physics 212 Lecture 7, Slide 25

Calculation cross-section +Q + + - - - - + + - -Q - + + - - metal - + + + + a 2 a 1 + a 4 a 3 A capacitor is constructed from two conducting cylindrical shells of radii a 1, a 2, a 3, and a 4 and length L (L >> ai). What is the capacitance C of this capacitor ? a 2 < r < a 3: metal r < a 2: E(r) = 0 since Qenclosed = 0 Q (2 pe 0 a 2 L) • What is V? • The potential difference between the conductors What is the sign of V = Vouter - Vinner? (A) Vouter-Vinner < 0 (B) Vouter-Vinner = 0 (C) Vouter-Vinner > 0 Physics 212 Lecture 7, Slide 26

Calculation cross-section A capacitor is constructed from two conducting cylindrical shells of radii a 1, a 2, a 3, and a 4 and length L (L >> ai). What is the capacitance C of this capacitor ? +Q + + - - - - + a 2 a 1 + a 4 a 3 + - -Q - + + - - metal - + + + a 2 < r < a 3: + metal What is V Vouter - Vinner? (A) (B) (C) (D) Q (2 pe 0 a 2 L) V proportional to Q, as promised Physics 212 Lecture 7, Slide 27