Physics 212. Lecture 7. Conductors and Capacitance. Physics 212 Lecture 7, Slide 1

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1 Physics 212 Lecture 7 Conductors and Capacitance Physics 212 Lecture 7, Slide 1

2 Conductors The Main Points Charges free to move E = 0 in a conductor Surface = Equipotential In fact, the entire conductor is an equipotential E perpendicular to surface E = / o Cavity inside a conductor: E = 0, unaffected by fields and charges outside. 5 Physics 212 Lecture 7, Slide 2

3 Checkpoint 1a Two spherical conductors are separated by a large distance. They each carry the same positive charge. 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. A > B B. A = B C. A < B A k B k r A r B 6 Physics 212 Lecture 7, Slide 3

4 Checkpoint 1b The two conductors are now connected by a wire. How do the potentials at the conductor surfaces compare now? A. A > B B. A = B C. A < B 7 Physics 212 Lecture 7, Slide 4

5 Checkpoint 1c What happens to the charge on conductor A after it is connected to conductor B by the wire? A. A increases B. A decreases C. A doesn t change A A k B k ra r B B 8 Physics 212 Lecture 7, Slide 5

6 Charging a Capacitor C - C =C This really means that the battery has moved charge from one plate to the other, so that one plate holds and the other -. 8 Physics 212 Lecture 8, Slide 6

7 Capacitance Capacitance is defined for any pair of spatially separated conductors C How do we understand this definition? Unit is the Farad 9 Consider two conductors, one with excess charge = and the other with excess charge = - d - 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! The ratio of to the potential difference is the capacitance and only depends on the geometry of the conductors E Physics 212 Lecture 7, Slide 7

8 Parallel plate capacitance y First determine E field produced by charged conductors: x d - E E o A A = area of plate Second, integrate E to find the potential difference d E dy 0 As promised, is proportional to. d ( Edy) d 0 0 E dy d A o 12 C d / o A C 0A d C determined by geometry! Physics 212 Lecture 7, Slide 8

9 uestion 0 Initial charge on capacitor = 0 d - 0 Insert uncharged conductor Charge on capacitor now = 1 d 1-1 t How is 1 related to 0?? A. 1 < 0 B. 1 = 0 C. 1 > 0 Plates not connected to anything CHARGE CANNOT CHANGE!! 14 Physics 212 Lecture 7, Slide 9

10 Parallel Plate Capacitor Two parallel plates of equal area carry equal and opposite charge 0. The potential difference between the two plates is measured to be 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 1 such that the potential difference between the plates remains the same. Physics 212 Lecture 7, Slide 10

11 Where to Start?? 0 d t - 0 What is the total charge induced on the bottom surface of the conductor? A. 0 B. - 0 C. 0 D. Positive but the magnitude unknown E. Negative but the magnitude unknown 17 Physics 212 Lecture 7, Slide 11

12 WHY? E = 0 E E - 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 12

13 Calculate Now calculate as a function of distance from the bottom conductor. ( y) d y E dy 0 0 y E t y -E 0 d E = 0 t 21-0 What is D = (d)? A) D = E 0 d B) D = E 0 (d t) C) D = E 0 (d t) The integral = area under the curve y Physics 212 Lecture 7, Slide 13

14 Back to Checkpoint 2a Two parallel plates of equal area carry equal and opposite charge 0. The potential difference between the two plates is measured to be 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 1 such that the potential difference between the plates remains the same. A) 1 < o B) 1 = o C) 1 > o 0 E0d d d A d t 1 0 E1 d t d t A 0 0 Physics 212 Lecture 7, Slide 14

15 Checkpoint 2b Two parallel plates of equal area carry equal and opposite charge 0. The potential difference between the two plates is measured to be 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 1 such that the potential difference between the plates remains the same. What happens to C 1 relative to C 0? A) C 1 > C o B) C 1 = C o C) C 1 < C o We store more charge, 1 = C 1 0 > 0 = C 0 0 for the same voltage difference. E = / 0 A Same : 0 = E 0 d 0 = E 1 (d t) C 0 = 0 /E 0 d C 1 = 1 /(E 1 (d t)) C 0 = 0 A/d C 1 = 0 A/(d t) Physics 212 Lecture 7, Slide 15

16 Energy in Capacitors 31 Physics 212 Lecture 7, Slide 16

17 cross-section Calculation a 4 3 a 2 a 1 A capacitor is constructed from two conducting cylindrical shells of radii a 1, a 2, a 3, and a 4 and length L (L >> a i ). metal What is the capacitance C of this device? metal Conceptual Analysis: C But what is and what is? Important Point: C is a property of the object! (concentric cylinders) Assume some (i.e., on one conductor and on the other) These charges create E field in region between conductors This E field determines a potential difference between the conductors should be proportional to ; the ratio / is the capacitance. 33 Physics 212 Lecture 7, Slide 17

18 cross-section Calculation a 4 3 a 2 a 1 metal 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 >> a i ). What is the capacitance C of this capacitor? C Where is on outer conductor located? (A) at r=a 4 (B) at r=a 3 (C) both surfaces (D) throughout shell Why? Gauss law: E da enclosed We know that E = 0 in conductor (between a 3 and a 4 ) o enclosed 0 enclosed 0 must be on inside surface (a 3 ), so that enclosed = = 0 Physics 212 Lecture 7, Slide 18

19 cross-section Calculation a 4 3 a 2 a 1 metal - 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 >> a i ). What is the capacitance C of this capacitor? C Where is - on inner conductor located? (A) at r=a 2 (B) at r=a 1 (C) both surfaces (D) throughout shell Why? Gauss law: E da enclosed We know that E = 0 in conductor (between a 1 and a 2 ) o enclosed 0 enclosed 0 must be on outer surface (a 2 ), so that enclosed = 0 Physics 212 Lecture 7, Slide 19

20 a 4 3 a 2 a 1 cross-section metal - metal a 2 < r < a 3 : What is E(r)? 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 >> a i ). What is the capacitance C of this capacitor? C (A) 0 (B) (C) (D) (E) 4 2 r 2 o Lr 2o Lr 4 2 o r o Why? Gauss law: E da enclosed o E 2rL o 1 E 2 Lr Direction: Radially In 0 Physics 212 Lecture 7, Slide 20

21 a 4 3 a 2 a 1 cross-section metal - metal r < a 2 : E(r) = 0 since enclosed = 0 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 >> a i ). What is the capacitance C of this capacitor? C (2 0 a 2 L) a 2 < r < a 3 : E Lr What is? The potential difference between the conductors What is the sign of = outer - inner? (A) outer - inner < 0 (B) outer - inner = 0 (C) outer - inner > 0 Physics 212 Lecture 7, Slide 21

22 cross-section Calculation a 4 3 a 2 a 1 metal - 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 >> a i ). What is the capacitance C of this capacitor? C a 2 < r < a 3 : What is outer - inner? a1 a ln 4 a3 a ln ln ln 2 2oL a4 2oL a1 2oL a2 2 o L a3 (A) (B) (C) (D) E 1 2 o rl (2 0 a 2 L) a 3 a o r 2 dr 2 L proportional to, as promised 2 L ln( / ) 0 C a 3 a 2 2 L o a a 3 2 dr r a3 ln 2 L a 0 2 Physics 212 Lecture 7, Slide 22