Electricity & Magnetism Lecture 9: Conductors and Capacitance

Transcription

1 Electricity & Magnetism Lecture 9: Conductors and Capacitance Today s Concept: A) Conductors B) Capacitance ( Electricity & Magne7sm Lecture 7, Slide 1

2 Some of your comments This chapter makes absolute 0 sense. I would really love an explana7on on how capacitors are used as temporary baheries for storing energy in them.

3 Capacitors vs. Batteries A bahery uses an electrochemical reac7on to separate charges. ΔV between terminals is called EMF, symbol:, E or E. E Capacitors store charge on its plates.

4 ! 0 Some capacitors

5 Our Comments WE BELIEVE THERE ARE ONLY THREE THINGS YOU NEED TO KNOW TO DO ALL OF HOMEWORK! 1. E = 0 within the material of a conductor: (at sta7c equilibrium) Charges move inside a conductor in order to cancel out the fields that would be there in the absence of the conductor. This principle determines the induced charge densi7es on the surfaces of conductors. 2. Gauss Law: If charge distribu7ons have sufficient symmetry (spherical, cylindrical, planar), then Gauss law can be used to determine the electric field everywhere. I surface ~E d ~A = Q enclosed 0 V a b U a b q = Z b a ~E d~l CONCEPTS DETERMINE THE CALCULATION! Electricity & Magne7sm Lecture 7, Slide 3

6 Conductors The Main Points " Charges free to move " E = 0 in a conductor " Surface = Equipoten7al " E at surface perpendicular to surface Electricity & Magne7sm Lecture 7, Slide 4

7 CheckPoint: Two Spherical Conductors 1 Two spherical conductors are separated by a large distance. They each carry the same posi7ve charge Q. Conductor A has a larger radius than conductor B. Compare the poten7al at the surface of conductor A with the poten7al at the surface of conductor B. A. V A > V B B. V A = V B C. V A < V B The poten7al is the same as from a point charge at the center of A or B. So, following V=kQ/r, A would have a smaller poten7al Electricity & Magne7sm Lecture 7, Slide 5

8 CheckPoint: Two Spherical Conductors 2 Two spherical conductors are separated by a large distance. They each carry the same posi7ve charge Q. Conductor A has a larger radius than conductor B. The two conductors are now connected by a wire. How do the poten7als at the conductor surfaces compare now? A. V A > V B B. V A = V B C. V A < V B The poten7als will become equal since the charges will want to go to places of lower poten7al, un7l it balances out. Electricity & Magne7sm Lecture 7, Slide 6

9 CheckPoint: Two Spherical Conductors 3 Two spherical conductors are separated by a large distance. They each carry the same posi7ve charge Q. Conductor A has a larger radius than conductor B. What happens to the charge on conductor A aber it is connected to conductor B by the wire? A. Q A increases B. Q A decreases C. Q A does not change Since B initially has a higher potential, charges move from B to A. Electricity & Magne7sm Lecture 7, Slide 7

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11 (a)1. Conservation of charge gives us one relation between the charges and Q 2 : Q 1 Q 1 Q 2 Q 2. Equating the potential of the spheres gives us a second relation for the charges and Q 2 : Q 1 kq 1 R 1 kq 2 R 2 Q 2 R 2 R 1 Q 1 3. Combine the results from steps 1 and 2 and solve for Q 1 and Q 2 : Q 1 R 1 R 2 Q 1 Q Q 1 R 1 R 1 R 2 Q so 6.0 cm (80 nc) 8.0 cm 60 nc Q 2 Q Q 1 20 nc (b) Use these results to calculate the electric field strengths at the surface of the spheres: E 1 kq 1 R 2 1 ( N # m2 >C 2 )( C) (0.060 m) kn>c E 2 kq 2 R 2 2 ( N # m2 >C 2 )( C) (0.020 m) kn>c (c) Calculate the common potential from kq>r for either sphere: V 1 kq 1 ( N # m2 >C 2 )( C) R m 9.0 kv >

12 CheckPoint: Charged Parallel Plates 1 Two parallel plates of equal area carry equal and opposite charge Q 0. The poten7al difference between the two plates is measured to be V 0. An uncharged conduc7ng 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 poten7al 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 preflight ques7ons. Compare Q 1 and Q 0. o Q 1 < Q 0 o Q 1 = Q 0 o Q 1 > Q 0 Electricity & Magne7sm Lecture 7, Slide 8

13 Capacitance Capacitance is defined for any pair of spa7ally separated conductors. How do we understand this defini7on?! Consider two conductors, one with excess charge = Q and the other with excess charge = Q d Q Q! These charges create an electric field in the space between them! We can integrate the electric field between them to find the poten7al difference between the conductor! This poten7al difference should be propor7onal to Q! The ra7o of Q to the poten7al difference is the capacitance and only depends on the geometry of the conductors E V Electricity & Magne7sm Lecture 7, Slide 9

14 Example (done in Prelecture 7) y First determine E field produced by charged conductors: Q x Second, integrate E to find the potenhal difference V V = Z d 0 d Q ~E d~y As promised, V is proporhonal to Q! What is σ? E A = area of plate C determined by geometry! Electricity & Magne7sm Lecture 7, Slide 10

15 Clicker Question Related to CheckPoint Q 0 IniHal charge on capacitor = Q 0 d Q 0 Insert uncharged conductor Charge on capacitor now = 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 d Q 1 Q 1 Plates not connected to anything CHARGE CANNOT CHANGE! t Electricity & Magne7sm Lecture 7, Slide 11

16 Where to Start? Q 1 d t Q 1 What is the total charge induced on the bohom surface of the conductor? A) Q 1 B) Q 1 C) 0 D) PosiHve but the magnitude unknown E) NegaHve but the magnitude unknown Electricity & Magne7sm Lecture 7, Slide 12

17 Why? Q 0 Q 0 Q 0 E = 0 E E Q 0 WHAT DO WE KNOW? E must be = 0 in conductor! Charges inside conductor move to cancel E field from top & bo8om plates. Electricity & Magne7sm Lecture 7, Slide 13

18 Calculate V Now calculate V as a funchon of distance from the bosom conductor. V( y) = d Z y 0 ~E d~y d Q 0 E = 0 t y E -E 0 t y V Q 0 What is ΔV = V(d)? A) ΔV = E 0 d B) ΔV = E 0 (d t) C) ΔV = E 0 (d t) The integral = area under the curve y Electricity & Magne7sm Lecture 7, Slide 14

19 CheckPoint Results: Charged Parallel Plates 1 Two parallel plates of equal area carry equal and opposite charge Q 0. The poten7al difference between the two plates is measured to be V 0. An uncharged conduc7ng 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 poten7al difference between the plates remains the same. Compare Q 1 and Q 0. A. Q 1 < Q 0 B. Q 1 = Q 0 C. Q 1 > Q 0 The distance for Q1 is smaller therefore the charge must decrease to compensate for the change in distance Since the poten?al remains the same, there is no change in the charge of Q. the field through the conductor is zero, so it has constant poten?al. Because of this it must have greater charge so the total V is that same. Electricity & Magne7sm Lecture 7, Slide 15

20 CheckPoint Results: Charged Parallel Plates 2 Two parallel plates of equal area carry equal and opposite charge Q 0. The poten7al difference between the two plates is measured to be V 0. An uncharged conduc7ng 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 poten7al difference between the plates remains the same. Compare the capacitance of the two configura7ons in the above problem. A. C1 > C0 B. C1 = C0 C. C1 < C0 The distance for Q1 is smaller therefore the charge must decrease to compensate for the change in distance Since the poten?al remains the same, there is no change in the charge of Q. the field through the conductor is zero, so it has constant poten?al. Because of this it must have greater charge so the total V is that same. Electricity & Magne7sm Lecture 7, Slide 16

21 CheckPoint Results: Charged Parallel Plates 2 Two parallel plates of equal area carry equal and opposite charge Q 0. The poten7al difference between the two plates is measured to be V 0. An uncharged conduc7ng 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 poten7al difference between the plates remains the same. Compare the capacitance of the two configura7ons in the above problem. A. C1 > C0 B. C1 = C0 C. C1 < C0 We can determine C from either case same V (preflight) same Q (lecture) C depends only on geometry! E 0 = Q 0 /ε 0 A C 0 = Q 0 /E 0 d C 1 = Q 0 /[E 0 (d t)] C 0 = ε 0 A /d C 1 = ε 0 A /(d t) Electricity & Magne7sm Lecture 7, Slide 17

22 Energy in Capacitors BANG Electricity & Magne7sm Lecture 7, Slide 18

23 Calculation a cross-sec7on 4 A capacitor is constructed from two conduc7ng 3 cylindrical shells of radii a 1, a 2, a 3, and a 4 and length a 2 a 1 metal L (L >> a i ). What is the capacitance C of this capacitor? 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 poten7al difference V between the conductors V should be propor7onal to Q; the ra7o Q/V is the capacitance. Electricity & Magne7sm Lecture 7, Slide 19

24 Calculation a 4 3 a 2 a 1 cross-sec7on A capacitor is constructed from two conduc7ng 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? metal 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 ra7o Q/V: should get expression only using geometric parameters (a i, L) Electricity & Magne7sm Lecture 7, Slide 20

25 a 4 3 a 2 a 1 metal metal Q cross-sec7on Q Calculation A capacitor is constructed from two conduc7ng 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? 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: I surface ~E d ~A = Q enclosed 0 We know that E = 0 in conductor (between a 3 and a 4 ) Q must be on inside surface (a 3 ), so that Q enclosed = Q Q = 0 Electricity & Magne7sm Lecture 7, Slide 21

26 a 4 3 a 2 a 1 Q metal metal cross-sec7on Q Calculation A capacitor is constructed from two conduc7ng 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? 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: I ~E d ~A = Q enclosed 0 surface We know that E = 0 in conductor (between a 1 and a 2 ) Q must be on outer surface (a 2 ), so that Q enclosed = 0 Electricity & Magne7sm Lecture 7, Slide 22

27 a 4 3 a 2 a 1 Q metal metal cross-sec7on Q a 2 < r < a 3 : What is E(r)? Calculation A capacitor is constructed from two conduc7ng 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? A) 0 B) C) D) E) Why? Gauss law: I surface ~E d ~A = Q enclosed 0 E(2 rl) = Q 0 Direc7on: Radially In Electricity & Magne7sm Lecture 7, Slide 23

28 a 4 3 a 2 a 1 metal r < a 2 : E(r) = 0 Q metal since Q enclosed = 0 cross-sec7on Q Q (2πε 0 a 2 L) Calculation A capacitor is constructed from two conduc7ng 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? a 2 < r < a 3 : What is V? The poten7al difference between the conductors. What is the sign of V = V outer V inner? A) V outer V inner < 0 B) V outer V inner = 0 C) V outer V inner > 0 Electricity & Magne7sm Lecture 7, Slide 24

29 a 4 3 a 2 a 1 Q metal metal cross-sec7on Q What is V V outer V inner? Calculation A capacitor is constructed from two conduc7ng 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? a 2 < r < a 3 : Q (2πε 0 a 2 L) (A) (B) (C) (D) V propor7onal to Q, as promised C Q V = 2 0L ln(a 3 /a 2 ) Electricity & Magne7sm Lecture 7, Slide 25

30 The Potential of a Ring of Charge (Example 29.11) A thin, uniformly charged ring of radius R has total charge Q. Find the potential at distance z on the axis of the ring. The distance r i is r i = p R 2 z 2 The potential is then the sum: NX 1 Q V = = r i 4 0 R 2 z 2 i=1 NX i=1 Q = Q R 2 z 2 Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring / 19