Physics 212 Lecture 7 Conductors and Capacitance Physics

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

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

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 6 Physics 212 Lecture 7, Slide 3

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 7 Physics 212 Lecture 7, Slide 4

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 8 Physics 212 Lecture 7, Slide 5

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

Capacitance is defined for any pair of spatially separated conductors Unit is the Farad 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 7

Parallel plate capacitance First determine E field produced by charged conductors: +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. C determined by geometry ! 12 Physics 212 Lecture 7, Slide 8

Question +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 9

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. Physics 212 Lecture 7, Slide 10

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 11

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 12

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 13

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 Physics 212 Lecture 7, Slide 14

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 store more charge, Q 1 = C 1 V 0 > Q 0 = C 0 V 0 for the same voltage difference. Same V: V 0 = E 0 d V 0 = E 1(d – t) C 0 = Q 0/E 0 d C 1 = Q 1/(E 1(d – t)) E = Q/ 0 A C 0 = 0 A/d C 1 = 0 A/(d – t) Physics 212 Lecture 7, Slide 15

Energy in Capacitors 31 Physics 212 Lecture 7, Slide 16

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? • Important Point: C is a property of the object! (concentric cylinders) • 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 17

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 18

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 19

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 20

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 p 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 21

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 p 0 a 2 L) V proportional to Q, as promised Physics 212 Lecture 7, Slide 22