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

Physics 212 Lecture 7, Slide 2 Conductors 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 5 The Main Points Cavity inside a conductor: E = 0, unaffected by fields and charges outside.

Physics 212 Lecture 7, Slide 3 Checkpoint 1a 6 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. V A > V B B. V A = V B C. V A < V B

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

Physics 212 Lecture 7, Slide 5 Checkpoint 1c 8 What happens to the charge on conductor A after it is connected to conductor B by the wire? A. Q A increasesB. Q A decreases C. Q A doesn’t change

Physics 212 Lecture 8, Slide 6 C V +Q+Q+Q+Q-Q Q 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. C V Q=VC Charging a Capacitor 8

Physics 212 Lecture 7, Slide 7 Capacitance Capacitance is defined for any pair of spatially separated conductors 9 How do we understand this definition ? +Q -Q d Consider two conductors, one with excess charge = +Q Consider two conductors, one with excess charge = +Q and the other with excess charge = -Q and the other with excess charge = -Q These charges create an electric field in the space between them These charges create an electric field in the space between them E This potential difference should be proportional to Q ! 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 The ratio of Q to the potential difference is the capacitance and only depends on the geometry of the conductors V We can integrate the electric field between them to find the potential difference between the conductors. We can integrate the electric field between them to find the potential difference between the conductors. Unit is the Farad

Physics 212 Lecture 7, Slide 8 First determine E field produced by charged conductors: Parallel plate capacitance 12+Q-Q d E Second, integrate E to find the potential difference V xy As promised, V is proportional to Q. C determined by geometry ! A = area of plate

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

Physics 212 Lecture 7, Slide 10 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 11 A.+Q 0 B.-Q 0 C.0 D.Positive but the magnitude unknown E.Negative but the magnitude unknown 17 t d +Q 0 -Q 0 What is the total charge induced on the bottom surface of the conductor? Where to Start??

Physics 212 Lecture 7, Slide 12 E +Q 0 -Q 0 E WHAT DO WE KNOW ? +Q 0 -Q 0 19 E = 0 E must be = 0 in conductor ! Charges inside conductor move to cancel E field from top & bottom plates WHY ?

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

Physics 212 Lecture 7, Slide 14 Back to Checkpoint 2a A) Q 1 < Q o B) Q 1 = Q o C) Q 1 > Q o 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 15 Checkpoint 2b A) C 1 > C o B) C 1 = C o C) C 1 < C o 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)) C 0 =  0 A/d C 1 =  0 A/(d – t) E = Q/  0 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. What happens to C 1 relative to C 0 ? We store more charge, Q 1 = C 1 V 0 > Q 0 = C 0 V 0 for the same voltage difference.

Physics 212 Lecture 7, Slide Energy in Capacitors

Physics 212 Lecture 7, Slide 17Calculation 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 device ? metal a1a1 a2a2 a3a3 a4a4 Conceptual Analysis: cross-section 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 18Calculation 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 ? metal +Q a1a1 a2a2 a3a3 a4a4 cross-section -Q Why? Where is +Q on outer conductor located? (A) (B) (C) (D) (A) at r=a 4 (B) at r=a 3 (C) both surfaces (D) throughout shell We know that E = 0 in conductor (between a 3 and a 4 ) Gauss’ law: +Q must be on inside surface (a 3 ), so that Q enclosed = + Q – Q = 0

Physics 212 Lecture 7, Slide 19Calculation 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 ? metal +Q a1a1 a2a2 a3a3 a4a4 cross-section -Q Why? Where is -Q on inner conductor located? (A) (B) (C) (D) (A) at r=a 2 (B) at r=a 1 (C) both surfaces (D) throughout shell  metal                   We know that E = 0 in conductor (between a 1 and a 2 ) Gauss’ law: +Q must be on outer surface (a 2 ), so that Q enclosed = 0

Physics 212 Lecture 7, Slide 20Calculation 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 ? metal +Q a1a1 a2a2 a3a3 a4a4 cross-section -Q  metal                   Why? Direction: Radially In Gauss’ law: a 2 < r < a 3 : What is E(r)? (A) (B) (C) (D) (E) (A) 0 (B) (C) (D) (E)

Physics 212 Lecture 7, Slide 21Calculation 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 ? metal +Q a1a1 a2a2 a3a3 a4a4 cross-section -Q r < a 2 : E(r) = 0 since Q enclosed =  metal                   a 2 < r < a 3 : What is V? What is V? The potential difference between the conductors The potential difference between the conductors What is the sign of V = V outer - V inner ? (A) (B) (C) (A) V outer -V inner 0 Q (2  0 a 2 L )

What is V  V outer - V inner ? (A) (B) (C) (D) (A) (B) (C) (D) Physics 212 Lecture 7, Slide 22Calculation 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 ? metal +Q a1a1 a2a2 a3a3 a4a4 cross-section -Q  metal                   a 2 < r < a 3 : Q (2  0 a 2 L ) V proportional to Q, as promised