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**Unformatted text preview: **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 E at surface perpendicular to surface 5 Physics 212 Lecture 7, Slide 2 Checkpoint 1a
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 1b 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 1c 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 Two parallel plates of equal area carry equal and opposite charge Q0. The potential difference between the two plates is measured to be V0. 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 Q1 such that the potential difference between the plates remains the same. Parallel Plate Capacitor Physics 212 Lecture 7, Slide 6 Capacitance is defined for any pair of spatially separated conductors Capacitance
Q V C
How do we understand this definition ??? Consider two conductors, one with excess charge = +Q and the other with excess charge = Q +Q d -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 conductor 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 E V Physics 212 Lecture 7, Slide 7 Example (done in Prelecture 7)
First determine E field produced by charged conductors: y +Q d
x E -Q E= o What is ?? Q = A A = area of plate Second, integrate E to find the potential difference V V = - E dy
0 d V = - (- Edy) = E dy =
0 0 d d Q d o A As promised, V is proportional to Q !! 12 Q Q C = V Qd / o A 0 A C= d C determined by geometry !!
Physics 212 Lecture 7, Slide 8 Question Related to Checkpoint 2
Initial charge on capacitor = Q0 +Q0 d -Q0 +Q1 Insert uncharged conductor Charge on capacitor now = Q1 How is Q1 related to Q0 ?? A. B. C.
14 d -Q1 t Q1 < Q0 Q1 = Q0 Q1 > Q0 Plates not connected to anything CHARGE CANNOT CHANGE !!
Physics 212 Lecture 7, Slide 9 Where to Start??
+Q0 d -Q0
What is the total charge induced on the bottom surface of the conductor? t A. B. C. D. E. +Q0 Q0 0 Positive but the magnitude unknown Negative but the magnitude unknown 17 Physics 212 Lecture 7, Slide 10 WHY ??
+Q0 -Q0 +Q0 E E=0 E -Q0
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 11 Now calculate V as a function of distance from the bottom ( y ) = - E dy V
conductor. conductor d +Q0 y t V -Q0 What is V = V(d)? A) V = E0d B) V = E0(d t) C) V = E (d + t) y
The integral = area under the curve
Physics 212 Lecture 7, Slide 12 Calculate V y 0 E
-E0 t y d E=0 21 Two parallel plates of equal area carry equal and opposite charge Q0. The potential difference between the two plates is measured to be V0. 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 Q1 such that the potential difference between the plates remains the same. Back to Checkpoint 2a A) Q1 < Qo B) Q1 = Qo C) Q1 > Qo Physics 212 Lecture 7, Slide 13 Two parallel plates of equal area carry equal and opposite charge Q0. The potential difference between the two plates is measured to be V0. 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 Q1 such that the potential difference between the plates remains the same. What happens to C1 relative to C0? Checkpoint 2b A) C1 > Co B) C1 = Co C) C1 < Co We can determine C from either case same V (Checkpoint) same Q (Prelecture) C depends only on geometry !! Same V: V0 = E0d V0 = E1(d t) C0 = Q0/E0d C1 = Q1/(E1(d t)) E = Q/0A
C0 = 0A/d
Physics 212 Lecture 7, Slide 14 C1 = 0A/(d t) Energy in Capacitors BANG 31 Physics 212 Lecture 7, Slide 15 cross-section
a4 a3 a2 a1 Calculation
A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this device ? metal metal Conceptual Analysis: Q C But what is Q a n d wh a t is V? T h e y a re no t g ive n? ? V 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 16 cross-section
a4 a3 a2 a1 Calculation
A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this capacitor ? metal metal Q C V 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 17 cross-section + + + +
metal metal Calculation
A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this capacitor ?
+ + + + + + + + + Where is +Q on outer conductor located? (A) at r=a4 (B) at r=a3 (C) both surfaces (D) throughout shell (A) (B) (C) (D) Why? Gauss' law: We know that E = 0 in conductor (between a3 and a4) Qenclosed E dA = o + + Q + a2 a1 + a4 a3 +Q + + Q C V Qenclosed = 0
+Q must be on inside surface (a3), so that Qenclosed = + Q Q = 0
Physics 212 Lecture 7, Slide 18 Qenclosed = 0 cross-section + + +
metal Calculation
A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this capacitor ?
+ + + Q - - + + - - + + - - metal - - - + + + + + Where is Q on inner conductor located? (A) at r=a2 (B) at r=a1 (C) both surfaces (D) throughout shell (A) (B) (C) (D) Why? Gauss' law: We know that E = 0 in conductor (between a1 and a2) Qenclosed E dA = o - - - - - + + a2 a1 - - - - + a4 a3 +Q + Q C V Qenclosed = 0
+Q must be on outer surface (a2), so that Qenclosed = 0
Physics 212 Lecture 7, Slide 19 Qenclosed = 0 cross-section + + +
metal Calculation
A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this capacitor ?
+ + + Q - - + + - - + + - - metal - - - + + + + + a2 < r < a3: What is E(r)?
1 2Q 1 Q 1 Q (A) 0 (B) (C) (D) (E) 2 Lr (A) 2o Lr 4o r 2 o
Why? Gauss' law: Qenclosed E dA = o - - - - - + + a2 a1 - - - - + a4 a3 +Q + Q C V
1 2Q 4o r 2 Q E 2rL = o E= 1 Q 2 Lr
0 Direction: Radially In
Physics 212 Lecture 7, Slide 20 cross-section + + +
metal Calculation
A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this capacitor ?
+ + + Q - - + + - - + + - - metal - - - + + + + + r < a2: E(r) = 0 since Qenclosed = 0 What is V? The potential difference between the conductors What is the sign of V = Vouter Vinner? (A) VouterVinner < 0 (B) VouterVinner = 0 (C) VouterVinner > 0 (A) (B) (C) Physics 212 Lecture 7, Slide 21 - - - - - + + a2 a1 - - - - + a4 a3 +Q + Q C V
Q (20a2L) a2 < r < a3: 1 Q E= 2 Lr
0 cross-section + + +
metal Calculation
A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this capacitor ?
+ + + Q - - + + - - + + - - metal - - - + + + + + What is V Vouter Vinner?
Q ln 2o L a1 a4 Q ln 2o L (A) (B) (C) (D) Q (20a2L) - - - - - + + a2 a1 - - - - + a4 a3 +Q + Q C V
a4 a1 a2 < r < a3: 1 Q E= 2o rL
a2 a3 Q ln 2o L a3 a2 Q ln 2o L - Q dr V =- a 2o L r
2 a3 Q a3 dr V= 2o L a r
2 V= Q a ln 2 L a
0 3 2 V proportional to Q, as promised C Q 2 L = V ln(a / a )
0 3 2 Physics 212 Lecture 7, Slide 22 ...

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