Unformatted text preview: Physics 212
Lecture 7
Today's Concept: Conductors and Capacitance
Go to Gradebook and choose Conflict Exam (5:15) if desired
50 40 30 20 10 0 Exam 1 in 8 days !! Confused Avg = 3.0 Confident Physics 212 Lecture 7, Slide 1 Physics Your comments
• The relationship between F,E,W,U,V and C. I feel like to mess them up sometimes. (Hey—what about Q?) • The part where an uncharged plate is added between two charges, like in the last question • I would like to understand the concept of capacitors better • PZombies, and the BanachTarski paradox. (by JDK) • Will there be an exam review? YES • What is a capacitor? What does it do? Also, how do you know the sign of potential? I had problems knowing the sign when doing the homework. • Please go over everything, I feel that I'm getting way too confused now. That exam is right around the corner =/ • How potential is affected by changing the shape of a conductor.
Physics 212 Lecture 7, Slide 2 Physics Capacitors are devices that store charge and energy: Consist of two metal plates (geometry can differ) Q C≡ V This is the definition of capacitance. This ratio will NOT depend on Q or V, but only on the GEOMETRY of the capacitor. This means that different geometries of plates will have different values of C. Physics 212 Lecture 7, Slide 3 Physics Conductors
You did well on the questions on charge distributions on 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 4 Physics Preflight 2 70 60 50 40 30 20 10 0 “Since V=kq/r, sphere A has a larger radius, and therefore has a smaller potential inversely proportional to the larger radius” (MRD)
Physics 212 Lecture 7, Slide 5 Physics 6 Preflight 4 100 80 60 40 20 0 “The potentials will be equal because the wire makes the two spheres into one conductor.” (AMS) 7 Physics 212 Lecture 7, Slide 6 Physics Preflight 6 80 60 40 20 0 “The lower potential at A causes charges from the higher potential from B to flow into A.” (CFL) 8 Physics 212 Lecture 7, Slide 7 Physics Preflight 8 A small positive charge is placed near a conducting sphere with net negative charge, Q. Which of the following figures best represents the distribution of distribution excess net negative charge as viewed from the center crosssection of the crosssphere.
70 • When +q moved near to conducting sphere, charges 60 50 will move 40 • How? • Negative charges in conductor will move toward +q 30 • What determines the final distribution? 20 • total E must be equal to zero in conductor 10 • C and D are nonstarters…. 0 • A and B better, but excess charge must be on surface
Physics 212 Lecture 7, Slide 8 Physics Preflight 10
BB THIS ONE WAS TOUGH THE PLAN: • I’ll show you some explanations • We’ll vote • We’ll do some work • We’ll vote again
“Since the distance between plates decreases and the potential remains the same, then the charge increases, so Q1>Q0. Therefore, since the potential remains the same but the charge increases, the capacitance also increases, so C1>C0. “ “Capacitance is only dependent on the geometry of the planes of the capacitator, therefore if the distance separating them and the area they cover are exactly the same, they will have the same capacitance.” “Capacitance will decrease because the distance over which the electric field acts will decrease” Physics 212 Lecture 7, Slide 9 Physics Capacitance
Capacitance is defined for any pair of spatially separated conductors How do we understand this definition ??? • Consider two conductors, one with excess charge = +Q and the other with excess charge = Q Q C≡ V +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 10 Physics Example (done in Prelecture 7) Example
First determine E field produced by charged conductors:
y +Q d
x E Q σ E= ε0 What is σ ?? ?? A = area of plate Q σ= A Second, integrate E to find the potential difference V dr d d Q r V = − ∫ E ⋅ dy V = − ∫ (− Edy ) =E ∫ dy = d ε0 A 0 0 0 As promised, V is proportional to Q !! 12 Q Q C≡ = V Qd /(ε 0 A) C= ε0 A
d C determined by geometry !!
Physics 212 Lecture 7, Slide 11 Physics Question Related to Preflight
+Q0
Initial charge on capacitor = Q0 d Q0 +Q1 BB Insert uncharged conductor into isolated capacitor Charge on capacitor now = Q1 How is Q1 related to Q0 ?? A. Q1 < Q0 B. Q1 = Q0 C. Q1 > Q0
14 d Q1 t Plates not connected to anything—different from Prefl 10 CHARGE CANNOT CHANGE !!
Physics 212 Lecture 7, Slide 12 Physics Where to Start??
+Q0 d Q0
What is the total charge induced on the bottom surface of the conductor? t BB A. B. C. D. E. +Q0 Q0 0 Positive but the magnitude unknown Negative but the magnitude unknown 17 Physics 212 Lecture 7, Slide 13 Physics WHY ??
+Q0 Q0 E=0 +Q0 Q0
WHAT DO WE KNOW ??? E must be = 0 in conductor !! E E Charges inside conductor move to cancel E field from top & bottom plates 19 Physics 212 Lecture 7, Slide 14 Physics Calculate V
Now calculate V as a function of distance as from the bottom conductor. conductor +Q0 +Q E
E0 rr V ( y ) = − ∫ E ⋅ dy
y 0 BB d t y y t V Q0 What is ΔV = V(d)? A) ΔV = E0d B) ΔV = E0(d – t) C) ΔV = E0(d + t) y
The integral = area under the curve
Physics 212 Lecture 7, Slide 15 Physics d E=0 21 Back to Preflight 10
BB What have we learned? C0 = Q0/V0 = ε0A/d If charges are the same Q1 = Q0 V0 = E0d V1 = E0(d – t)
50 40 What do these results tell us about how C1 is related to C0 ?? is Preflight Results 30 20 10 0 Physics 212 Lecture 7, Slide 16 Physics Back to Preflight 10
BB Same V: What is Q1 interms of Q0? Leave as exercise! We can determine C from either case same V (preflight) same Q (lecture) V0 = E0d C depends only on geometry !!
Same Q: V1 = E0(d – t) C0 = Q0/E0d E0 = Q0/ε0A C0 = ε0A/d C1 = Q0/(E0(d – t)) C1 = ε0A/(d – t)
Physics 212 Lecture 7, Slide 17 Physics Energy in Capacitors BANG 31 Physics 212 Lecture 7, Slide 18 Physics crosssection cross
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). metal What is the capacitance C of this device ? metal • Conceptual Analysis: Q C≡ But what is Q and what is V? They are not given?? 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 19 Physics crosssection
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). metal metal What is the capacitance C of this capacitor ? 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) Note: Many of you email me and ask “prof—what equation should I use for homework problem 3?” I say: Ppttuy, yukk! Never ask which equation! Ask what concepts should I apply and equations will become obvious. Physics 212 Lecture 7, Slide 20 Physics crosssection + + + +
metal metal Calculation
A capacitor is constructed from two BB conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this capacitor ? Q C≡ V
+ + + + + + + + + Where is +Q on outer conductor located? (B) at r=a3 (C) both surfaces (A) at r=a4
rrQ Gauss’ law: ∫ E dA = Gauss
Why?
enclosed We know that E = 0 in conductor (between a3 and a4) + + Q + a2 a1 + a4 a3 +Q + + (D) throughout shell ε 0 Q enclosed =0 Q enclosed =0 +Q must be on inside surface (a3), so that Qenclosed = + Q – Q = 0
Physics 212 Lecture 7, Slide 21 Physics crosssection + + +
metal Calculation
A capacitor is constructed from two BB conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this capacitor ? Q C≡ V
+ + + Q −+ − + − − + + − − metal − − − + + +++ Where is Q on inner conductor located? (B) at r=a1 (C) both surfaces (A) at r=a2
rrQ Gauss’ law: ∫ E dA = Gauss
Why?
enclosed We know that E = 0 in conductor (between a1 and a2) − − − −−− − −+ −+ a2 a1 + a4 a3 +Q + (D) throughout shell ε 0 Q enclosed =0 Q enclosed =0 Q must be on outside surface (a2), so that Qenclosed = 0
Physics 212 Lecture 7, Slide 22 Physics crosssection + + +
metal Calculation
A capacitor is constructed from two conducting BB cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this capacitor ?
+
0 Q −+ − + − − + + − − metal − − − + + +++ + a2 < r < a3: What is E(r)? (A) 0
Why? 1Q (B) 4πε r
2 rrQ Gauss’ law: ∫ E dA = Gauss − − − −−− − −+ −+ a2 a1 + + a4 a3 +Q + C≡ Q V (C) Q 2πε Lr
0 1 1 2Q (D) 2πε Lr
0 1Q (E) 4πε r
2 0 rrQ ∫ E dA =
0 ε enclosed ε enclosed 0 E= E 2πrL = Q Q 2πε Lr
0 1 ε0 Direction: Radially In
Physics 212 Lecture 7, Slide 23 Physics crosssection + + +
metal Calculation
A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and BB 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? (B) VouterVinner = 0 (A) VouterVinner < 0 − − − −−− − −+ −+ a2 a1 + a4 a3 +Q + C≡ Q V a2 < r < a3: E = Q 2πε Lr
0 1 Q (2πε0a2L) (C) VouterVinner > 0
Physics 212 Lecture 7, Slide 24 Physics crosssection + + +
metal
1 4 Calculation
A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and BB length L (L >> ai). What is the capacitance C of this capacitor ?
+ + Q −+ − + − − + + − − metal − − − + + +++ + What is V ≡ Vouter  Vinner? Q a Q a V= ln V= ln 2πε L a 2πε L a
0 0 (A)
Q (2πε0a2L) − − − −−− − −+ −+ a2 a1 + a4 a3 +Q + C≡ Q V a2 < r < a3: E = 1Q 2πε r
0 4 V= Q 2πε L
0 ln 1 a a 3 V= Q 2πε L (D)
0 ln 2 a a 2 3 (B)
a3 (C)
−Q dr V = −∫ 2πε L r
a2 0 V= dr ∫ 2πε L r
a3 0 a2 Q V= Q 2πε L
0 ln a a 3 2 V proportional to Q, as promised C≡ Q 2πε L = V ln(a / a )
0 3 2 Physics 212 Lecture 7, Slide 25 Physics ...
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 Spring '08
 Kim
 Physics, Capacitance, Electric charge

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