Lect08 - Physics 212 Lecture 8 Today's Concept Capacitors...

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Unformatted text preview: Physics 212 Lecture 8 Today's Concept: Capacitors Capacitors in a circuits, Dielectrics, Energy in capacitors 40 30 20 10 0 Confused Avg = 3.4 Confident Physics 212 Lecture 8, Slide 1 Music Who is the Artist? BB A) B) C) D) E) Michael McDonald Van Morrison Glenn Frey Robert Palmer David Bowie My favorite Van the Man album… Many of Many my friends agree that this one is on the list of 10 albums you must have on a desert island.. What does that say about my friends?? does Did you know he wrote “Gloria”.. He did! Did From Last time: Really.. Incredible “Send in the Clowns” By Van the Man A Classic, Of Course ! Physics 212 Lecture 8, Slide 2 Physics 212 Lecture 8 Today's Concept: Capacitors Capacitors in a circuits, Dielectrics, Energy in capacitors 40 30 20 10 0 Confused Avg = 3.4 Confident Physics 212 Lecture 8, Slide 3 Some Exam Stuff • Exam next Wednesday at 7:00 – – – – Covers material through lecture 8 Bring your ID Conflict exam at 5:15 – sign up in your gradebook if you need to If you have conflicts with both of these contact us immediately online.physics.uiuc.edu/courses/phys212/fall09/ExamPrepHe1.html Physics 212 Lecture 8, Slide 4 Simple Capacitor Circuit Q V C +Q -Q V C Q=VC Q This “Q” really means that the battery has This really moved charge Q from one plate to the other, moved so that one plate holds +Q and the other -Q. so 8 Physics 212 Lecture 8, Slide 5 Dielectrics V C0 Q0=VC0 V C1=κC0 Q1=VC1 By adding a dielectric you are just making a By new capacitor with larger capacitance (factor of κ) new 11 Physics 212 Lecture 8, Slide 6 Qtotal Parallel Capacitor Circuit V C1 Q1 = C1V Qtotal C2 Q2 = C2V Key point: V is the same for both capacitors Key Point: Qtotal = Q1 + Q2 = VC1 + VC2 = V(C1 + C2) Ctotal = C1 + C2 14 Physics 212 Lecture 8, Slide 7 Series Capacitor Circuit Q Q=VCtotal +Q -Q V V1 C1 Q +Q -Q V V2 Q C2 Q Q Key point: Q is the same for both capacitors Key point: Q = VCtotal = V1C1 = V2C2 VC Also: V = V1 + V2 Q/Ctotal = Q/C1 + Q/C2 17 1 1 1 = + C2 Ctotal C1 Physics 212 Lecture 8, Slide 8 Preflight 2 Which has lowest total capacitance: C C C C C Ctotal = C 1/Ctotal = 1/C + 1/C = 2/C Ctotal = C/2 70 60 50 40 30 20 10 Ctotal = 2C 18 0 Physics 212 Lecture 8, Slide 9 Preflight 4 Which has lowest total capacitance: C C C C C Cleft = C/2 Ctotal = C Ctotal = Cleft + Cright 70 60 40 30 20 10 0 Cright = C/2 Same 50 Ctotal = C 20 Physics 212 Lecture 8, Slide 10 10 Similar to Preflight 6 Q2 C2 BB V0 V1 V2 C1 Q1 V3 C3 Q3 Ctotal Which of the following is NOT necessarily true: A) B) C) D) E) 24 V0 = V1 Ctotal > C1 V2 = V3 Q2 = Q3 V1 = V2 + V3 Physics 212 Lecture 8, Slide 11 11 Preflight 6 BB NO YES NO YES NO YES V1 = V2 + V3 Q2 = Q3 V1 > V2 Q1 Q2 Q2 = + C1 C2 C3 Q1 = Q2 V1 < V2 Q2 = C2C3 Q1 C1 (C2 + C3 ) Q1 = Q2 V1 < V2 Q1 = Q2 V1 < V2 25 Q1 = Q2 V1 < V2 (A) (B) (C) (D) Physics 212 Lecture 8, Slide 12 12 Energy in a Capacitor In pre-llecture 7 we calculated the work done to move charge Q from one ecture plate to another: plate C V +Q U = 1/2QV = 1/2CV2 =1/2Q2/C -Q Since Q = VC This is potential energy waiting to be used… 26 Physics 212 Lecture 8, Slide 13 13 Messing w/ Capacitors • If connected to a battery V stays constant V1 = V V κ C1 = κ C Q1 = C1V1 = κ CV = κQ • If isolated then total Q stays constant κ Q1 = Q C1 = κ C V1 = Q1/C1 = Q/κC = V / κ Physics 212 Lecture 8, Slide 14 14 Preflight 8 BB 60 “Q is fixed in this case and we have just increased the capacitance on C2. Therefore, the voltage difference on C2 must be lower.” “Dielectrics increase Capacitance. They do not increase or decrease potential difference” “The dielectric increases the energy stored so it must also increase the voltage” 33 50 40 30 20 10 0 Physics 212 Lecture 8, Slide 15 15 Messing w/ Capacitors ACT Two identical parallel plate capacitors are connected to identical batteries. Then a dielectric is inserted between the plates of capacitor C1. Compare the energy stored in the two capacitors. BB BB V C0 V κ=2 C1 A) U1 < U0 B) U0 = U1 Compare using U = 1/2CV2 C) U1 > U0 U1/U0 = κ 35 Potential Energy goes UP Potential UP Physics 212 Lecture 8, Slide 16 16 Preflight 10 BB A B C 50 40 30 20 10 0 “We know that U = 0.5*(Q^2)/C as in our case Q did not We change and C has increased in C2 due to the dielectric, therefore U1>U2” therefore “Charge doesn't change so neither does U. .” “When a dielectic iis placed in a capacitor, the When dielectic s capacitance increases, which means that the potential energy increases” potential 33 Physics 212 Lecture 8, Slide 17 17 Preflight 12 BB V must be the same ! Q: 50 40 Q1 Q2 = C1 C2 C1 Q1 = Q2 C2 U1 = C1 U2 C2 30 20 10 0 U1 = 1 C1V 2 2 U: U 2 = 1 C2V 2 2 Physics 212 Lecture 8, Slide 18 18 Calculation V C0 x0 V κ x0/4 BB having capacitance C 0 An air-gap capacitor, and width x0 is connected to a battery of voltage V. A dielectric (κ) of width x0/4 is inserted into the gap as shown. What is Qf, the final • Conceptual Analysis: charge on the Q capacitor? C≡ V What changes when the dielectric added? (A) Only C (B) only Q (B) (C) only V (C) (D) C and Q (E) V and Q C changes Q changes Adding dielectric changes the physical capacitor V does not change and C changes 38 Physics 212 Lecture 8, Slide 19 19 Calculation V C0 x0 V κ x0/4 BB An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. A dielectric (κ) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? • Strategic Analysis: – – Calculate new capacitance C Apply definition of capacitance to determine Q To calculate C, let’s first look at: Vleft κ Vright (A) Vleft < Vright (A) (B) Vleft = Vright (B) (C) Vleft > Vright (C) Physics 212 Lecture 8, Slide 20 20 The conducting plate is an equipotential !! 40 Calculation V C0 x0 An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. V κ x0/4 BB A dielectric (κ) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? Can consider capacitor to be two capacitances, C1 and C2, in parallel κ What is C1 ? = C1 κ C2 (A) C1 = C0 (B) C1 = 3/4C0 (A) (B) (C) C1 = 4/3C0 (C) (D) C1 = 1/4C0 (D) 43 In general. For parallel plate capacitor: C = ε0A/d In A = 3/4A0 C1 = 3/4 (ε0A0/d0) d = d0 C1 = 3/4C0 Physics 212 Lecture 8, Slide 21 21 Calculation V C0 x0 V An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. κ x0/4 BB A dielectric (κ) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? κ What is C2 ? = C1 κ C2 C1 = 3/4C0 (A)C2 = κC0 (B) C2 = 3/4 κC0 (B) (C) C2 = 4/3 κC0 (C) (D) C2 = 1/4 κC0 (D) In general. For parallel plate capacitor filled with dielectric: C = κε0A/d A = 1/4A0 d = d0 45 C = ¼(κε0A0/d0) C2 = 1/4 κC0 Physics 212 Lecture 8, Slide 22 22 Calculation V C0 x0 V An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. κ BB x0/4 A dielectric (κ) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? κ What is C? What C = C1 κ C2 C1 = 3/4C0 C2 = 1/4 κC0 (A) C = C1 + C2 (B) C = C1 + κ C2 (C) C = 1 + 1 (B) C C 1 2 C = parallel combination of C1 and C2: C = C1 + C2 −1 C = C0 (3/4 + 1/4 κ) 46 Physics 212 Lecture 8, Slide 23 23 Calculation V C0 x0 An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. V κ BB x0/4 A dielectric (κ) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? κ C = C1 κ C2 C1 = 3/4C0 C2 = 1/4 κC0 C = C0 (3/4 + 1/4 κ) What is Q? C≡ Q V Q = VC 3 1 Q f = VC0 + κ 4 4 50 Physics 212 Lecture 8, Slide 24 24 Different Problem V Q0 x0 C0 V κ BB BB An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V and then battery is disconnected. A dielectric (κ) of width 1/4x0 is inserted into the gap as shown. What is Vf, the final voltage on the capacitor? 1/4x0 (A) Vf < V (B) Vf = V (B) (C) Vf > V (C) Q stays same: no way to add or subtract stays We know C: (property of capacitor) We Q = Q0 = C0V C = C0 (3/4 + 1/4 κ) Vf = Q/C = V/(C0 (3/4 + 1/4 κ)) Physics 212 Lecture 8, Slide 25 25 Different Problem V Q0 x0 C0 V BB An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V and then battery is disconnected. A dielectric (κ) of width 1/4x0 is inserted into the gap as shown. What is Vf, the final voltage on the capacitor? κ 1/4x0 Vf = Q/C = V/(C0 (3/4 + 1/4 κ)) How did energy stored in capacitor How change when dielectric inserted? change (A) U increased (B) U stayed same (C) U decreased (B) (C) U = ½ Q2/C Q remained same C increased U decreased Physics 212 Lecture 8, Slide 26 26 ...
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