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Lectures_Lect08

# Lectures_Lect08 - Physics 212 Lecture 8 Today's Concept...

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Unformatted text preview: Physics 212 Lecture 8 Today's Concept: Capacitors Capacitors in a circuits, Dielectrics, Energy in capacitors Physics 212 Lecture 8, Slide 1 Simple Capacitor Circuit Q V C Q +Q Q V C Q=VC 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 2 Dielectrics V C0 Q0=VC0 V C1=C0 Q1=VC1 By adding a dielectric you are just making a new capacitor with larger capacitance (factor of ) 11 Physics 212 Lecture 8, Slide 3 Qtotal Parallel Capacitor Circuit C2 V C1 Q1 = C1V Qtotal 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 4 Series Capacitor Circuit Q Q=VCtotal Q +Q Q +Q Q V V V1 V2 C1 C2 Q Q Q Key point: Q is the same for both capacitors Key point: Q = VCtotal = V1C1 = V2C2 Also: V = V1 + V2 17 Q/Ctotal = Q/C1 + Q/C2 Physics 212 Lecture 8, Slide 5 1 1 1 = + C1 C2 Ctotal Which has lowest total capacitance: C C Checkpoint 1 C C C Ctotal = C 1/Ctotal = 1/C + 1/C = 2/C Ctotal = C/2 Ctotal = 2C 18 Physics 212 Lecture 8, Slide 6 Which has lowest total capacitance: Checkpoint 2 C C C C C Cleft = C/2 Ctotal = C Ctotal = Cleft + Cright Same Cright = C/2 Ctotal = C 20 Physics 212 Lecture 8, Slide 7 Similar to Checkpoint 3 C2 Q2 V0 V1 C1 V2 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 8 A circuit consists of three unequal capacitors C1, C2, and C3 which are connected to a battery of voltage V0. The capacitance of C2 is twice that of C1. The capacitance of C3 is three times that of C1. The capacitors obtain charges Q1, Q2, and Q3. Checkpoint 3 X Compare Q1, Q2, and Q3. A. Q1 > Q3 > Q2 B. Q1 > Q2 > Q3 C. Q1 > Q2 = Q3 D. Q1 = Q2 = Q3 E. Q1 < Q2 = Q3 X 1. See immediately: Q2 = Q3 (capacitors in series) 2. How about Q1 vs. Q2 and Q3? Calculate C23 first. 1 1 1 1 1 5 6 = + = + = C23 = C1 C23 C2 C3 2C1 3C1 6C1 5 Q1 = C1V0 6 Q23 = Q2 = Q3 = C23V0 = C1V0 5 25 Physics 212 Lecture 8, Slide 9 Energy in a Capacitor In Prelecture 7 we calculated the work done to move charge Q from one plate to another: 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 10 Messing w/ Capacitors If connected to a battery V stays constant V1 = V Q1 = C1V1 = CV = Q V C1 = C If isolated then total Q stays constant Q1 = Q C1 = C V1 = Q1/C1 = Q/C = V / Physics 212 Lecture 8, Slide 11 Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C2 has been charged and disconnected, it is filled with a dielectric. Checkpoint 4a Compare the voltages of the two capacitors. A. V1 > V2 B. V1 = V2 C. V1 < V2 Consider first the effect on the capacitance, then, does Q or V stay constant? 33 Physics 212 Lecture 8, Slide 12 Messing w/ Capacitors 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. V C0 V C1 =2 A) U1 < U0 B) U0 = U1 Compare using U = 1/2CV2 C) U1 > U0 U1/U0 = 35 Potential Energy goes UP Physics 212 Lecture 8, Slide 13 Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C2 has been charged and disconnected, it is filled with a dielectric. Checkpoint 4b Compare the potential energy stored by the two capacitors. A. U1 > U2 B. U1 = U2 C. U1 < U2 Consider first the effect on the capacitance, then, does Q or V stay constant? 33 Physics 212 Lecture 8, Slide 14 Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C2 has been charged and disconnected, it is filled with a dielectric. Checkpoint 4c A. The charges will flow so that the charge on C1 will become equal to the charge on C2. B. The charges will flow so that the energy stored in C1 will become equal to the energy stored in C2. C. The charges will flow so that the potential difference across C1 will become the same as the potential difference across C2. D. No charges will flow. The charge on the capacitors will remain what it was before they were connected. V must be the same !! Q: Q1 Q2 = C1 C2 C Q1 = 1 Q2 C2 U1 = 1 C1V 2 2 U: U 2 = 1 C2V 2 2 U1 = C1 U2 C2 Physics 212 Lecture 8, Slide 15 Calculation V x0 C0 V x0/4 An airgap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. A dielectric () of width x0/4 is inserted into the gap as shown. Conceptual Analysis: Q C V What is Qf, the final charge on the capacitor? What changes when the dielectric added? (A) Only C (B) only Q (C) only V (D) C and Q (E) V and Q (A) (B) (C) (D) (E) Adding dielectric changes the physical capacitor V does not change and C changes 38 C changes Q changes Physics 212 Lecture 8, Slide 16 Calculation V x0 C0 V x0/4 An airgap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. A dielectric () 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 Vright (A) Vleft < Vright (B) Vleft = Vright (C) Vleft > Vright (A) (B) (C) 40 The conducting plate is an equipotential !! Physics 212 Lecture 8, Slide 17 Calculation V x0 An airgap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. C0 V x0/4 A dielectric () of width x0/4 i 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 What is C1 ? = C1 C2 (A) C1 = C0 (B) C1 = 3/4C0 (C) C1 = 4/3C0 (D) C1 = 1/4C0 (A) (B) (C) (D) In general. For parallel plate capacitor: C = 0 A/d In general. For parallel plate capacitor: C = A = 3/4A0 C1 = 3/4 (0A0/d0) d = d0 C1 = 3/4C0 Physics 212 Lecture 8, Slide 18 43 Calculation V x0 C0 V An airgap capacitor, having capacitance C0 and width x0 connected to a battery of voltage V. x0/4 A dielectric () of width x0/4 inserted into the gap as show What is Qf, the final charge o the capacitor? What is C2 ? = C1 C2 C1 = 3/4C0 (A) C2 = C0 (B) C2 = 3/4 C0 (C) C2 = 4/3 C0 (D) C2 = 1/4 C0 (B) (C) (D) In general. For parallel plate capacitor filled with dielectric: C = 0 A/d A = 1/4A0 d = d0 45 C = (0A0/d0) C2 = 1/4 C0 Physics 212 Lecture 8, Slide 19 Calculation V x0 C0 An airgap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. V x0/4 A dielectric () of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? C2 = 1/4 C0 What is C? C = C1 C2 C1 = 3/4C0 (A) (B) (C) C 2 C = C1 + C2 (B) C = C 1 (C) + 1 1 C = + C 1 C2 -1 C = parallel combination of C1 and C2: C = C1 + C2 C = C0 (3/4 + 1/4 ) 46 Physics 212 Lecture 8, Slide 20 Calculation V x0 An airgap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. C0 V x0/4 A dielectric () of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? C2 = 1/4 C0 C = C1 C2 C1 = 3/4C0 C = C0 (3/4 + 1/4 ) What is Q? C Q V Q = VC 3 1 Q f = VC 0 + 4 4 50 Physics 212 Lecture 8, Slide 21 Different Problem V Q 0 x0 C0 V An airgap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V and then battery is disconnected. 1/4x0 A dielectric () of width 1/4x0 is inserted into the gap as shown. What is Vf, the final voltage on the capacitor? (A) Vf < V (B) Vf = V (C) Vf > V (B) (C) Q stays same: no way to add or subtract We know C: (property of capacitor) Q = Q0 = C0V C = C0 (3/4 + 1/4 ) Vf = Q/C = V/(3/4 + 1/4 ) Physics 212 Lecture 8, Slide 22 Different Problem V Q0 x0 C0 V An airgap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V and then battery is disconnected. 1/4x0 A dielectric () of width 1/4x0 is inserted into the gap as shown. What is Vf, the final voltage on the capacitor? Vf = Q/C = V/(3/4 + 1/4 ) How did energy stored in capacitor change when dielectric inserted? (A) U increased (B) U stayed same (C) U decreased (B) (C) U = Q2/C Q remained same C increased U decreased Physics 212 Lecture 8, Slide 23 ...
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