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Physics 212 Lecture 8

Physics 212 Lecture 8. Today's Concept: Capacitors Capacitors in a circuits, Dielectrics, Energy in capacitors. Music. Who is the Artist? Robert Plant Lyle Lovett Alison Krauss Mark Knopfler John Prine. Mark Knopfler , OBE Dire Straits to country. Physics 212 Lecture 15.

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Physics 212 Lecture 8

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  1. Physics 212 Lecture 8 Today's Concept: Capacitors Capacitors in a circuits, Dielectrics, Energy in capacitors

  2. Music • Who is the Artist? • Robert Plant • Lyle Lovett • Alison Krauss • Mark Knopfler • John Prine • Mark Knopfler, OBE • Dire Straits to country Physics 212 Lecture 15

  3. Some Exam Stuff • Exam next Wednesday at 7:00 (Feb. 15) • Covers material through lecture 8 • Bring your ID • Conflict exam at 5:15 – sign up in your gradebook by 10:00 Mon. Feb. 13 • If you have class conflicts with both of these, please email Dr. Park (wkpark@illinois.edu) TODAY! • Where do I go? EXAM ROOMS (by discussion section) • Preparation: • Practice exams • Exam prep exercises • Worked examples • Extra Office Hours

  4. “this one was probably one of the easiest yet, but it would be useful to do more examples“ “Can we go over the last check point and go over the effects of a dielectric in more detail?“ Your Comments “Wow this was so much easier than the beginning of the course. It was nice to feel as though I actually understood something for once.” “I feel like I understood the stuff, but some of the later checkpoint questions had me a little stumped.” “I don't understand the last question in the checkpoint” We’ll summarize main points. “Had a lot of trouble understanding this stuff ” “Duuuuude Gauss' Law is soooo much easier than potential difference and capacitance!!!!! I'm lost :( Please help me, pleeeeeease!!!” Will Do.. They were the hardest WILL DO See Calculation at End “When the capacitors are placed in series is it best of 5 games or best of 7?” 05

  5. = 1/2CV2 Since Q = VC =1/2Q2/C Energy in a Capacitor “I still don't understand what voltage is exactly.” In Prelecture 7 we calculated the work done to move charge Q from one plate to another: +Q C V U = 1/2QV -Q This is potential energy waiting to be used… 26

  6. Q V +Q C Q=VC -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. Simple Capacitor Circuit V C 8

  7. Qtotal Q1 = C1V Q2 = C2V Qtotal Parallel Capacitor Circuit C2 C1 V Key point:V is the same for bothcapacitors Key Point: Qtotal = Q1 + Q2 = VC1 + VC2 = V(C1 + C2) Ctotal = C1 + C2 14

  8. Q Q=VCtotal +Q C1 -Q Q Q V1 V +Q C2 -Q Q V2 Q Q/Ctotal = Q/C1 + Q/C2 1 1 1 = + Ctotal C1 C2 Series Capacitor Circuit V Key point:Q is the same for both capacitors Key point:Q = VCtotal = V1C1 = V2C2 Also: V = V1 + V2 17

  9. Checkpoint 1 Which has lowest total capacitance: A) B) C) C C C C C 1/Ctotal = 1/C + 1/C = 2/C Ctotal = 2C Ctotal = C/2 Ctotal = C 18

  10. Checkpoint 2 Cleft = C/2 Cright = C/2 Ctotal = C Which has lowest total capacitance: A) B) C) Same C C C C C Ctotal = C Ctotal = Cleft + Cright 20

  11. Similar to Checkpoint 3 Q2 C2 V0 V2 C1 C3 V3 Q3 V1 Q1 Ctotal • Which of the following is NOT necessarily true: • V0 = V1 • Ctotal > C1 • V2 = V3 • Q2 = Q3 • V1 = V2 + V3 24

  12. Checkpoint 3 (HW 3,4) 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. Compare Q1, Q2, and Q3. A.Q1 > Q3 > Q2B.Q1 > Q2 > Q3C.Q1 > Q2 = Q3D.Q1 = Q2 = Q3E.Q1 < Q2 = Q3 “C1 receive the same amount of charges as C2 AND C3 combined. Since C3's capacity is larger than C2, more charges are stored in C3.” “Q1 comes before Q2 comes before Q3” “C1 and C2,C3 are in parallel. C2 and C3 are in series. Q1=C1V0. Q2=Q3=(3/5)C1V0” “C23's equivalent capacitance is greater than C1. Therefore, it attracts more charge.” 25

  13. Checkpoint 3 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. Compare Q1, Q2, and Q3. A.Q1 > Q3 > Q2B.Q1 > Q2 > Q3C.Q1 > Q2 = Q3D.Q1 = Q2 = Q3E.Q1 < Q2 = Q3 X X 1. See immediately: Q2 = Q3 (capacitors in series) 2. How about Q1 vs. Q2 and Q3? Calculate C23 first. 25

  14. C0 C1=kC0 V V Q0=VC0 Q1=VC1 Dielectrics By adding a dielectric you are just making a new capacitor with larger capacitance (factor ofk) 11

  15. Messing w/ Capacitors Q1 = C1V1 =kCV =kQ V V1 = Q1/C1 = Q/kC= V/k • If connected to a battery V stays constant • If isolated then total Q stays constant V1 = V C1 =kC k Q1 = Q k C1 =kC

  16. Checkpoint 4a 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. Compare the voltages of the two capacitors. A. V1 > V2B. V1 = V2C. V1 < V2 “q doesn't change but capacitance increases. q/c=v and if c is bigger v is smaller.“ “Same charge, same V.” “The have the same charge and C2 is larger than C1, so V1 must be smaller than C2.” Consider first the effect on the capacitance, then, does Q or V stay constant? 33

  17. C0 V Compare using U = 1/2CV2 U1/U0 = k Potential Energy goes UP 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. C1 V k=2 A) U1 < U0 B) U0 = U1 C) U1 > U0 35

  18. Checkpoint 4b 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. Compare the potential energy stored by the two capacitors. A. U1 > U2B. U1 = U2C. U1 < U2 “The potential energy is directly proportional to the voltage.” “independent of capacitance” “An increase in capacitance causes an increase in the potential energy stored in the capacitor.” Consider first the effect on the capacitance, then, does Q or V stay constant? 33

  19. Checkpoint 4c 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. The two capacitors are now connected to each other by wires as shown. How will the charge redistribute itself, if at all? 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. “Charges on a conductor distribute themselves evenly” “Equal energy throughout a system.” “When placed in parallel, capacitors have the same potential difference” “no charges will flow because Q is the same. ”

  20. Checkpoint 4c V must be the same !! Q: U: 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. The two capacitors are now connected to each other by wires as shown. How will the charge redistribute itself, if at all? 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.

  21. Calculation (HW 2, 5) C0 V x0 k What changes when the dielectric added? (A) Only C(B) only Q (C) only V (D) C and Q (E) V and Q C changes Adding dielectric changes the physical capacitor V does not change and C changes Q changes An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. A dielectric (k) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? V x0/4 • Conceptual Analysis: 38

  22. Calculation C0 V x0 k k To calculate C, let’s first look at: Vleft Vright V An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. A dielectric (k) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? x0/4 • Strategic Analysis: • Calculate new capacitance C • Apply definition of capacitance to determine Q (A) Vleft < Vright(B) Vleft = Vright(C) Vleft > Vright The conducting plate is an equipotential !! 40

  23. Calculation An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. A dielectric (k) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? C0 V x0 k k k C2 C1 = What is C1 ? (A) C1 = C0(B) C1 = 3/4C0(C) C1 = 4/3C0(D) C1 = 1/4C0 A = 3/4A0 C1 = 3/4C0 C1 = 3/4 (e0A0/d0) d = d0 V x0/4 Can consider capacitor to be two capacitances, C1 and C2, in parallel In general. For parallel plate capacitor: C = e0A/d 43

  24. Calculation C0 V x0 k k k What is C2 ? • C2 = kC0(B) C2 = 3/4kC0(C) C2 = 4/3 kC0(D) C2 = 1/4kC0 A = 1/4A0 C2 = 1/4kC0 C = ¼(ke0A0/d0) d = d0 An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. A dielectric (k) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? V x0/4 C2 C1 = C1 = 3/4C0 In general. For parallel plate capacitor filled with dielectric: C = ke0A/d 45

  25. Calculation C0 V x0 k k k What is C? • (B) (C) C = C0 (3/4 + 1/4k) An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. A dielectric (k) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? V x0/4 C2 C C1 = C1 = 3/4C0 C2 = 1/4kC0 C = parallel combination of C1 and C2: C = C1 + C2 46

  26. Calculation C0 V x0 k k k C = C0 (3/4 + 1/4k) An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. A dielectric (k) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? V x0/4 C2 C C1 = C1 = 3/4C0 C2 = 1/4kC0 What is Q? 50

  27. Calculation C0 V x0 k k k C = C0 (3/4 + 1/4k) An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. A dielectric (k) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor? V x0/4 C2 C C1 = C1 = 3/4C0 C2 = 1/4kC0 Suppose C0 = 4.6 mF, k = 3.2 (mylar), V = 3 V 50

  28. Different Problem k Q = Q0 = C0V Vf = Q/C = V/(3/4 + 1/4k) C = C0 (3/4 + 1/4k) 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 (k) of width 1/4x0 is inserted into the gap as shown. What is Vf, the final voltage on the capacitor? Q0 C0 V V x0 1/4x0 Q (A) Vf < V(B) Vf = V(C) Vf > V Q stays same: no way to add or subtract We know C: (property of capacitor)

  29. 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 (k) of width 1/4x0 is inserted into the gap as shown. What is Vf, the final voltage on the capacitor? Different Problem k U decreased Q0 C0 V V x0 1/4x0 Vf = Q/C = V/(3/4 + 1/4k) How did energy stored in capacitor change when dielectric inserted? • U increased(B) U stayed same(C) U decreased U = ½ Q2/C Q remained same C increased

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