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

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Presentation on theme: "Physics 212 Lecture 8, Slide 1 Physics 212 Lecture 8 Today's Concept: Capacitors Capacitors in a circuits, Dielectrics, Energy in capacitors."— Presentation transcript:

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

2 Music Who is the Artist? A)Robert Plant B)Lyle Lovett C)Alison Krauss D)Mark Knopfler E)John Prine Mark Knopfler, OBE Dire Straits to country Dire Straits to country Physics 212 Lecture 15

3 Physics 212 Lecture 8, Slide 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!wkpark@illinois.edu Where do I go? EXAM ROOMS (by discussion section)EXAM ROOMS Preparation: – –Practice exams – –Exam prep exercises – –Worked examples – –Extra Office HoursExtra Office Hours

4 Physics 212 Lecture 8, Slide 4 Your Comments 05 “ 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 ” WILL DO See Calculation at End Will Do.. They were the hardest “ 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!!!” “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.” “ 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?“ We’ll summarize main points. “When the capacitors are placed in series is it best of 5 games or best of 7?”

5 Physics 212 Lecture 8, Slide 5 In Prelecture 7 we calculated the work done to move charge Q from one plate to another: This is potential energy waiting to be used… Energy in a Capacitor +Q V -Q U = 1 / 2 QV = 1 / 2 CV 2 Since Q = VC = 1 / 2 Q 2 /C C 26 “ I still don't understand what voltage is exactly. ”

6 Physics 212 Lecture 8, Slide 6 C V +Q+Q+Q+Q-Q QQ 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. C V Q=VC Simple Capacitor Circuit 8

7 Physics 212 Lecture 8, Slide 7 V Q 1 = C 1 V Q 2 = C 2 V C2C2C2C2 C1C1C1C1 Parallel Capacitor Circuit Key point: V is the same for both capacitors Key Point: Q total = Q 1 + Q 2 = VC 1 + VC 2 = V(C 1 + C 2 ) C total = C 1 + C 2 Q total 14

8 Physics 212 Lecture 8, Slide 8 Series Capacitor Circuit QQ Q +Q+Q -Q -Q V C1C1C1C1 C2C2C2C2 V1V1V1V1 V2V2V2V2 Q V Q Key point: Q is the same for both capacitors Also: V = V 1 + V 2 Q=VC total Key point: Q = VC total = V 1 C 1 = V 2 C 2 Q/C total = Q/C 1 + Q/C 2 1 C total 1 C1C1C1C11 C2C2C2C2 + = 17

9 Physics 212 Lecture 8, Slide 9 Checkpoint 1 C C C C C 1/C total = 1/C + 1/C = 2/C C total = C/2 C total = C C total = 2C Which has lowest total capacitance : 18 A)B)C)

10 Physics 212 Lecture 8, Slide 10 Checkpoint 2 C total = C Which has lowest total capacitance: C C C C C C left = C/2 C right = C/2 C total = C left + C right C total = C 20 A)B) C) Same

11 Physics 212 Lecture 8, Slide 11 Similar to Checkpoint 3 C1C1 C2C2 C3C3 V0V0 V1V1 Q1Q1 V2V2 Q2Q2 V3V3 Q3Q3 C total Which of the following is NOT necessarily true: A)V 0 = V 1 B)C total > C 1 C)V 2 = V 3 D)Q 2 = Q 3 E)V 1 = V 2 + V 3 24

12 Physics 212 Lecture 8, Slide 12 Checkpoint 3 25 A circuit consists of three unequal capacitors C 1, C 2, and C 3 which are connected to a battery of voltage V 0. The capacitance of C 2 is twice that of C 1. The capacitance of C 3 is three times that of C 1. The capacitors obtain charges Q 1, Q 2, and Q 3. Compare Q 1, Q 2, and Q 3. A. Q 1 > Q 3 > Q 2 B. Q 1 > Q 2 > Q 3 C. Q 1 > Q 2 = Q 3 D. Q 1 = Q 2 = Q 3 E. Q 1 < Q 2 = Q 3 “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.” (HW 3,4)

13 Physics 212 Lecture 8, Slide 13 Checkpoint 3 25 1. See immediately: Q 2 = Q 3 (capacitors in series) 1. See immediately: Q 2 = Q 3 (capacitors in series) A circuit consists of three unequal capacitors C 1, C 2, and C 3 which are connected to a battery of voltage V 0. The capacitance of C 2 is twice that of C 1. The capacitance of C 3 is three times that of C 1. The capacitors obtain charges Q 1, Q 2, and Q 3. Compare Q 1, Q 2, and Q 3. A. Q 1 > Q 3 > Q 2 B. Q 1 > Q 2 > Q 3 C. Q 1 > Q 2 = Q 3 D. Q 1 = Q 2 = Q 3 E. Q 1 < Q 2 = Q 3 XX 2. How about Q 1 vs. Q 2 and Q 3 ? Calculate C 23 first. 2. How about Q 1 vs. Q 2 and Q 3 ? Calculate C 23 first.

14 Physics 212 Lecture 8, Slide 14 By adding a dielectric you are just making a new capacitor with larger capacitance (factor of  ) Dielectrics C0C0C0C0 V Q 0 =VC 0 C1=C0C1=C0C1=C0C1=C0 V Q 1 =VC 1 11

15 Physics 212 Lecture 8, Slide 15 Messing w/ Capacitors If connected to a battery V stays constant If isolated then total Q stays constant  V C 1 =  C V 1 = V Q 1 = C 1 V 1 =  CV =  Q  C 1 =  C Q 1 = Q V 1 = Q 1 /C 1 = Q/  C = V / 

16 Physics 212 Lecture 8, Slide 16 Checkpoint 4a “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. ” 33 Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C 2 has been charged and disconnected, it is filled with a dielectric. Compare the voltages of the two capacitors. A. V 1 > V 2 B. V 1 = V 2 C. V 1 < V 2 Consider first the effect on the capacitance, then, does Q or V stay constant?

17 Physics 212 Lecture 8, Slide 17 Messing w/ Capacitors U 1 /U 0 =   =2 C0C0C0C0 V C1C1C1C1 V U = 1 / 2 CV 2 Compare using Potential Energy goes UP 35 A) U 1 U 0 Two identical parallel plate capacitors are connected to identical batteries. Then a dielectric is inserted between the plates of capacitor C 1. Compare the energy stored in the two capacitors.

18 Physics 212 Lecture 8, Slide 18 Checkpoint 4b 33 “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. ” Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C 2 has been charged and disconnected, it is filled with a dielectric. Compare the potential energy stored by the two capacitors. A. U 1 > U 2 B. U 1 = U 2 C. U 1 < U 2 Consider first the effect on the capacitance, then, does Q or V stay constant?

19 Physics 212 Lecture 8, Slide 19 Checkpoint 4c Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C 2 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 C 1 will become equal to the charge on C 2. B. The charges will flow so that the energy stored in C 1 will become equal to the energy stored in C 2. C. The charges will flow so that the potential difference across C 1 will become the same as the potential difference across C 2. 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 Physics 212 Lecture 8, Slide 20 Checkpoint 4c Q: U: Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C 2 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 C 1 will become equal to the charge on C 2. B. The charges will flow so that the energy stored in C 1 will become equal to the energy stored in C 2. C. The charges will flow so that the potential difference across C 1 will become the same as the potential difference across C 2. D. No charges will flow. The charge on the capacitors will remain what it was before they were connected. V must be the same !!

21 Physics 212 Lecture 8, Slide 21Calculation An air-gap capacitor, having capacitance C 0 and width x 0 is connected to a battery of voltage V. A dielectric (  ) of width x 0 /4 is inserted into the gap as shown. What is Q f, the final charge on the capacitor? Conceptual Analysis:  V C0C0 x0x0 V x 0 /4 What changes when the dielectric added? (A) (B) (C) (D) (E) (A) Only C (B) only Q (C) only V (D) C and Q (E) V and Q Adding dielectric changes the physical capacitor C changes V does not change and C changes Q changes 38 (HW 2, 5)

22 Physics 212 Lecture 8, Slide 22Calculation Strategic Analysis: – – Calculate new capacitance C – –Apply definition of capacitance to determine Q  V C0C0 x0x0 V (A) (B) (C) (A) V left V right To calculate C, let’s first look at:  V left V right The conducting plate is an equipotential !! x 0 /4 40 An air-gap capacitor, having capacitance C 0 and width x 0 is connected to a battery of voltage V. A dielectric (  ) of width x 0 /4 is inserted into the gap as shown. What is Q f, the final charge on the capacitor?

23 Physics 212 Lecture 8, Slide 23Calculation Can consider capacitor to be two capacitances, C 1 and C 2, in parallel  V C0C0 x0x0 V (A) (B) (C) (D) (A) C 1 = C 0 (B) C 1 = 3 / 4 C 0 (C) C 1 = 4 / 3 C 0 (D) C 1 = 1 / 4 C 0 What is C 1 ? In general. For parallel plate capacitor: C = In general. For parallel plate capacitor: C =   A/d  =  C1C1 C2C2 A = 3 / 4 A 0 d = d 0 C 1 = 3 / 4 (  0 A 0 /d 0 ) C 1 = 3 / 4 C 0 x 0 /4 43 An air-gap capacitor, having capacitance C 0 and width x 0 is connected to a battery of voltage V. A dielectric (  ) of width x 0 /4 is inserted into the gap as shown. What is Q f, the final charge on the capacitor?

24 Physics 212 Lecture 8, Slide 24Calculation  V C0C0 x0x0 V (A) (B) (C) (D) (A)C 2 =  C 0 (B) C 2 = 3 / 4  C 0 (C) C 2 = 4 / 3  C 0 (D) C 2 = 1 / 4  C 0 What is C 2 ?  In general. For parallel plate capacitor filled with dielectric: C =  In general. For parallel plate capacitor filled with dielectric: C =   A/d =  C1C1 C2C2 A = 1 / 4 A 0 d = d 0  C = ¼(  0 A 0 /d 0 ) C 1 = 3 / 4 C 0 C 2 = 1 / 4  C 0 x 0 /4 45 An air-gap capacitor, having capacitance C 0 and width x 0 is connected to a battery of voltage V. A dielectric (  ) of width x 0 /4 is inserted into the gap as shown. What is Q f, the final charge on the capacitor?

25 Physics 212 Lecture 8, Slide 25Calculation  V C0C0 x0x0 V (A) (B) (C) What is C?  C = parallel combination of C 1 and C 2 : C = C 1 + C 2 =  C1C1 C2C2 C 2 = 1 / 4  C 0 C C = C 0 ( 3 / 4 + 1 / 4  ) x 0 /4 C 1 = 3 / 4 C 0 46 An air-gap capacitor, having capacitance C 0 and width x 0 is connected to a battery of voltage V. A dielectric (  ) of width x 0 /4 is inserted into the gap as shown. What is Q f, the final charge on the capacitor?

26 Physics 212 Lecture 8, Slide 26Calculation  V C0C0 x0x0 V What is Q? =  C1C1 C2C2 C x 0 /4 C 2 = 1 / 4  C 0 C 1 = 3 / 4 C 0 C = C 0 ( 3 / 4 + 1 / 4  ) 50 An air-gap capacitor, having capacitance C 0 and width x 0 is connected to a battery of voltage V. A dielectric (  ) of width x 0 /4 is inserted into the gap as shown. What is Q f, the final charge on the capacitor?

27 Physics 212 Lecture 8, Slide 27Calculation  V C0C0 x0x0 V Suppose C 0 = 4.6  F,  = 3.2 (mylar), V = 3 V =  C1C1 C2C2 C x 0 /4 C 2 = 1 / 4  C 0 C 1 = 3 / 4 C 0 C = C 0 ( 3 / 4 + 1 / 4  ) 50 An air-gap capacitor, having capacitance C 0 and width x 0 is connected to a battery of voltage V. A dielectric (  ) of width x 0 /4 is inserted into the gap as shown. What is Q f, the final charge on the capacitor?

28 Physics 212 Lecture 8, Slide 28 Different Problem An air-gap capacitor, having capacitance C 0 and width x 0 is connected to a battery of voltage V and then battery is disconnected. A dielectric (  ) of width 1/4x 0 is inserted into the gap as shown. What is V f, the final voltage on the capacitor?Q  V C0C0 x0x0 V 1/4x 0 Q stays same: no way to add or subtract (B) (C) (A) V f V Q = Q 0 = C 0 V Q0Q0 We know C: (property of capacitor) C = C 0 ( 3 / 4 + 1 / 4  )V f = Q/C = V/( 3 / 4 + 1 / 4  )

29 Physics 212 Lecture 8, Slide 29 Different Problem  V C0C0 x0x0 V 1/4x 0 Q0Q0 How did energy stored in capacitor change when dielectric inserted? (A) (B) (C) (A) U increased (B) U stayed same (C) U decreased U = ½ Q 2 /C Q remained same C increased U decreased An air-gap capacitor, having capacitance C 0 and width x 0 is connected to a battery of voltage V and then battery is disconnected. A dielectric (  ) of width 1/4x 0 is inserted into the gap as shown. What is V f, the final voltage on the capacitor? V f = Q/C = V/( 3 / 4 + 1 / 4  )


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