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Physics 2112 Unit 8: Capacitors Today’s Concept:  Capacitors in a circuits  Dielectrics  Energy in capacitors Unit 8, Slide 1.

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Presentation on theme: "Physics 2112 Unit 8: Capacitors Today’s Concept:  Capacitors in a circuits  Dielectrics  Energy in capacitors Unit 8, Slide 1."— Presentation transcript:

1 Physics 2112 Unit 8: Capacitors Today’s Concept:  Capacitors in a circuits  Dielectrics  Energy in capacitors Unit 8, Slide 1

2 Where we are……. Unit 8, Slide 2

3 C V 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 C V Q  VC Electricity & Magnetism Lecture 8, Slide 3 Direction of arrows is opposite of the direction of electron motion Sorry. It’s historical. There is nothing I can do.

4 V Q1  C1VQ1  C1V Q2  C2VQ2  C2V C2C2 C1C1 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 Parallel Capacitor Circuit Electricity & Magnetism Lecture 8, Slide 4

5 Q V Key point: Q is the same for both capacitors Key point: Q  VC total  V 1 C 1  V 2 C 2 1 C total 1 C1C1 1 C2C2   Series Capacitor Circuit Q Q Q  VC total C1C1 C2C2 V1V1 V2V2 V Also: V  V 1  V 2 Q  C total  Q  C 1  Q  C 2 Electricity & Magnetism Lecture 8, Slide 5

6 Voltage Current Capacitance Series Parallel Different for each capacitor. V total  V 1  V 2 Increases C eq  C 1  C 2 Same for each capacitor I total  I 1  I 2 Same for each capacitor. V total  V 1  V 2 Decreases 1  C eq  1  C 1  1  C 2 Wiring Every electron that goes through one must go through the other Each resistor on a different wire. Different for each capacitor I total  I 1  I 2 C1C1 C2C2 C1C1 C2C2 Capacitor Summary Electricity & Magnetism Lecture 9, Slide 6

7 Example 8.1 (Capacitors in Series) Unit 8, Slide 7 C 1 =3uF C 2 =9uF V =12V Given the circuit to the left: What is Q 1 ? What is V 1 ?   Conceptual Idea:   Plan: Find equivalent capacitance Use knowledge that in series Q 1 = Q 2 = Q tot Find V using Q = CV

8 Example 8.2 (Capacitors in Parallel) Unit 8, Slide 8 C 1 =3uF C 2 =9uF V =12V Given the circuit to the left: What is Q 1 ? What is Q 2 ?   Conceptual Idea:   Plan: Find equivalent capacitance Use knowledge that in parallel V 1 = V 2 Find Q using Q = CV

9 C total  C  2 1  C total  1  C  1  C  2  C C total  C C total  2C CheckPoint: Three Capacitor Configurations C C C C C Electricity & Magnetism Lecture 8, Slide 9 The three configurations shown below are constructed using identical capacitors. Which of these configurations has lowest total capacitance? A B C

10 C total  C C total  C left  C right C total  C CheckPoint: Two Capacitor Configurations C C C C C C left  C  2 C right  C  2 Electricity & Magnetism Lecture 8, Slide 10 The two configurations shown below are constructed using identical capacitors. Which of these configurations has the lowest overall capacitance? A B A. A. A B. B. B C. C. Both configurations have the same capacitance

11 CheckPoint: Capacitor Network Unit 8, Slide 11 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. C1C1 C2C2 C3C3 V0V0 V1V1 Q1Q1 V2V2 Q2Q2 V3V3 Q3Q3 Compare Q 1, Q 2, and Q 3. A. A. Q 1 > Q 3 > Q 2 B. B. Q 1 > Q 2 > Q 3 C. C. Q 1 > Q 2 = Q 3 D. D. Q 1 = Q 2 = Q 3 E. E. Q 1 < Q 2 = Q 3

12 Example 8.3 (Capacitor Network) Unit 8, Slide 12 C 3 =12uF C 4 =6uF C 2 =11uF C 1 =3uF C 5 =9uF V =12C Given the circuit to the left: What is Q 1 ? What is Q 3 ?   Conceptual Idea:   Plan: Break circuit down into elements that are in parrallel or in series. Find equivalent capacitance Work backwards to find  V across each one Fine Q  using Q = CV Find V at each capacitor and then Q.

13 Example 8.1 (Capacitor Network) Unit 8, Slide 13 C3C3 C4C4 C2C2 C1C1 C5C5 C 34 C2C2 C1C1 C5C5 C3C3 C 1234 C5C5 C 12345

14 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… QQ V QQ U  1  2 QV  1  2 CV 2 Since Q  VC  12Q2C 12Q2C C Energy in a Capacitor Electricity & Magnetism Lecture 8, Slide 14

15 If connected to a battery V stays constant If isolated then total Q stays constant Messing with Capacitors C1   CC1   C V1  VV1  V C 1  C Q1  QQ1  Q V Q 1  C 1 V 1   CV    Q V 1  Q 1  C 1  Q   C  V   Electricity & Magnetism Lecture 8, Slide 15

16 By adding a dielectric you are just making a new capacitor with larger capacitance (factor of  ) C0C0 V Q 0  VC 0 Dielectrics C1 C0 C1 C0 V Q 1  VC 1 Electricity & Magnetism Lecture 8, Slide 16

17 CheckPoint: Capacitors and Dielectrics 1 Electricity & Magnetism Lecture 8, Slide 17 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. A. V 1 > V 2 B. B. V 1 = V 2 C. C. V 1 < V 2

18 CheckPoint: Capacitors and Dielectrics 2 Electricity & Magnetism Lecture 8, Slide 18 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. A. U 1 > U 2 B. B. U 1 = U 2 C. C. U 1 < U 2

19 CheckPoint: Capacitors and Dielectrics 3 Electricity & Magnetism Lecture 8, Slide 19 The two capacitors are now connected to each other by wires as shown. How will the charge redistribute itself, if at all? A. A.The charges will flow so that the charge on C1 will become equal to the charge on C2. B. B.The charges will flow so that the energy stored in C1 will become equal to the energy stored in C2 C. C.The charges will flow so that the potential difference across C1 will become the same as the potential difference across C2. D. D.No charges will flow. The charge on the capacitors will remain what it was before they were connected.

20 Example 8.4 (Partial Dielectric) Electricity & Magnetism Lecture 8, Slide 20 An air-gap capacitor, having capacitance C 0 and width x 0 is connected to a battery of voltage V. Conceptual Analysis: V C0C0 x0x0  V x04x04 Strategic Analysis:   Think of new capacitor as two capacitors in parallel   Calculate new capacitance C   Apply definition of capacitance to determine Q A dielectric    of width x 0  4 is inserted into the gap as shown. What is Q f, the final charge on the capacitor?

21 Calculation Electricity & Magnetism Lecture 8, Slide 21   A 1 =3/4 A o A 2 =1/4 A o


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