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

Physics 212 Lecture 8, Slide 2 C V +Q+Q+Q+Q-Q QQ Battery has moved charge Q from one plate to the other. C V Q=VC Simple Capacitor Circuit 8

Physics 212 Lecture 8, Slide 3 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

Physics 212 Lecture 8, Slide 4 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

Physics 212 Lecture 8, Slide 5 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

Physics 212 Lecture 8, Slide 6 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

Physics 212 Lecture 8, Slide 7 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

Physics 212 Lecture 8, Slide 8 Checkpoint 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.

Physics 212 Lecture 8, Slide 9 Charge capacitor – store a logical “1” Discharge capacitor – store a logical “0” Row select : close the switch 1 Gigabyte ~ cells Column select : charge or discharge C

Physics 212 Lecture 8, Slide 10 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

Physics 212 Lecture 8, Slide 11 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

Physics 212 Lecture 8, Slide 12 Capacitor with dielectric 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 /  If connected to a battery V stays constant. Isolated capacitor: total Q stays constant.

Physics 212 Lecture 8, Slide 13 Checkpoint 4a 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?

Physics 212 Lecture 8, Slide 14 Effects of dielectric 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.

Physics 212 Lecture 8, Slide 15 Checkpoint 4b 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 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?

Physics 212 Lecture 8, Slide 16 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. 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 !

Physics 212 Lecture 8, Slide 17Calculation 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?  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

Physics 212 Lecture 8, Slide 18Calculation  V C0C0 x0x0 V (A) (B) (C) (A) V left V right How does potential difference change ?  V left V right The conducting plate is an equipotential ! x 0 /4 40

Physics 212 Lecture 8, Slide 19Calculation 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

Physics 212 Lecture 8, Slide 20Calculation  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

Physics 212 Lecture 8, Slide 21Calculation  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  ) x 0 /4 C 1 = 3 / 4 C 0 46

Physics 212 Lecture 8, Slide 22Calculation  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  ) 50

Physics 212 Lecture 8, Slide 23 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  )V f = Q/C = V/( 3 / / 4  )

Physics 212 Lecture 8, Slide 24 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  )