Physics 212 Lecture 8 Today's oncept: apacitors apacitors in a circuits, Dielectrics, Energy in capacitors Physics 212 Lecture 8, Slide 1
Simple apacitor ircuit Q +Q -Q Q= Q Battery has moved charge Q from one plate to the other. 8 Physics 212 Lecture 8, Slide 2
Q total Parallel apacitor ircuit 1 Q 1 = 1 2 Q 2 = 2 Q total Key point: is the same for both capacitors Key Point: Q total = Q 1 + Q 2 = 1 + 2 = ( 1 + 2 ) total = 1 + 2 14 Physics 212 Lecture 8, Slide 3
Series apacitor ircuit Q Q= total Q +Q -Q +Q -Q 1 2 1 2 Q Q Q Key point: Q is the same for both capacitors Key point: Q = total = 1 1 = 2 2 17 Also: = 1 + 2 Q/ total = Q/ 1 + Q/ 2 1 1 1 = + total 1 2 Physics 212 Lecture 8, Slide 4
hecpoint 1 Which has lowest total capacitance: total = 1/ total = 1/ + 1/ = 2/ total = /2 total = 2 18 Physics 212 Lecture 8, Slide 5
Which has lowest total capacitance? left = /2 right = /2 total = total = left + right total = 20 Physics 212 Lecture 8, Slide 6
Similar to hecpoint 3 2 Q 2 0 2 1 1 Q 1 3 3 Q 3 total Which of the following is NOT necessarily true: A) 0 = 1 B) total > 1 ) 2 = 3 D) Q 2 = Q 3 E) 1 = 2 + 3 24 Physics 212 Lecture 8, Slide 7
hecpoint 3 A circuit consists of three unequal capacitors 1, 2, and 3 which are connected to a battery of voltage 0. The capacitance of 2 is twice that of 1. The capacitance of 3 is three times that of 1. The capacitors obtain charges Q 1, Q 2, and Q 3. ompare Q 1, Q 2, and Q 3. A. Q 1 > Q 3 > Q 2 B. Q 1 > Q 2 > Q 3. Q 1 > Q 2 = Q 3 D. Q 1 = Q 2 = Q 3 E. Q 1 < Q 2 = Q 3 X X 1. See immediately: Q 2 = Q 3 (capacitors in series) 2. How about Q 1 vs. Q 2 and Q 3? alculate 23 first. 1 1 1 1 1 5 6 2 3 6 23 2 3 1 1 1 23 1 5 Q 1 1 0 6 Q Q Q 5 23 2 3 23 0 1 0 25 Physics 212 Lecture 8, Slide 8
harge capacitor store a logical 1 Discharge capacitor store a logical 0 Row select : close the switch olumn select : charge or discharge 1 Gigabyte ~ 10 10 cells Physics 212 Lecture 8, Slide 9
Energy in a apacitor In Prelecture 7 we calculated the wor done to move charge Q from one plate to another: +Q -Q U = 1 / 2 Q = 1 / 2 2 Since Q = = 1 / 2 Q 2 / This is potential energy waiting to be used 26 Physics 212 Lecture 8, Slide 10
Dielectrics 0 Q 0 = 0 1 = 0 Q 1 = 1 By adding a dielectric you are just maing a new capacitor with larger capacitance (factor of ) 11 Physics 212 Lecture 8, Slide 11
apacitor with dielectric If connected to a battery stays constant. 1 = Q 1 = 1 = 1 1 = = Q Isolated capacitor: total Q stays constant. Q 1 = Q 1 = 1 = Q 1 / 1 = Q/ = / Physics 212 Lecture 8, Slide 12
hecpoint 4a Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After 2 has been charged and disconnected, it is filled with a dielectric. ompare the voltages of the two capacitors. A. 1 > 2 B. 1 = 2. 1 < 2 onsider first the effect on the capacitance, then, does Q or stay constant? 33 Physics 212 Lecture 8, Slide 13
Effects of dielectric Two identical parallel plate capacitors are connected to identical batteries. Then a dielectric is inserted between the plates of capacitor 1. ompare the energy stored in the two capacitors. 0 =2 1 A) U 1 < U 0 B) U 0 = U 1 ) U 1 > U 0 ompare using U = 1 / 2 2 U 1 /U 0 = 35 Potential Energy goes UP Physics 212 Lecture 8, Slide 14
hecpoint 4b Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After 2 has been charged and disconnected, it is filled with a dielectric. ompare the potential energy stored by the two capacitors. A. U 1 > U 2 B. U 1 = U 2. U 1 < U 2 onsider first the effect on the capacitance, then, does Q or stay constant? 33 Physics 212 Lecture 8, Slide 15
hecpoint 4c Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After 2 has been charged and disconnected, it is filled with a dielectric. A. The charges will flow so that the charge on 1 will become equal to the charge on 2. B. The charges will flow so that the energy stored in 1 will become equal to the energy stored in 2.. The charges will flow so that the potential difference across 1 will become the same as the potential difference across 2. D. No charges will flow. The charge on the capacitors will remain what it was before they were connected. must be the same! Q: Q 1 Q2 Q 1 1 Q 1 2 2 2 U: 1 2 1 2 1 1 2 2 2 2 U U U 1 1 U 2 2 Physics 212 Lecture 8, Slide 16
alculation x 0 0 x 0 /4 An air-gap capacitor, having capacitance 0 and width x 0 is connected to a battery of voltage. 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? What changes when the dielectric added? (A) Only (B) only Q () only (D) and Q (E) and Q Adding dielectric changes the physical capacitor does not change and changes changes Q changes 38 Physics 212 Lecture 8, Slide 17
alculation x 0 0 x 0 /4 How does potential difference change? left right (A) left < right (B) left = right () left > right 40 The conducting plate is an equipotential! Physics 212 Lecture 8, Slide 18
alculation x 0 0 x 0 /4 onsider capacitor to be two capacitances, 1 and 2, in parallel: What is 1? 1 2 = (A) 1 = 0 (B) 1 = 3 / 4 0 () 1 = 4 / 3 0 (D) 1 = 1 / 4 0 In general. For parallel plate capacitor: = e 0 A/d 43 A = 3 / 4 A 0 d = d 0 1 = 3 / 4 (e 0 A 0 /d 0 ) 1 = 3 / 4 0 Physics 212 Lecture 8, Slide 19
alculation x 0 0 x 0 /4 1 2 = 1 = 3 / 4 0 What is 2? (A) 2 = 0 (B) 2 = 3 / 4 0 () 2 = 4 / 3 0 (D) 2 = 1 / 4 0 In general. For parallel plate capacitor filled with dielectric: = e 0 A/d A = 1 / 4 A 0 d = d 0 = ¼(e 0 A 0 /d 0 ) 2 = 1 / 4 0 45 Physics 212 Lecture 8, Slide 20
alculation x 0 0 x 0 /4 = 1 2 2 = 1 1 = 3 / 4 0 / 4 0 What is? (A) (B) () 1 2 1 2 1 1 1 2 1 = parallel combination of 1 and 2 : = 1 + 2 = 0 ( 3 / 4 + 1 / 4 ) 46 Physics 212 Lecture 8, Slide 21
alculation x 0 0 x 0 /4 1 2 = 1 = 3 / 4 0 2 = 1 / 4 0 = 0 ( 3 / 4 + 1 / 4 ) What is Q? Q 50 3 1 Q f 0 4 4 Physics 212 Lecture 8, Slide 22
Q 0 x 0 Different Problem 0 1/4x 0 An air-gap capacitor, having capacitance 0 and width x 0 is connected to a battery of voltage and then battery is disconnected. A dielectric () of width 1/4x 0 is inserted into the gap as shown. What is f, the final voltage on the capacitor? (A) f < (B) f = () f > Q stays same: no way to add or subtract We now : (property of capacitor) Q = Q 0 = 0 = 0 ( 3 / 4 + 1 / 4 ) f = Q/ = /( 3 / 4 + 1 / 4 ) Physics 212 Lecture 8, Slide 23
Q 0 x 0 Different Problem 0 1/4x 0 An air-gap capacitor, having capacitance 0 and width x 0 is connected to a battery of voltage and then battery is disconnected. A dielectric () of width 1/4x 0 is inserted into the gap as shown. What is f, the final voltage on the capacitor? f = Q/ = /( 3 / 4 + 1 / 4 ) How did energy stored in capacitor change when dielectric inserted? (A) U increased (B) U stayed same () U decreased U = ½ Q 2 / Q remained same increased U decreased Physics 212 Lecture 8, Slide 24