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Capacitors PowerPoint Presentation

Capacitors

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Capacitors

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  1. Physics 102:Lecture 04 Capacitors

  2. Recall from last lecture…..Electric Fields, Electric Potential

  3. q A B Comparison:Electric Potential Energy vs. Electric Potential DVAB: the difference in electric potential between points B and A DUAB : the change in electric potential energy of a charge q when moved from A to B DUAB = q DVAB

  4. Electric Potential: Summary • E field lines point from higher to lower potential • For positive charges, going from higher to lower potential is “downhill” • For negative charges, going from lower to higher potential is “downhill” • For a battery, the + terminal is at a higher potential than the – terminal Positive charges tend to go “downhill”, from + to - Negative charges go in the opposite direction, from - to + DUAB = q DVAB

  5. Important Special CaseUniform Electric Field - - - - - + + + + + • Two large parallel conducting plates of area A • +Q on one plate • - Q on other plate • Then E is • uniform between the two plates: E=4kQ/A • zero everywhere else • This result is independent of plate separation • This is call a parallel plate capacitor

  6. A B Parallel Plate Capacitor:Potential Difference Charge Q on plates Charge 2Q on plates V =VA – VB = +E0 d V =VA – VB = +2E0 d E=2 E0 E=E0 - - - - - + + + + + - - - - - + + + + + - - - - - + + + + + A B d d Potential difference is proportional to charge: Double Q  Double V • E0=4kQ/A 10

  7. - - - - - + + + + + E d Capacitance: The ability to store separated charge CQ/V • Any pair conductors separated by a small distance. (e.g. two metal plates) • Capacitor stores separated charge • Positive Q on one conductor, negative Q on other • Net charge is zero Q=CV • Stores Energy U =(½) Q V Units: 1 C/Volt = 1 Farad (F) 12

  8. Why Separate Charge? • Camera Flash • Defibrillator—see ex. 17.12 in text • AC -> DC • Tuners • Radio • Cell phones

  9. Example Capacitance Practice How much charge is on a 0.9 F capacitor which has a potential difference of 200 Volts? How much energy is stored in this capacitor? 14

  10. Example Capacitance Practice How much charge is on a 0.9 F capacitor which has a potential difference of 200 Volts? Q = CV = (0.9)(200) = 180 Coulombs How much energy is stored in this capacitor? U = ½ Q V = ½ (180) (200) = 18,000 Joules! 14

  11. Capacitance of Parallel Plate Capacitor V V = Ed E=4kQ/A (Between two large plates) So: V = 4kQd/A Recall: CQ/V So: C = A/(4kd) Define: e0=1/(4pk)=8.85x10-12 C2/Nm2 + E - A A d C =e0A/d Parallel plate capacitor 16

  12. Parallel Plate CapacitorC = e0A/d Calculate the capacitance of a parallel plate capacitor made from two large square metal sheets 1.3 m on a side, separated by 0.1 m. Example A A d 18

  13. Parallel Plate CapacitorC = e0A/d Calculate the capacitance of a parallel plate capacitor made from two large square metal sheets 1.3 m on a side, separated by 0.1 m. Example A A d 18

  14. Dielectric constant (k > 1) Capacitance without dielectric Capacitance with dielectric Dielectric • Placing a dielectric between the plates increases the capacitance. C = k C0 19

  15. + + + + - - - - -q +q d pull pull ACT: Parallel Plates A parallel plate capacitor given a charge q. The plates are then pulled a small distance further apart. What happens to the charge q on each plate of the capacitor? 1) Increases 2) Constant 3) Decreases Remember charge is real/physical. There is no place for the charges to go. 22

  16. + + + + - - - - -q +q d pull pull Preflight 4.1 A parallel plate capacitor given a charge q. The plates are then pulled a smalldistance further apart. Which of the following apply to the situation after the plates have been moved? 66% 59% 52% 35% 1)The capacitance increases True False 2)The electric field increases True False 3)The voltage between the plates increases True False 4)The energy stored in the capacitor increases True False C = e0A/d C decreases! E= Q/(e0A) Constant V= Ed :24

  17. ACT/Preflight 4.1 + + + + - - - - -q +q d pull pull A parallel plate capacitor given a charge q. The plates are then pulled a small distance further apart. Which of the following apply to the situation after the plates have been moved? The energy stored in the capacitor A) increases B) constant C) decreases U= ½ QV Q constant, V increased Plates are attracted to each other, you must pull them apart, so the potential energy of the plates increases. :25

  18. ACT/Preflight 4.2 Two identical parallel plate capacitors are shown in end-view in A) of the figure. Each has a capacitance of C. B ) A ) If the two are joined as in (B) of the figure, forming a single capacitor, what is the final capacitance? “More area, means more capacitance .” 63% 7% 30% 1) 2C 2) C 3) C/2 “because C=EA/d. none of these variables are changing so C must be the same when the 2 are joined to form a single capacitor” “2 capacitors become 1 which means that the total capacitance got halved ” 26

  19. Vwire 1= 0 V Vwire 2= 5 V Vwire 3= 12 V Vwire 4= 15 V Voltage in Circuits • Elements are connected by wires. • Any connected region of wire has the same potential. • The potential difference across an element is the element’s “voltage.” Example C1 C2 C3 VC1= 5-0 V= 5 V VC2= 12-5 V= 7 V VC3= 15-12 V= 3 V 28

  20. 15 V Ceq 10 V Capacitors in Parallel • Add Areas: Ceq = C1+C2 remember C=0A/d • Share Charge: Qeq = Q1+Q2 • Both ends connected together by wire • Same voltage: V1 = V2 = Veq 15 V 15 V C1 C2 10 V 10 V 30

  21. Parallel Practice Example A 4 mF capacitor and 6 mF capacitor are connected in parallel and charged to 5 volts. Calculate Ceq, and the charge on each capacitor. = 4 mF+6 mF = 10 mF Ceq = C4+C6 Q4 = C4 V4 = (4 mF)(5 V) = 20 mC = (6 mF)(5 V) = 30 mC Q6 = C6 V6 = (10 mF)(5 V) = 50 mC Qeq = Ceq Veq = Q4+Q6 V = 5 V 5 V 5 V 5 V C4 C6 Ceq 0 V 0 V 0 V :34

  22. +Q -Q + + - - Ceq +Q -Q Capacitors in Series • Add d: • Connected end-to-end with NO other exits • Same Charge: Q1 = Q2 = Qeq • Share Voltage:V1+V2=Veq + + + +Q + + C1 + + - + + + + C2 -Q - + - + + 36 +

  23. +Q 5 V +Q -Q + +Q -Q - 0 V -Q Series Practice Example A 4 mF capacitor and 6 mF capacitor are connected in series and charged to 5 volts. Calculate Ceq, and the charge on the 4 mF capacitor. Q = CV 5 V + C4 - Ceq + C6 - 0 V :38

  24. C1 C1 C2 C2 Comparison:Series vs. Parallel Series • Can follow a wire from one element to the other with no branches in between. Parallel • Can find a loop of wire containing both elements but no others (may have branches).

  25. Electromotive Force + • Battery • Maintains potential difference V • Not constant power • Not constant current • Does NOT produce or supply charges, just “pushes” them. - 40

  26. C 2 +q2 -q2 + V2 E +q1 +q3 C C V1 V3 - 1 3 -q1 -q3 Preflight 4.4 A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf E. The capacitors obtain charges q1, q2,q3, and have voltages across their plates V1, V2, and V3.Which of these are true? • q1 = q2 • q2 = q3 • V2 = V3 • E = V1 • V1 < V2 • Ceq > C1 43

  27. Preflight 4.4 C 2 +q2 -q2 + V2 E +q1 +q3 C C V1 V3 - 1 3 -q1 -q3 A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf E. The capacitors obtain charges q1, q2,q3, and have voltages across their plates V1, V2, and V3. 1) q1 = q2 Not necessarily C1 and C2 are NOT in series. 2) q2 = q3 Yes! C2 and C3 are in series. 45

  28. Preflight 4.4 10V 7V?? C 2 0V +q2 -q2 + V2 E +q1 +q3 C C V1 V3 - 1 3 -q1 -q3 A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf E. The capacitors obtain charges q1, q2,q3, and have voltages across their plates V1, V2, and V3. 3) V2 = V3 Not necessarily, only if C1 = C2 4) E = V1 Yes! Both ends are connected by wires 48

  29. Preflight 4.4 10V 7V?? C 2 0V +q2 -q2 + V2 E +q1 +q3 C C V1 V3 - 1 3 -q1 -q3 A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf E. The capacitors obtain charges q1, q2,q3, and have voltages across their plates V1, V2, and V3. 5) V1 < V2 Nope, V1 > V2. (E.g. V1 = 10-0, V2 =10-7 6) Ceq > C1 Yes! C1 is in parallel with C23 48

  30. Recap of Today’s Lecture • Capacitance C = Q/V • Parallel Plate: C = 0A/d • Capacitors in parallel: Ceq = C1+C2 • Capacitors in series: Ceq = 1/(1/C1+1/C2) • Batteries provide fixed potential difference